ZetaTalk and Spaceguard UK (D8)
Nancy Lieder
> In sci.astro Nancy Lieder <zeta...@zetatalk.com> wrote:
>> while out where the Concord flies or our satellites float,
>> while moving at twice the speed of the Concord, or only
>> 1/4 the speed of those satellites.
>
> The Concord flies, at maximum, at less than 12 miles above
> the surface. The lowest satellites orbit at around 100 miles.
> The geostationary satellites are at over 12,000 miles - maybe
> 18,000 (I forget). The moon is in orbit at over 200,000 miles
> distance. The Concord is not in orbit when it flies - so it
> cannot be used for comparison.
And using some other figures quoted (below), we've established, then,
for the
CONCORD:
533.33 m/s
12 miles high
SATELLITES:
7897.873415 m/s
100 miles high (lowest)
Geo Stationary
4,635.7155 m/s
12,000 - 18,000 miles high
MOON:
1023 m/s
200,000 miles high
In Article <6ko22i$1...@bgtnsc01.worldnet.att.net> Eric George writes:
>> Eric has kindly calculated the pace of your Moon at some
>> 1023 meters per second.. ...
> satellite .. velocity would be 7897.873415 m/s.
> This is almost 15 times faster then the concord.
In Article <6ko22i$1...@bgtnsc01.worldnet.att.net> Eric George writes:
> circular orbit velocity is: V = C/P = sqrt(G*Me/a) ... So yes,
> geosync satellites do go faster then the moon, they are much
> closer.
In Article <6kl07u$g...@dfw-ixnews7.ix.netcom.com> Nancy Lieder wrote:
> the Britannica states to be the supersonic speed that the Concord
> flies, when breaking the sound barrier. 1,200 mph, or 1,920 km/hr,
> which is 1,920,000 m/hr, which is 32,000 m/minute or 533.33 m/s,
> is it not? Second, if the diameter of the Earth is 12,756.27 km, then
> stationary satellites must travel 4,635.7155 m/second.
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Magnus Nyborg
"Nancy Lieder" <zeta...@zetatalk.com> wrote in message
news:3B5C59D4...@zetatalk.com...
[...]
> And using some other figures quoted (below), we've established, then,
> for the
>
> CONCORD:
> 533.33 m/s
> 12 miles high
Non-orbital body - the orbitlal velocity would (if orbit was possible) be a
little higher than for satelites.
>
> SATELLITES:
> 7897.873415 m/s
> 100 miles high (lowest)
Lowest orbit - highest speed...
Note that 100 milse high refers to distance from surface, and that the
orbital velovity is determined from the center of the gravitating body
(Earth).
>
> Geo Stationary
> 4,635.7155 m/s
> 12,000 - 18,000 miles high
Higher orbit, lower speed...
>
> MOON:
> 1023 m/s
> 200,000 miles high
Highest orbit (of the three) - lowest speed (again of the three)
Orbital speed for ideal circular motion of a low-mass object circling a
high-mass object M (which refers to it's mass) is determined by the formula
v = sqrt( G*M / r )
which gives the following...
Ground orbit (if possible) - v = sqrt( 6.67E-11 * 5.976E24 / 6.378E6 ) =
7905 m/s
Satellite orbit - v = sqrt( 6.67E-11 * 5.976E24 / 6.478E6 ) = 7844 m/s
Geostationary orbit - v = sqrt( 6.67E-11 * 5.976E24 / ? ) = ? m/s (didn't
find an accurate number for distance in the hurry)
Moon orbit - v = sqrt( 6.67E-11 * 5.976E24 / 3.844E8 ) = 1018 m/s
reservation made for constants (G only with three significant digits
effecting the accuracy of the calculations negatively).
Aside from the absent value for geostationary orbit, the fit is very good -
don't you agree Nancy !?
Clear Skies,
Magnus
john Latala
> > The Concord is not in orbit when it flies - so it cannot be used for
> > comparison.
Whic part of the above line didn't you understand?
> And using some other figures quoted (below), we've established, then,
> for the
>
> CONCORD:
> 533.33 m/s
> 12 miles high
When the Concord lands it's travelling well under 200 miles per hour or
100 m/s. So? At this moment I'm sitting at my computer in my apartment so
I'm doing 0 m/s at 50 ft. So?
--
john R. Latala
jrla...@golden.net
Nancy Lieder
> Orbital speed for ideal circular motion of a low-mass object
> circling a high-mass object M (which refers to it's mass) is
> determined by the formula
>
> v = sqrt( G*M / r )
>
> which gives the following...
> Ground orbit (if possible) -
> v = sqrt( 6.67E-11 * 5.976E24 / 6.378E6 ) = 7905 m/s
> Satellite orbit -
> v = sqrt( 6.67E-11 * 5.976E24 / 6.478E6 ) = 7844 m/s
> Geostationary orbit -
> v = sqrt( 6.67E-11 * 5.976E24 / ? ) = ? m/s
> (didn't find an accurate number for distance in the hurry)
> Moon orbit -
> v = sqrt( 6.67E-11 * 5.976E24 / 3.844E8 ) = 1018 m/s
>
> reservation made for constants (G only with three significant
> digits effecting the accuracy of the calculations negatively).
You used 5.976E-11 for Earth Mass where Eric George computed this to be
5.9763e+24 kg. I assume this to be essentially the same. I'm assuming r
is the distance between the objects, per MC Harrison post (below).
You've got this divisor progressing from a ground orbit to the Moon's
orbit. I don't know how to read these numbers, but assume you've got
the distance, from center of Earth or whatever, per points made in your
post.
In article <353CFF...@spammers.of.the.world.unite.etc> M.C. Harrison
writes.
> The force of gravity is an inverse square law, which means
> a mass will experience a force due to another mass according to
> the equation F=M1*M2/r^2 where M1 is one mass, M2 is the
> other mass, and r is the separation of the masses. ...
But Magnus, how can the gravity pull be THE SAME for a satellite and the
Moon! Gravity is a factor of BOTH objects, the pull between them. Not
sure what a satellite weighs, but Eric George computed the Mass of the
Moon to be 7.3508e+22 kg , or 73,696,438,000,000,000,000 Metric Tons!
The formula you used, for a "low-mass object circling a high-mass object
M" may work for a satellite, but the Moon is NOT a low-mass object.
Give me a velosity equation that has a spot of the Mass of the MOON, as
well as the Earth being orbited! Lets plug in an equivaltne Mass of the
Moon, using the same basis as we do for the Mass of the Earth in your
computations above!
In article <MPG.fc0bb991...@news.connect.ab.ca> Paul Campbell
writes:
> Mass is not based on speed but mass can be solved by
> F=GMm/r^2 and solving for m. The weight of the moon is
> zero, the mass can be calculated by the above formula. If you
> change the mass of the moon then the Moon's orbit as
> presently observed would not happen. Therefore the moon's
> mass must be what it is regardless of it's composition.
Greg Neill
> In Article <8u_67.10984$e5.16...@newsb.telia.net> Magnus Nyborg wrote:
>
> > Orbital speed for ideal circular motion of a low-mass object
> > circling a high-mass object M (which refers to it's mass) is
> > determined by the formula
> >
> > v = sqrt( G*M / r )
> >
> > which gives the following...
> > Ground orbit (if possible) -
> > v = sqrt( 6.67E-11 * 5.976E24 / 6.378E6 ) = 7905 m/s
> > Satellite orbit -
> > v = sqrt( 6.67E-11 * 5.976E24 / 6.478E6 ) = 7844 m/s
> > Geostationary orbit -
> > v = sqrt( 6.67E-11 * 5.976E24 / ? ) = ? m/s
> > (didn't find an accurate number for distance in the hurry)
> > Moon orbit -
> > v = sqrt( 6.67E-11 * 5.976E24 / 3.844E8 ) = 1018 m/s
> >
> > reservation made for constants (G only with three significant
> > digits effecting the accuracy of the calculations negatively).
>
> You used 5.976E-11 for Earth Mass where Eric George computed this to be
> 5.9763e+24 kg.
No, he used 5.976E24. The E-11 exponeent is on the value for G, the
gravitational constant.
[snip]
>
> In article <353CFF...@spammers.of.the.world.unite.etc> M.C. Harrison
> writes.
> > The force of gravity is an inverse square law, which means
> > a mass will experience a force due to another mass according to
> > the equation F=M1*M2/r^2 where M1 is one mass, M2 is the
> > other mass, and r is the separation of the masses. ...
>
> But Magnus, how can the gravity pull be THE SAME for a satellite and the
> Moon! Gravity is a factor of BOTH objects, the pull between them. Not
> sure what a satellite weighs, but Eric George computed the Mass of the
> Moon to be 7.3508e+22 kg , or 73,696,438,000,000,000,000 Metric Tons!
> The formula you used, for a "low-mass object circling a high-mass object
> M" may work for a satellite, but the Moon is NOT a low-mass object.
> Give me a velosity equation that has a spot of the Mass of the MOON, as
> well as the Earth being orbited! Lets plug in an equivaltne Mass of the
> Moon, using the same basis as we do for the Mass of the Earth in your
> computations above!
Trivially done. Replace M in the formula with (M+m), where M is the
mass of the Earth and m the mass of the Moon. Since the mass of the
Moon is about 1/81 the mass of the Earth, that's equivalent to 1.012M.
So the results would differ by about 1.2 percent.
Magnus Nyborg
>
> > Orbital speed for ideal circular motion of a low-mass object
> > circling a high-mass object M (which refers to it's mass) is
> > determined by the formula
> >
> > v = sqrt( G*M / r )
> >
> > which gives the following...
> > Ground orbit (if possible) -
> > v = sqrt( 6.67E-11 * 5.976E24 / 6.378E6 ) = 7905 m/s
> > Satellite orbit -
> > v = sqrt( 6.67E-11 * 5.976E24 / 6.478E6 ) = 7844 m/s
> > Geostationary orbit -
> > v = sqrt( 6.67E-11 * 5.976E24 / ? ) = ? m/s
> > (didn't find an accurate number for distance in the hurry)
> > Moon orbit -
> > v = sqrt( 6.67E-11 * 5.976E24 / 3.844E8 ) = 1018 m/s
> >
> > reservation made for constants (G only with three significant
> > digits effecting the accuracy of the calculations negatively).
>
> You used 5.976E-11 for Earth Mass where Eric George computed this to be
> 5.9763e+24 kg. I assume this to be essentially the same. I'm assuming r
> is the distance between the objects, per MC Harrison post (below).
> You've got this divisor progressing from a ground orbit to the Moon's
> orbit. I don't know how to read these numbers, but assume you've got
> the distance, from center of Earth or whatever, per points made in your
> post.
Simply put - you are a moron Nancy, unable to read even the simplest of
texts!
...I used 5.976E24 as mass for Earth, and nothing else!
>
> In article <353CFF...@spammers.of.the.world.unite.etc> M.C. Harrison
> writes.
> > The force of gravity is an inverse square law, which means
> > a mass will experience a force due to another mass according to
> > the equation F=M1*M2/r^2 where M1 is one mass, M2 is the
> > other mass, and r is the separation of the masses. ...
>
> But Magnus, how can the gravity pull be THE SAME for a satellite and the
> Moon! Gravity is a factor of BOTH objects, the pull between them. Not
That's because the "gravity pull" (acceleration) is caused by the same
Earth, thus creating the same pull. Had you had even the smallest
understanding on these matters, you would understand this! But you don't...
Typicall misunderstandings in these cases are
F = GMm / r^2 (which I assume you would term "gravity pull", and that we
call force of gravitation)
a = GM / r ^2 (since a = F / m) a is of course "acceleration", unless you
missed that (again)
that's the betarelease of gravitation 101... ;o)
> sure what a satellite weighs, but Eric George computed the Mass of the
> Moon to be 7.3508e+22 kg , or 73,696,438,000,000,000,000 Metric Tons!
> The formula you used, for a "low-mass object circling a high-mass object
> M" may work for a satellite, but the Moon is NOT a low-mass object.
Compared to Earth, the Moon _is_ a low-mass object. Just like comparing your
brain to an ants would make your brain a low-mass object...
> Give me a velosity equation that has a spot of the Mass of the MOON, as
> well as the Earth being orbited! Lets plug in an equivaltne Mass of the
> Moon, using the same basis as we do for the Mass of the Earth in your
> computations above!
Simple!
v = sqrt( G*(M+m) / r )
where
M = Mass of Earth
m = mass of Moon
G = constant of gravity
r = orbital mean distance between Moon and Earth
plug in the numbers, and what do you get, Nancy ?
I bet you will get "arithmetic overflow" if you try to think about it!
Now take this message back to ZetaTalk - you haven't the slightest clue
about how physics works, and your misunderstandings are rediculus in
absurdum...
Clear Skies,
Magnus
Bob Officer
<zeta...@zetatalk.com> wrote:
>In Article <9jgg0s$8c4$3...@sevenofnine.peak.org> Bill Nelson wrote:
>> In sci.astro Nancy Lieder <zeta...@zetatalk.com> wrote:
>>> while out where the Concord flies or our satellites float,
>>> while moving at twice the speed of the Concord, or only
>>> 1/4 the speed of those satellites.
>>
>> The Concord flies, at maximum, at less than 12 miles above
>> the surface. The lowest satellites orbit at around 100 miles.
>> The geostationary satellites are at over 12,000 miles - maybe
>> 18,000 (I forget). The moon is in orbit at over 200,000 miles
>> distance. The Concord is not in orbit when it flies - so it
>> cannot be used for comparison.
>
>And using some other figures quoted (below), we've established, then,
>for the
>
>CONCORD:
> 533.33 m/s
> 12 miles high
Why are you putting this here. the concord is not in 'orbit'.
Apples and oranges.
>SATELLITES:
> 7897.873415 m/s
> 100 miles high (lowest)
>
> Geo Stationary
> 4,635.7155 m/s
> 12,000 - 18,000 miles high
>
>MOON:
> 1023 m/s
> 200,000 miles high
--
Uyelvha`i
Nancy Lieder
> the "gravity pull" (acceleration) is caused by the same Earth,
>
> F = GMm / r^2 (which I assume you would term "gravity
> pull", and that we call force of
gravitation)
>
> a = GM / r ^2 (since a = F / m) a is of course "acceleration",
> unless you missed that (again)
I'm sorry, what happened to the force of gravity being a FACTOR of the
mass of the two objects? Inverse Square law. Or can't we put the two
formulas on the same page, as the Zetas have stated?
And in doing this, we should NOT be retreating to an abstract "mass" for
the Moon, as a cop-out! So as long as we use the same UNIT of measure
for the Earth, Moon, and Satellite, we are comparing apples-to-apples in
the forumulas. So sayeth Mr. Tholen (who works for NASA under various
hand-off arrangements but just won't admit it).
In Article <Ayc77.12138$0s2.1...@typhoon.hawaii.rr.com> David Tholen
wrote:
>> QUESTION: How to compute the MASS of the Earth and Moon,
>> using something REAL like granite (not an abstract number
>> computed to make Newton's formula work)?
>
> one can determine that the Earth is 81 times more massive than
> the Moon, but the mass in absolute units depends on the definition
> of the unit.
The cop-out, described:
In article <MPG.fc0bb991...@news.connect.ab.ca> Paul Campbell
writes:
>> You're saying that your equations balance, but then only
>> balance because you've CALCULATED the weight of the
>> Moon using the orbital mechanics formula, right? So if
>> you would enter any other weight for the Moon into those
>> equations, then the Moon either plummets or ejects into
>> space?
> Nancy wrote
Nancy Lieder
>> basis as we do for the Mass of the Earth in your computations above!
>
>
> where
> M = Mass of Earth
> m = mass of Moon
> G = constant of gravity
> r = orbital mean distance between Moon and Earth
In Article <0qh77.26780$CM3.1...@weber.videotron.net> Greg Neill
wrote:
> Replace M in the formula with (M+m), where M is the mass
> of the Earth and m the mass of the Moon. Since the mass of
> the Moon is about 1/81 the mass of the Earth, that's equivalent
> to 1.012M. So the results would differ by about 1.2 percent.
Why are we ADDING the two masses together here, instead of an Inverse
Square computation? There is obviously a lot more MASS when the two
multiply each other! Can't get you math to fit on the same page, eh?
Just as the Zetas said!
(M+m) = 5.9763e+24 kg + 7.3508e+22 kg = cop out
(M1*M2) = 5.9763e+24 kg * 7.3508e+22 kg = same page
In article <353CFF...@spammers.of.the.world.unite.etc> M.C. Harrison
writes.
> The force of gravity is an inverse square law, which means
> a mass will experience a force due to another mass according to
> the equation F=M1*M2/r^2 where M1 is one mass, M2 is the
> other mass, and r is the separation of the masses. ...
Nancy Lieder
> gravitational CONSTANT.
We have a gravitational CONSTANT? Gravity is a factor of both masses,
and increases in proportion to the size of the masses involved, per the
Inverse Square law! Or can't we put both these equations on the same
page, as the Zetas stated.
In article <353CFF...@spammers.of.the.world.unite.etc> M.C. Harrison
writes.
> The force of gravity is an inverse square law, which means
> a mass will experience a force due to another mass according to
> the equation F=M1*M2/r^2 where M1 is one mass, M2 is the
> other mass, and r is the separation of the masses. ...
So if we put these two "laws" on the same page, then G in the equation
for a "low mass object" might be a constant for a satellite, but should
be COMPUTED for the Moon. Computing this, then, gives us the following
equation (below) for the correct velosity of the Moon. Lets plug in the
PROPER force of gravity between the Earth and Moon in the velosity
equation (which after all should work if your math can be put on the
same page), and see what we get! Or can't we put both your equations on
the same page, as the Zetas stated.
In Article <Lsi77.11081$e5.16...@newsb.telia.net> Magnus Nyborg
> v = sqrt( G*M / r )
>
> Ground orbit (if possible) -
> v = sqrt( 6.67E-11 * 5.976E24 / 6.378E6 ) = 7905 m/s
> Satellite orbit -
> v = sqrt( 6.67E-11 * 5.976E24 / 6.478E6 ) = 7844 m/s
> Moon orbit -
> v = sqrt( 6.67E-11 * 5.976E24 / 3.844E8 ) = 1018 m/s
<============ WRONG FORMULA FOR HIGH MASS MOON
But the Moon's velosity should be:
v = sqrt( G*M / r )
v = sqrt( (Inverse Square) / 3.844E8) = ?
v = sqrt( (F=M1*M2/r^2) / 3.844E8) = ?
v = sqrt( (F=5.9763e+24 kg * 7.3508e+22 kg) / 3.844E8) = ?
<============ RIGHT FORMULA FOR HIGH MASS MOON
Where
M1 = Earth = 5.9763e+24 kg
M2 = Moon = 7.3508e+22 kg
r = 200,000 miles = 3.844E8
G*M = Gravity Constant of Earth = 6.67E-11 * 5.976E24
(only valid for low-mass orbiters)
And I'll bet the resulting m/s are no where near the 1023 m/s or 1018
m/s of the Moon's actual rate.
Nancy Lieder
Inverse Square law (F=M1*M2/r^2) and
Newton's laws (F = mA)
and find that F = F NOT when applies to the Moon, I suggest to compute
the Repulsion Force per the Zetas description of when and how fast this
clicks in, and see if THIS computes not only for the Earth and Moon, but
explains why Planet do NOT perturb closer into the Sun, but stay in
their orbit paths when passing each other! There is, after all, no
explanation for this in current human theory. Unless everyone runs away
when it is shown that human math cannot be put on the same page, and you
have NO explanation for why the Moon is UP there, and why planets do NOT
perturb TOWARD the Sun.
Why would the planets not drift into the Sun? Are the orbits
all that swift so that centrifugal force is extreme? ... The reason
Mankind is Unaware of a repulsive force, also inherent in
gravity, is that for this to become evident there must be a
semblance of equality in size and weight, i.e. the mass of the
objects, and freedom of movement such as exists in space,
and lack of undue influence from other nearby objects. ... The
repulsion force is generated as a result of two bodies exerting a
gravitational force on each other. ... Where the repulsion force
comes to equal the force of gravity by the time the objects in
play would make contact, it builds at a rate that differs from
gravity. ... The repulsion force is infinitesimally smaller than
the force of gravity, but has a sharper curve so that it equals
the force of gravity at the point of contact.
ZetaTalk, Repulsion Force s34
(http://www.zetatalk.com/science/s34.htm)
Gravity particles produce a flow but produce no discernible
flow, and have no irregularities in the pattern. Does your
Earth not pull evenly from all parts of its surface? And if
there is a flow, then at what point does the flow reverse, such
that surface particles are pushed away? In fact there is a
reversal, but the outward streams are propelled, with a force
and at a speed so much greater than the downward drafts that
this occurs over less of a surface area and without engaging
the mass of the object. A laser of gravity particles, versus a
floodlight upon the return. So why would the weight of
returning particles be the only ones mankind is aware of,
and why would they not feel the violent lift of the updrafts?
The updrafts blast through, tearing a hole as it were, where
the returning particles do not tear what they press upon,
and so have the greater effect.
Gravity particles, in their motion, do not affect what they
move against or through, the effect being in essence
mechanical. The upward drafts push aside other matter,
letting it return upon completion of the updraft, leaving no
trace of the temporary tear. The downward push of gravity
particles returning to the large mass they are attracted to,
the core of the Earth for instance, spread out upon objects
they encounter, taking some time to drift through these
object and with a constant downward press during the
motion of this drift. Thus, returning particles, due to the
time they spend upon and within the surface objects, and
due to their continual direction of motion, are a mechanical
force that is stronger, overall, than the updraft of particles
that quickly pass through the surface objects, essentially
pushing them aside rather than engaging them.
The nature of this gravity flow is what determines the
repulsion force we speak of. It is a complement of gravity
only when large bodies are close to each other. The
updrafts, when encountering a large body also exuding
updrafts of gravity particles, hold the bodies apart. This
occurs at what humans would call a distance from each
other, as small objects such as satellites do not exude
updrafts and if far enough from the surface of a
gravitational giant such as a planet, find a down-draft
and updraft of gravity particles in balance, what humans
might term in their ignorance a zero gravity field,
weightlessness. At this point the updrafts are still tearing
through, but at a slower rate, so that a mechanical push
upward is involved, and the down-drafts are more thinly
dispersed over the surface as they work their way through
the density of these objects in space. Large bodies,
exuding their own updrafts of gravity particles, create a
situation where their updrafts and the updrafts from
another sun or planet bump against each other, creating a
buffer and preventing the gravity masses from touching or
even approaching each other except at great distances.
ZetaTalk, Gravity Flow
(http://www.zetatalk.com/science/s96.htm)
Gravity is particles, moving, just as magnetic fields are, and
there is a polarization in gravity, which we have explained
as the repulsion force. Before mankind discovered that
magnetism was polarized, they discovered it as an attractive
force. Metallic items stuck to the sides of magnetized rock -
how curious. After centuries of digging about in this
phenomena, humans have satisfied their curiosity to the
extent that they understand that magnetism is a force field,
has a flow out from one pole and in at the other pole, that
the Sun and some other planets are magnetized and line up
with each other. They still don't understand the cause of
this force field, or its nature. Magnetism is caused by a
particle, in motion, as we have explained. What other
explanation is there for a force that reaches out and affects
another? Magic?
The bi-polar aspect of magnetism is only apparent when
what occurs in nature can be countered in the laboratory.
You force magnetized objects to do what they do not want to
do - touch north pole to north pole or vice versa. Then you
can observe the bi-polar nature. In gravity, you are seeing
but one aspect in the positioning of the planets, and dealing
with a phenomena that does not lend itself to easy
experimentation. However, experimentation is possible, in
space and away from the surface of the planet. The repulsion
force fills the gap in some of your other theories where you
have no explanation for discrepancies.
In magnetism, the simple flow of particles creates more than
a force for alignment, it creates an attraction. The gap is filled.
Like water in a stream, where flotsam eventually lines up in
the center, evenly spaced, just so magnetized objects do not
keep their distance when free to move. They approach each
other, and attach like a string of pearls. Likewise the
phenomena of gravity, where the desire to fill the gap causes
objects to approach one another. It is only where this gap is
overfilled, by the presence of two large objects coming near,
that the repulsion force is expressed. There is no room for the
flow of gravity particles, so the objects stay apart!
ZetaTalk, Gravity Particles
(http://www.zetatalk.com/science/s87.htm)
Greg Neill
> In Article <0qh77.26780$CM3.1...@weber.videotron.net> Greg Neill
> wrote:
> > The [6.67E-11 ] exponent is on the value for G, the
> > gravitational CONSTANT.
>
> We have a gravitational CONSTANT? Gravity is a factor of both masses,
> and increases in proportion to the size of the masses involved, per the
> Inverse Square law! Or can't we put both these equations on the same
> page, as the Zetas stated.
Idiot squared.
Newton's formula for the gravitational force between two masses:
F = G*M1*M2/r^2
where G is the Gravitational constant, the constant of proportionality
that makes the units all work out and sets the magnitude of the strength of
the gravitational force.
So yes, silly girl, we have a gravitational CONSTANT.
Greg Neill
> In Article <Lsi77.11081$e5.16...@newsb.telia.net> Magnus Nyborg
> >> Lets plug in an equivalent Mass of the Moon, using the same
> >> basis as we do for the Mass of the Earth in your computations above!
> >
> > v = sqrt( G*(M+m) / r )
> >
> > where
> > M = Mass of Earth
> > m = mass of Moon
> > G = constant of gravity
> > r = orbital mean distance between Moon and Earth
>
> In Article <0qh77.26780$CM3.1...@weber.videotron.net> Greg Neill
> wrote:
> > Replace M in the formula with (M+m), where M is the mass
> > of the Earth and m the mass of the Moon. Since the mass of
> > the Moon is about 1/81 the mass of the Earth, that's equivalent
> > to 1.012M. So the results would differ by about 1.2 percent.
>
> Why are we ADDING the two masses together here, instead of an Inverse
> Square computation? There is obviously a lot more MASS when the two
> multiply each other! Can't get you math to fit on the same page, eh?
> Just as the Zetas said!
Blithering idiot.
Suppose you take a given lump of matter and cut it in half. Do you
expect the resulting gravitational energy to suddenly multiply?
Mass does not multiply to yield more mass. Mass adds. If you
add 10 kilograms of sugar to another 10 kilograms of sugar, you
do not end up with 100 kilograms of sugar. You end up with 20.
When the equations are derived for one very heavy central mass with
an insignificant (by comparison) mass in orbit, then the frame of
reference taken is usually that of the center of the heavy central
mass. In reality, no matter how tiny the secondary mass is, both
masses actually orbit about their mutual center of gravity, but for
a puny secondary that's as close to the center of the massive primary
as makes no difference. In this case we find that
F = G*M*m/r^2
and since a = F/m
a = G*M/r^2
However, when the second mass (m in this case) starts to become of
non-negligible size (as you seem to think is the case here, where the
mass of the Earth is only some 81 time the mass of the Moon), then one
has to consider moving ones frame of reference to the center of mass
of the system. The center of mass is a stable point around which both
the bodies will orbit, kind of like two people holding hands and spining
around.
In this case the equations retain the same form, namely a = G*M/r^2, if
the mass M is replaced with M+m. It is trivially derivable, and is
usually done in the first chapter of any astrodynamics book.
You should be aware that in the case of the Earth-Moon system, despite
the seemingly large mass of the Moon at least when considered as an
isolated number rather than in comparison to the that of the Earth,
the center of gravity is still well below the surface of the Earth. In
this case accuracy does not suffer significantly by ignoring the mass
of the Moon.
Nancy Lieder
> But the Moon's velosity should be:
> v = sqrt( G*M / r )
> v = sqrt( (Inverse Square) / 3.844E8) = ?
> v = sqrt( (F=M1*M2/r^2) / 3.844E8) = ?
> v = sqrt( (F=5.9763e+24 kg * 7.3508e+22 kg) / 3.844E8) = ?
> <============ RIGHT FORMULA FOR HIGH MASS MOON
>
> Where
> M1 = Earth = 5.9763e+24 kg
> M2 = Moon = 7.3508e+22 kg
> r = 200,000 miles = 3.844E8
> G*M = Gravity Constant of Earth = 6.67E-11 * 5.976E24
> (only valid for low-mass orbiters)
I forgot a divisor by the square of r. That line should be:
v = sqrt( (F=5.9763e+24 kg * 7.3508e+22 kg)/ 3.844E8^2 / 3.844E8) = ?
<============ RIGHT FORMULA FOR HIGH MASS MOON
Sorry.
Nancy Lieder
wrote:
>>> gravitational CONSTANT.
>>
>> We have a gravitational CONSTANT?
>
> F = G*M1*M2/r^2
> where G is the Gravitational constant, the constant of proportionality
> that makes the units all work out and sets the magnitude of the
> strength of the gravitational force.
Then lets plug that CONSTANT into both equations, which should come up
with the same VELOSITY for the Moon (which it won’t and Greg will go off
blustering). Restated, then, the problem is to put it all on the same
page, using same units of measure for MASS per Mr. Tholen (who actually
works for NASA under various hand-off arrangements but won’t admit it).
The Moon's velosity should be:
v = sqrt( G*M / r )
v = sqrt( (Inverse Square) / 3.844E8) = ?
v = sqrt( (G*M1*M2/r^2) / 3.844E8) = ?
v = sqrt( (6.67E-11*5.9763e+24 kg * 7.3508e+22 kg/3.844E8^2) /
3.844E8) = ?
<============ RIGHT FORMULA FOR HIGH MASS MOON
Where
M1 = Earth = 5.9763e+24 kg
M2 = Moon = 7.3508e+22 kg
r = 200,000 miles = 3.844E8
G = Gravity Constant of Earth = 6.67E-11
And where the force of gravity between two objects is another Newton
law:
F = G*M1*M2/r^2
Should the velosity not be countering the force of gravity, to keep the
Moon from plunging to Earth? But it’s out there sadately poking along!
Floating up there! Almost as though it were floating on .... a bed of
gravity particles jammed between the Earth and Moon, as in the REPULSION
FORCE!
Michael Davis
> In Article <0qh77.26780$CM3.1...@weber.videotron.net> Greg Neill
> wrote:
> > The [6.67E-11 ] exponent is on the value for G, the
> > gravitational CONSTANT.
>
> We have a gravitational CONSTANT?
ROTFLMAO!!! Classic stupidity!
--- Snip remainder of stupidity ---
--
The Evil Michael Davis™
http://mdavis19.tripod.com
http://www.mdpub.com/ufo/skeptic.html
http://skepticult.org Member #264-70198-536
Flaggy random killfile member #33 1/3
"Goddammit! The world is just filling up with more and more idiots! And the
computer is giving them access to the world! They're Spreading their
stupidity! At least they were contained before - now they're on the loose
everywhere!" - Harlan Ellison
Greg Neill
> In Article <6iD77.35516$CM3.2...@weber.videotron.net> Greg Neill
> wrote:
> > In Article <3B5EFB62...@zetatalk.com> Nancy Lieder wrote:
> >> In Article <0qh77.26780$CM3.1...@weber.videotron.net> Greg Neill
> wrote:
> >>> The [6.67E-11 ] exponent is on the value for G, the
> >>> gravitational CONSTANT.
> >>
> >> We have a gravitational CONSTANT?
> >
> > Newton's formula for the gravitational force between two masses:
> > F = G*M1*M2/r^2
> > where G is the Gravitational constant, the constant of proportionality
>
> > that makes the units all work out and sets the magnitude of the
> > strength of the gravitational force.
>
> Then lets plug that CONSTANT into both equations, which should come up
> with the same VELOSITY for the Moon (which it won't and Greg will go off
> blustering). Restated, then, the problem is to put it all on the same
> page, using same units of measure for MASS per Mr. Tholen (who actually
> works for NASA under various hand-off arrangements but won't admit it).
>
Again you go off in all directions as though incapable of holding
a single thought captive without it wiggling away. Please also
learn to spell "velocity".
Now, about your little math problem.
The simplest way to approach the problem is to equate the two forces
that are in balance in such a system, namely the gravitational force
and the centrifugal force (that pseudo-force which is the result of
inertia for a body experiencing uniform circular motion). Thus:
Force due to gravity is F1 = G*M1*M2/r^2
Centrifugal force is F2 = M2*v^2/r
Note that the gravitational force includes both masses, which should
make you happy. Note that the centrifugal force involves just the
Moon's mass, which makes sense if you consider that for a given
accelerated path (in this case a circular path) this force depends
only on how massive the body is and how its direction of motion is
changing.
Fine, now equate the two, in order to place things in balance. We have:
F1 = F2
G*M1*M2/r^2 = M2*v^2/r
G*M1/r = v^2
v = sqrt(G*M1/r)
In the case of the Earth-Moon system we have:
M1 = 5.9736*10^24 kg
M2 = 0.07349*10^24 kg
r = 0.3844*10^6 km (384,400 km)
so that v = 1.018 km/sec.
In point of fact the Moon's orbit is not perfectly circular, and
there will be some variation in the orbital velocity. It varies
from about 0.964 km/sec to about 1.076 km/sec.
> The Moon's velosity should be:
> v = sqrt( G*M / r )
Okay.
> v = sqrt( (Inverse Square) / 3.844E8) = ?
What? What is "inverse square"? What are its units?
> v = sqrt( (G*M1*M2/r^2) / 3.844E8) = ?
What? Now you've got the distance cubed in the denominator
(assuming that the 3.844E8 item is really in meters). This
formula is incorrect because the units don't work out. Where did
you get this formula? Tell your source he his confused.
> v = sqrt( (6.67E-11*5.9763e+24 kg * 7.3508e+22 kg/3.844E8^2) /
> 3.844E8) = ?
> <============ RIGHT FORMULA FOR HIGH MASS MOON
>
> Where
> M1 = Earth = 5.9763e+24 kg
> M2 = Moon = 7.3508e+22 kg
> r = 200,000 miles = 3.844E8
> G = Gravity Constant of Earth = 6.67E-11
That's the Universal Gravitational Constant, which applies to *all* masses,
not just the Earth.
>
> And where the force of gravity between two objects is another Newton
> law:
> F = G*M1*M2/r^2
>
> Should the velosity not be countering the force of gravity, to keep the
> Moon from plunging to Earth? But it's out there sadately poking along!
> Floating up there! Almost as though it were floating on .... a bed of
> gravity particles jammed between the Earth and Moon, as in the REPULSION
> FORCE!
Who says that the velocity is not sufficient to maintain the orbit? By the
very derivation of the velocity above, the inward and outward forces were
assumed to balance, so the velocity obtained *must* balance things. Of
course, it is also nice to know that the _actual_ velocity of the Moon
can be measured empirically, and that it happens to match the result.
Now, what was you problem? Try to focus.
tho...@antispam.ham
> but explains why Planet do NOT perturb closer into the Sun,
But they do perturb "closer into the Sun". Consider the osculating
semimajor axis for Pluto as of epoch 2444800.5 = 1981 July 15
39.9165 AU
and as of epoch 2451800.5 = 2000 September 13
39.2353 AU
From DE403.
> but stay in their orbit paths when passing each other!
No, the planets do NOT "stay in their orbit paths when passing each
other". They perturb each other, according to the laws of gravity.
Go ahead, compare the osculating orbits for Saturn before and after
Jupiter passes it. You'll find that they are different.
> There is, after all, no explanation for this in current human theory.
You're erroneously presupposing that they "stay in their orbit paths".
> Unless everyone runs away when it is shown that human math cannot be
> put on the same page, and you have NO explanation for why the Moon
> is UP there,
I have an explanation for why the Moon is up there, and that explanation
uses human math. It's called Newton's laws.
> and why planets do NOT perturb TOWARD the Sun.
But they do. They also perturb away from the Sun. It just so happens
that the two average out to zero close enough for the Solar System to
be stable for billions of years.
tho...@antispam.ham
> I'm sorry, what happened to the force of gravity being a FACTOR of the
> mass of the two objects?
Nothing happened to it.
> Inverse Square law.
Non sequitur. You were talking about the mass. Suddenly you've switched
to talking about the distance.
> Or can't we put the two formulas on the same page, as the Zetas have
> stated?
Sure you can. Your "Zetas" are wrong.
> And in doing this, we should NOT be retreating to an abstract "mass" for
> the Moon, as a cop-out!
What is "an abstract 'mass'"?
> So as long as we use the same UNIT of measure for the Earth, Moon,
> and Satellite, we are comparing apples-to-apples in the forumulas.
> So sayeth Mr. Tholen
I said nothng about "an abstract 'mass'". I said that relative mass
can be computed for two objects independently of any adopted absolute
unit of mass. Neither of the words "relative" and "absolute" is the
same as "abstract".
> (who works for NASA under various hand-off arrangements
What alleged "hand-off arrangements"?
> but just won't admit it).
You're the one making the claim. Now substantiate it.
>>> QUESTION: How to compute the MASS of the Earth and Moon,
>>> using something REAL like granite (not an abstract number
>>> computed to make Newton's formula work)?
>> one can determine that the Earth is 81 times more massive than
>> the Moon, but the mass in absolute units depends on the definition
>> of the unit.
> The cop-out, described:
What alleged cop-out? You didn't provide any description.
tho...@antispam.ham
> Why are we ADDING the two masses together here, instead of an Inverse
> Square computation?
Inverse square applies to the distance, not the mass.
> There is obviously a lot more MASS when the two multiply each other!
Incorrect; there is the same amount of mass. Once you multiply them,
you no longer have mass, but rather mass-squared, which is a different
unit altogether, one that cannot be compared to mass.
For example, if you have a room 10 feet wide and 15 feet long, you
don't go to the carpet store and ask for 25 feet of carpeting, do you?
Rather, you ask for 150 square feet of carpeting. Square feet is not
the same unit as feet.
> Can't get you math to fit on the same page, eh?
I can; you obviously can't.
Still waiting for your answer to the questions:
] Nancy Lieder writes:
]
] > I was told by an astronomer who works for a large California observatory
] > the following "We were looking for it, then we found it, and now we're
] > tracking it.
]
] Which "astronomer"? Which "large California observatory"?
]
] Recall that you have repeatedly and erroneously accused me of evading
] answers. Will you act hypocritically and evade the answers to these
] questions?
Bob May
mean! Not only that, she doesn't know what she's looking at.
--
Bob May
Remember that computers do exactly what you tell them to do, not what you
think you told them to do.
tho...@antispam.ham
> We have a gravitational CONSTANT?
A constant of proportionality.
> Gravity is a factor of both masses,
So what? You need a constant of proportionality to make the units
work out. For example, if you have two masses each of 1 kilogram,
the product of those two masses has the numerical value of 1, but
those masses can also be expressed as 1000 grams each, which becomes
1 million after you multiply them. Yet the force must be the same.
The masses haven't changed, only the units used to measure them.
When you change the units, you have to change the constant of
proportionality accordingly.
Let's use another carpeting example. Suppose a carpet store
advertises a great deal of carpeting for a dollar a square yard.
Your room is 15 feet by 25 feet. Will you gladly pay $375 for
your carpeting, because 15 x 25 = 375? How about $54,000 for
the carpeting, because 15 feet = 180 inches, 25 feet = 300 inches,
and 180 x 300 = 54000? Or will you finally realize that the cost
per unit area is the constant of proportionality that needs to be
changed for the units in which you're working?
> and increases in proportion to the size of the masses involved,
> per the Inverse Square law!
Wrong. The inverse square law applies to the distance between the
masses, not the masses themselves.
> Or can't we put both these equations on the same page, as the
> Zetas stated.
You obviously can't. That's your problem.
Bob May
been around for centuries here on the Earth! Not only that, but the zetan
math doesn't work out in the real world.
tho...@antispam.ham
> Restated, then, the problem is to put it all on the same page,
> using same units of measure for MASS per Mr. Tholen (who actually
> works for NASA under various hand-off arrangements but won’t admit it).
Put up your evidence, or shut up, Nancy.
Bob May
corrected herself with a little more in an attempt to make reality and her
math correspond.
John Shakespeare
tho...@AntiSpam.ham wrote:
For Nancy's benefit, the requisite term is "stable in the sense of Lyapunov".
In other words, their orbits do vary, but the variation is bounded, so that
the orbits remain within certain finite envelopes. It's a sure thing that she
won't understand the mathematics involved.
Best Regards,
John.
Magnus Nyborg
> wrote:
> > In Article <3B5EFB62...@zetatalk.com> Nancy Lieder wrote:
> >> In Article <0qh77.26780$CM3.1...@weber.videotron.net> Greg Neill
> wrote:
> >>> The [6.67E-11 ] exponent is on the value for G, the
> >>> gravitational CONSTANT.
> >>
> >> We have a gravitational CONSTANT?
This is not equal to a CONSTANT gravitation...*sigh*
> >
> > Newton's formula for the gravitational force between two masses:
> > F = G*M1*M2/r^2
> > where G is the Gravitational constant, the constant of proportionality
>
> > that makes the units all work out and sets the magnitude of the
> > strength of the gravitational force.
>
> Then lets plug that CONSTANT into both equations, which should come up
> with the same VELOSITY for the Moon (which it won't and Greg will go off
> blustering). Restated, then, the problem is to put it all on the same
> page, using same units of measure for MASS per Mr. Tholen (who actually
> works for NASA under various hand-off arrangements but won't admit it).
>
> The Moon's velosity should be:
> v = sqrt( G*M / r )
> v = sqrt( (Inverse Square) / 3.844E8) = ?
Woooaaah! Please learn to use the correct formula, which is nothing other
than v = sqrt( GM / r )...period!
[incorrect stuff snipped]
> Where
> M1 = Earth = 5.9763e+24 kg
> M2 = Moon = 7.3508e+22 kg
> r = 200,000 miles = 3.844E8
> G = Gravity Constant of Earth = 6.67E-11
>
> And where the force of gravity between two objects is another Newton
> law:
> F = G*M1*M2/r^2
,,,and the centrifugal force is...
F = M2*v^2
solve for equality F1 = F2 (if there is a solution, there must be a balance
between the two formulas or the Moon would flie out of orbit) and you get...
G*M1*M2/r^2 = M2*v^2 or...
G*M1/r = v^2, which solved for v gives
v = sqrt(GM1 / r), which has been stated before - and which have now
been proven to you beyond any doubt!
>
> Should the velosity not be countering the force of gravity, to keep the
> Moon from plunging to Earth? But it's out there sadately poking along!
> Floating up there! Almost as though it were floating on .... a bed of
> gravity particles jammed between the Earth and Moon, as in the REPULSION
> FORCE!
Put the right number into the right formula in the right way - THEN you
might get the correct answer!
Clear Skies,
Magnus
john Latala
On Wed, 25 Jul 2001 tho...@AntiSpam.ham wrote:
> Nancy Lieder writes:
> > Unless everyone runs away when it is shown that human math cannot be
> > put on the same page, and you have NO explanation for why the Moon
> > is UP there,
>
> I have an explanation for why the Moon is up there, and that explanation
> uses human math. It's called Newton's laws.
I think there's an even easier explanation even the Zetans can understand.
It doesn't even need any math. The moon is up there and has been for all
of recorded human history because nothing's knocked it down yet. If your
math says it shouldn't be up there then it's easy to double-check your
math by just going outside and seeing if it's still there. If it is then
your math is wrong.
Bill Nelson
:> I have an explanation for why the Moon is up there, and that explanation
:> uses human math. It's called Newton's laws.
: I think there's an even easier explanation even the Zetans can understand.
: It doesn't even need any math. The moon is up there and has been for all
: of recorded human history because nothing's knocked it down yet. If your
: math says it shouldn't be up there then it's easy to double-check your
: math by just going outside and seeing if it's still there. If it is then
: your math is wrong.
But Nancy will not accept our mathematics. She tries to use incorrect
equations - then assumes there is a "repulsive force packed between the
Earth and Moon" (not an exact quote - but close).
The task will be to convince here that no repulsive force is necessary,
that the momentum of the Moon in orbit is sufficient.
--
Bill Nelson (bi...@peak.org)
da...@pebble.org
>Yuppie, the nitwit doesn't even know the basic formulas or what the letters
>mean! Not only that, she doesn't know what she's looking at.
It's funnier than that Bob. She doesn't even understand scientific
notation (as taught in grade school). JosX was an absolute riot
today as well. Nancy has JosX as a pet to do "stupid pet tricks."
Michael L Cunningham
> Yuppie, the nitwit doesn't even know the basic formulas or what the letters
> mean! Not only that, she doesn't know what she's looking at.
>
What's funny, her Zetas say the human math doesn't work, yet she uses
it to try and prove she's right!
Which is it Nancy? Either the Zeta's are right or wrong. You shouldn't be using
the human math because it doesn't work. Use Zeta math as anything is possible
with it!
--
Michael L. Cunningham
e-mail boge...@earthlink.net
web site http://home.earthlink.net/~bogeystar/
Visit the LX50 Web Site and join in our Discussion Forum!
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Nancy Lieder
wrote:
>> And where the force of gravity between two objects is another
>> Newton law:
>> F = G*M1*M2/r^2
>>
>> the Moon from plunging to Earth? But it's out there sadately poking
>> along! Floating up there! Almost as though it were floating on ....
>> a bed of gravity particles jammed between the Earth and Moon, as
>> in the REPULSION FORCE!
>
Because in your velocity formula, you EXCLUDE the mass the gravity
attraction between the Earth and Moon! This is, per Newton, a factor of
BOTH masses.
The challenge was to put Newton's Law on the force of gravity together
on the same page with his his other laws, such as velocity and rate of
acceleration required to keep a body in orbit. If the force of gravity
between the Earth and Moon is HUGE, then you cannot simply pretend the
Moon is another orbiting body like a satellite. In the velocity and
acceleration formulas, plug in the ACTUAL gravity attraction between
these two objects, and the point the Zetas are trying to drag you to
admitting is that the Moon is too massive, and moving too slowly, to be
up there PER NEWTON. Newton does not work. There is a repulsion force
in play, keeping the Moon aloft.
If you were to plug in the TRUE force of gravity between the Earth and
Moon, not a single M, into the Velocity formula, what would that formula
look like?
>> v = sqrt((6.67E-11*5.9763E24*7.3508E22 /3.844E8^2)/3.844E8)
>> <============ RIGHT FORMULA FOR HIGH MASS MOON
Nancy Lieder
>> it won't and Greg will go off blustering). Restated, then, the
>> problem is to put it all on the same page, using same units of
>
> The simplest way to approach the problem is to equate the two
> forces that are in balance in such a system, namely the
> gravitational force and the centrifugal force (that pseudo-force
> which is the result of inertia for a body experiencing uniform
> circular motion). Thus:
>
> Force due to gravity is F1 = G*M1*M2/r^2
>
> Centrifugal force is F2 = M2*v^2/r
But you're not doing what you SAID you were doing. If the force of
gravity is a factor of BOTH gravitational masses, increasing as the size
of these masses increases and the distance between then decreases, per
Newton, then why should the Centrifugal force NOT have to consider that
force? Put BOTH those Newton laws together. Plug in not a SINGLE mass
factor into Newton's Centrifugal force formula, but the force of gravity
per Newton! The Moon is having to overcome, with its velocity, a
greater gravity pull than the satellite.
Is there a Newton formula for velocity for a HIGH mass orbiting object?
Magnus mentioned the equation provided for satellites as a "low-mass
object". What's the formula for a high-mass orbiting object?
In Article <8u_67.10984$e5.16...@newsb.telia.net> Magnus Nyborg wrote:
> Orbital speed for ideal circular motion of a low-mass object
> circling a high-mass object M (which refers to it's mass) is
> determined by the formula
>
> v = sqrt( G*M / r )
Nancy Lieder
> should make you happy. Note that the centrifugal force
> involves just the Moon's mass, which makes sense if you
> consider that for a given accelerated path (in this case a
> circular path) this force depends only on how massive the
> body is and how its direction of motion is changing.
>
> Fine, now equate the two, in order to place things in balance. We
have:
>
> F1 = F2
> G*M1*M2/r^2 = M2*v^2/r
> G*M1/r = v^2
>
> v = sqrt(G*M1/r)
>
> In the case of the Earth-Moon system we have:
>
> M1 = 5.9736*10^24 kg
> M2 = 0.07349*10^24 kg
> r = 0.3844*10^6 km (384,400 km)
>
> so that v = 1.018 km/sec.
That looks lovely and I'm sure you sleep soundly at night knowing that
mathematically, your world is in order. Except the math is WRONG, as it
does not consider the gravity attraction between the Earth and Moon when
computing velocity. In the 1998 exercise, it was shown that we have a
massive Moon, per the Inverse Square law, of millions of trillions of
metric tons of equivalent weight if on the surface of the Earth, moving
at a rate of only 1023 m/s (or 1018 m/s if you prefer). As the Zetas
stated:
Your Moon, adjusted for the distance it is, and going at the rate it
does while carrying this adjusted weight [reduced for distance], is
going at only 1/4 THE SPEED OF YOUR SATATIONARY
SATELLITES! ... Your astronomers, unable to bring all their
equations and physics together on one page, are telling you that
your Moon, at an adjusted weight of 20,228,796,000,000,000 metric
tons, could cruise along at 1/4 the speed of your stationary
satellites
... and maintain it's place DUE TO CENTRIFUGAL FORCE! The
theatre of the absurd is about to open.
ZetaTalk™
The figures used during this discussion, were:
Earth: 5.9763e+24 kg ( 5.9763e+21 Metric Tons)
Moon: 7.3508e+22 kg ( 7.3508e+19 Metric Tons)
gravitational equation F = G*m1*m2/r^2
acceleration of the moon due to gravity toward the Earth
F = 2.20228796E+16 metric ton force
or = 20,228,796,000,000,000 metric tons-force
C = Pi*2*R = Pi*D
gives a circular orbital speed of: 1023.183 m/s
Nancy Lieder
wrote:
>> Where
>> M1 = Earth = 5.9763e+24 kg
>> M2 = Moon = 7.3508e+22 kg
>> r = 200,000 miles = 3.844E8
>> G = Gravity Constant of Earth = 6.67E-11
>>
>> Newton law:
>> F = G*M1*M2/r^2
>
> F = M2*v^2
>
> solve for equality F1 = F2 (if there is a solution, there must be a
> balance between the two formulas or the Moon would flie out of
> orbit) and you get...
>
> G*M1*M2/r^2 = M2*v^2 or...
> G*M1/r = v^2, which solved for v gives
> v = sqrt(GM1 / r), which has been stated before
How can centrifugal force consider ONLY the mass of the orbiting body?
Doesn't the Moon have a greater gravity pull to deal with, per the
Inverse Square law? We just don't put these together! Can't! They
don't work together as Newton is WRONG!
The Inverse Square law reduces the mass according to the distance. We
did that, in the 1998 exercise, to compute the EQUIVALENT mass of the
Moon if on the surface of the Earth, if orbiting that close. We kept
the velocity the same, as the mass had already been reduced. If you saw
an elephant floating slowly by at inches per hour, 10 feet off the
ground, and someone said it was staying aloft due to the centrifugal
force generated by its great speed, you'd KNOW something was wrong,
intuitively. Yet this is what this exercise demonstrated! No one wants
to look at that elephant, and DEAL with the inability of Newton's
formulas to deal with it either! You CANNOT put his Inverse Square law
together with his Centrifugal Force or Velocity laws.
Nancy Lieder
wrote:
>> but explains why Planet do NOT perturb closer into the Sun,
>
> But they do perturb "closer into the Sun". Consider the osculating
> semimajor axis for Pluto as of epoch 2444800.5 = 1981 July 15
> 39.9165 AU
> and as of epoch 2451800.5 = 2000 September 13
> 39.2353 AU
> From DE403.
Then why not STAY closer in to the Sun, after having arrived there? You
can argue that a planet speeds up in its orbit when approaching another
body, and then slow down due to the drag now behind it, perhaps
canceling each other out so the orbit pace remains overall the same.
But if an object being passed [DE403] is CLOSER IN toward the Sun,
perturbing Pluto to orbit closer to the Sun by their combined gravity
pull, then what is it that pulls Pluto BACK AWAY from the Sun? DE403's
dead twin, on the outside of Pluto's orbit (just kidding here, I'm not
asserting there is a DE403 dead twin, Dave, so don't go off on a
tangent, please).
In Article <RMG77.353$Kr3....@typhoon.hawaii.rr.com> David Tholen
wrote:
>> but stay in their orbit paths when passing each other!
>
> No, the planets do NOT "stay in their orbit paths when passing
> each other". They perturb each other, according to the laws of
> gravity. Go ahead, compare the osculating orbits for Saturn
> before and after Jupiter passes it. You'll find that they are
> different.
>
>> and why planets do NOT perturb TOWARD the Sun.
>
> But they do. They also perturb away from the Sun. It just so
> happens that the two average out to zero close enough for the
> Solar System to be stable for billions of years.
What pulls Saturn OUT away from the Sun and Jupiter, after this
passage? It's reverence for Newton? We have a gravity Push-Away Law?
Saturn finds itself floating along in its orbit, now closer to the Sun,
and decides to JUMP AWAY from the Sun to honor Newton? To keep you'all
feeling comfy and smug? It's now in the same position as satellites
when their orbits degrade or decay or whatever the proper term is, and
start spiraling down to Earth. Why would it NOT be? (The answer is, of
course, the Repulsion Force which is the push-away.)
Greg Neill
> In Article <VHF77.35833$CM3.2...@weber.videotron.net> Greg Neill
> wrote:
> >> And where the force of gravity between two objects is another
> >> Newton law:
> >> F = G*M1*M2/r^2
> >>
> >> Should the velosity not be countering the force of gravity, to keep
> >> the Moon from plunging to Earth? But it's out there sadately poking
> >> along! Floating up there! Almost as though it were floating on ....
> >> a bed of gravity particles jammed between the Earth and Moon, as
> >> in the REPULSION FORCE!
> >
> > Who says that the velocity is not sufficient to maintain the orbit?
>
> Because in your velocity formula, you EXCLUDE the mass the gravity
> attraction between the Earth and Moon! This is, per Newton, a factor of
> BOTH masses.
>
Look above. F = G*M1*M2/r^2.
What do you think M2 is?
Greg Neill
> In Article <VHF77.35833$CM3.2...@weber.videotron.net> Greg Neill
> wrote:
> >> Then lets plug that CONSTANT into both equations, which
> >> should come up with the same VELOCITY for the Moon (which
> >> it won't and Greg will go off blustering). Restated, then, the
> >> problem is to put it all on the same page, using same units of
> >> measure for MASS.
> >
> > The simplest way to approach the problem is to equate the two
> > forces that are in balance in such a system, namely the
> > gravitational force and the centrifugal force (that pseudo-force
> > which is the result of inertia for a body experiencing uniform
> > circular motion). Thus:
> >
> > Force due to gravity is F1 = G*M1*M2/r^2
> >
> > Centrifugal force is F2 = M2*v^2/r
>
> But you're not doing what you SAID you were doing. If the force of
> gravity is a factor of BOTH gravitational masses, increasing as the size
> of these masses increases and the distance between then decreases, per
> Newton, then why should the Centrifugal force NOT have to consider that
> force?
The centrifugal force experienced by a body depends only upon
the radius of curvature of the path it takes, and its own
mass. That's why.
Centrifugal force is an inertial effect, that is, it is the
result of a mass' tendency to want to continue in uniform
motion (straight line). By applying a force to a mass, you
make it change direction. The resistance to this change is
called "inertia".
In this case the force being applied is the gravitational
force between Earth and Moon, F1 above.
> Put BOTH those Newton laws together. Plug in not a SINGLE mass
> factor into Newton's Centrifugal force formula, but the force of gravity
> per Newton! The Moon is having to overcome, with its velocity, a
> greater gravity pull than the satellite.
Clearly you do not comprehend (or refuse to). F1 is the
gravitational force, while F2 is the centrifugal force. The
two are equal and opposite in magnitude. Simple.
>
> Is there a Newton formula for velocity for a HIGH mass orbiting object?
Yes. It is v = sqrt(G*(M1 + M2)/r)
It is simple to derive this using a bit of vector algebra and
Newton's laws.
Greg Neill
> In Article <VHF77.35833$CM3.2...@weber.videotron.net> Greg Neill
> wrote:
> > Note that the gravitational force includes both masses, which
> > should make you happy. Note that the centrifugal force
> > involves just the Moon's mass, which makes sense if you
> > consider that for a given accelerated path (in this case a
> > circular path) this force depends only on how massive the
> > body is and how its direction of motion is changing.
> >
> > Fine, now equate the two, in order to place things in balance. We
> have:
> >
> > F1 = F2
> > G*M1*M2/r^2 = M2*v^2/r
> > G*M1/r = v^2
> >
> > v = sqrt(G*M1/r)
> >
> > In the case of the Earth-Moon system we have:
> >
> > M1 = 5.9736*10^24 kg
> > M2 = 0.07349*10^24 kg
> > r = 0.3844*10^6 km (384,400 km)
> >
> > so that v = 1.018 km/sec.
>
> That looks lovely and I'm sure you sleep soundly at night knowing that
> mathematically, your world is in order. Except the math is WRONG, as it
> does not consider the gravity attraction between the Earth and Moon when
> computing velocity.
Please then explain what F = G*M1*M2/r^2 is saying.
Greg Neill
> In Article <B%G77.11236$e5.17...@newsb.telia.net> Magnus Nyborg
> wrote:
> >> Where
> >> M1 = Earth = 5.9763e+24 kg
> >> M2 = Moon = 7.3508e+22 kg
> >> r = 200,000 miles = 3.844E8
> >> G = Gravity Constant of Earth = 6.67E-11
> >>
> >> And where the force of gravity between two objects is another
> >> Newton law:
> >> F = G*M1*M2/r^2
> >
> > and the centrifugal force is...
> > F = M2*v^2
> >
> > solve for equality F1 = F2 (if there is a solution, there must be a
> > balance between the two formulas or the Moon would flie out of
> > orbit) and you get...
> >
> > G*M1*M2/r^2 = M2*v^2 or...
> > G*M1/r = v^2, which solved for v gives
> > v = sqrt(GM1 / r), which has been stated before
>
> How can centrifugal force consider ONLY the mass of the orbiting body?
Imagine a body all alone in space with no other nearby masses.
Tie a string to it and whirl it around, measuring the tension
on the string. If you whirl it faster, you find that the force
goes up with the square of the speed. If you vary the mass, you
find that the force varies proportionally. If you lengthen the
string, you find that the force decreases, shorten it and the
force increases. Now write down the effects as equations:
F ~ v^2
F ~ M
F ~ 1/r
Combine all three effects:
F = M*v^2/r
Notice that nowhere did we need to know the mass at the pivot
point of the string.
> Doesn't the Moon have a greater gravity pull to deal with, per the
> Inverse Square law? We just don't put these together! Can't! They
> don't work together as Newton is WRONG!
Gravity is the force pulling inwards, and is F = G*M1*M2/r^2.
Centrifugal force is the outward force, and is equal to F = M*v^2/r
as shown above. Got it?
>
> The Inverse Square law reduces the mass according to the distance.
The mass stays the same. The distance changes, and so does the
resuting force.
> We
> did that, in the 1998 exercise, to compute the EQUIVALENT mass of the
> Moon if on the surface of the Earth, if orbiting that close.
The mass is the same no matter where it is. And a body in orbit
(free fall) does not have weight. Please learn the difference
between mass and weight. Weight is the force that is required
to hold an object motionless against the sum total of other
effects, like gravity and centrifugal force. Since the
gravitational force and centrifugal force on a body in free-fall
are equal and opposite, the net weight is zero.
> We kept
> the velocity the same, as the mass had already been reduced.
Mass is not reduced.
Remainder of rant based upon faulty premis above snipped.
john Latala
Hmmm ... I wonder if that's why the Zetans 'talk' to Nancy?
tho...@antispam.ham
>>> but explains why Planet do NOT perturb closer into the Sun,
>> But they do perturb "closer into the Sun". Consider the osculating
>> semimajor axis for Pluto as of epoch 2444800.5 = 1981 July 15
>> 39.9165 AU
>> and as of epoch 2451800.5 = 2000 September 13
>> 39.2353 AU
>> From DE403.
> Then why not STAY closer in to the Sun, after having arrived there?
Because perturbations can work both ways. Six years later, Jupiter
will be on the other side of the Sun, causing the barycenter to
shift, thus the heliocentric osculating orbit for Pluto will now be
different. It's called indirect perturbations.
> You can argue that a planet speeds up in its orbit when approaching
> another body, and then slow down due to the drag now behind it, perhaps
> canceling each other out so the orbit pace remains overall the same.
Why would I want to argue that?
> But if an object being passed [DE403] is CLOSER IN toward the Sun,
DE403 is not an object.
> perturbing Pluto to orbit closer to the Sun by their combined gravity
> pull, then what is it that pulls Pluto BACK AWAY from the Sun?
Perturbations can work both ways. Six years later, Jupiter will be
on the other side of the Sun, causing the barycenter to shift, thus
the heliocentric osculating orbit for Pluto will now be different.
It's called indirect perturbations.
> DE403's dead twin,
DE403 does not have a dead twin.
> on the outside of Pluto's orbit (just kidding here, I'm not
> asserting there is a DE403 dead twin, Dave, so don't go off on a
> tangent, please).
You obviously don't even know what DE403 is.
>>> but stay in their orbit paths when passing each other!
>> No, the planets do NOT "stay in their orbit paths when passing
>> each other". They perturb each other, according to the laws of
>> gravity. Go ahead, compare the osculating orbits for Saturn
>> before and after Jupiter passes it. You'll find that they are
>> different.
>>> and why planets do NOT perturb TOWARD the Sun.
>> But they do. They also perturb away from the Sun. It just so
>> happens that the two average out to zero close enough for the
>> Solar System to be stable for billions of years.
> What pulls Saturn OUT away from the Sun and Jupiter, after this
> passage?
Perturbations can work both ways. Six years later, Jupiter will be
on the other side of the Sun, causing the barycenter to shift, thus
the heliocentric osculating orbit for Pluto will now be different.
It's called indirect perturbations.
> It's reverence for Newton?
Illogical.
> We have a gravity Push-Away Law?
> Saturn finds itself floating along in its orbit, now closer to the Sun,
> and decides to JUMP AWAY from the Sun to honor Newton?
Still illogical.
> To keep you'all feeling comfy and smug?
Still illogical.
> It's now in the same position as satellites
> when their orbits degrade or decay or whatever the proper term is, and
> start spiraling down to Earth.
Irrelevant, given that we're not dealing with atmospheric drag here.
> Why would it NOT be? (The answer is, of
> course, the Repulsion Force which is the push-away.)
Incorrect.
Still waiting for answers to the questions:
Bill Nelson
: The Inverse Square law reduces the mass according to the distance. We
It does no such thing. The mass remains constant. What does change is
the gravitational attraction, which decreases inversely with distance.
: did that, in the 1998 exercise, to compute the EQUIVALENT mass of the
: Moon if on the surface of the Earth, if orbiting that close. We kept
: the velocity the same, as the mass had already been reduced. If you saw
See above. Your analysis is silly.
Nothing can orbit the Earth at surface level at the velocity the Moon
has at 200,000+ miles distance. It has to move much faster. And the
mass is relatively immaterial - there is a minimum velocity that is
due to the mass of the Earth itself. For an object as massive as the
Moon, that velocity would be much higher than the minimum.
: to look at that elephant, and DEAL with the inability of Newton's
: formulas to deal with it either! You CANNOT put his Inverse Square law
: together with his Centrifugal Force or Velocity laws.
Let's put it more truthfully - You cannot put the two together because
you cannot understand how and why it is necessary when calculating orbital
speeds.
I can demonstrate centrifugal force, right here on Earth. And the results
do not depend on the gravitational attraction of the Earth, only on the
mass of the object that is generating the centrifugal force.
--
Bill Nelson (bi...@peak.org)
Nancy Lieder
wrote:
>>>> And where the force of gravity between two objects is another
>>>> Newton law:
>>>> F = G*M1*M2/r^2
>>>> keep the Moon from plunging to Earth? But it's out there
>>>> it were floating on .... a bed of gravity particles jammed
>>>> between the Earth and Moon, as in the REPULSION FORCE!
>
>>> Who says that the velocity is not sufficient to maintain the orbit?
>
>> Because in your velocity formula, you EXCLUDE the mass
>> per Newton, a factor of BOTH masses.
>
> Look above. F = G*M1*M2/r^2.
> What do you think M2 is?
The Moon, but this is the Inverse Square law.
My complaint was that you're not putting the laws TOGETHER, as the
velocity and acceleration laws do NOT compute the force of gravity in
the same way. You cannot put them TOGETHER, as the Zetas have been
saying.
Nancy Lieder
wrote:
>>>> Where
>>>> M1 = Earth = 5.9763e+24 kg
>>>> M2 = Moon = 7.3508e+22 kg
>>>> r = 200,000 miles = 3.844E8
>>>> G = Gravity Constant of Earth = 6.67E-11
>>>> another Newton law:
>>>> F = G*M1*M2/r^2
>
>>> F = M2*v^2
>>> solve for equality F1 = F2 (if there is a solution, there must
>>> be a balance between the two formulas or the Moon would
>>> G*M1*M2/r^2 = M2*v^2 or...
>>> G*M1/r = v^2, which solved for v gives
>>> v = sqrt(GM1 / r), which has been stated before
>
>> How can centrifugal force consider ONLY the mass of the
>> to deal with, per the Inverse Square law? We just don't put
>> these together! Can't! They don't work together as Newton
>> is WRONG!
>
> Gravity is the force pulling inwards, and is
> F = G*M1*M2/r^2.
> Centrifugal force is the outward force, and is equal to
> F = M*v^2/r
> force on a body in free-fall are equal and opposite, the net weight
> is zero.
Equal and opposite? If Centrifugal force has to EQUAL the force of
gravity pulling inward, it does NOT in this math. The force inward
takes into consideration both masses. The force outward is only dealing
with the mass of the secondary. How can they NOT both consider the same
factors!
Nancy Lieder
wrote:
>> Is there a Newton formula for velocity for a HIGH mass
>> orbiting object?
>
> Yes. It is v = sqrt(G*(M1 + M2)/r)
This is again stating that velocity must only take into consideration a
LESSER force of gravity than the Inverse Square law pronounces. You are
NOT putting your math together, and this is because it is WRONG and DOES
NOT WORK. The force of gravity between the Earth and a satellite is
trivial, because the satellite is a nit.
In Article <8u_67.10984$e5.16...@newsb.telia.net> Magnus Nyborg wrote:
> Ground orbit (if possible) -
> v = sqrt( 6.67E-11 * 5.976E24 / 6.378E6 ) = 7905 m/s
> Satellite orbit -
> v = sqrt( 6.67E-11 * 5.976E24 / 6.478E6 ) = 7844 m/s
> Moon orbit -
> v = sqrt( 6.67E-11 * 5.976E24 / 3.844E8 ) = 1018 m/s
But the mass of the Moon, adjusted by the Inverse Square rule to be the
mass AT THE DISTANCE IT IS, is huge! The satellite is a piece of dust
floating past, at a high speed. The Moon is an elephant, floating past
at a very slow speed.
Nancy Lieder
> Nothing can orbit the Earth at surface level at the
> velocity the Moon has at 200,000+ miles distance. It
> has to move much faster.
Here's the figures, again.
Earth density: 5.5170 g/cm^3
Moon density: 3.3411 g/cm^3
Earth: 5.9763e+24 kg ( 5.9763e+21 Metric Tons)
or 2.70789475e+21 Metric Tons
Moon: 7.3508e+22 kg ( 7.3508e+19 Metric Tons)
or 5.49973424e+19 Metric Tons
or 73,696,438,000,000,000,000 Metric Tons
gravitational equation F = G*m1*m2/r^2
acceleration of the moon due to gravity toward the Earth
F = 2.20228796E+16 metric ton force
or = 20,228,796,000,000,000 metric tons-force
C = Pi*2*R = Pi*D
C = 2415768.45 km
gives a circular orbital speed of: 1023.183 m/s
Note that the mass of the Moon has been REDUCED by the distance to be
only 20,228,796,000,000,000 metric tons-force. So if you bluster away
saying I cannot put this REDUCED Moon on the surface of the Earth,
floating by slowly at only 1023 m/s, because now it is closer to the
Earth, you are not dealing with the fact that it has ALREADY BEEN
REDUCED by the distance factor.
Mass of Moon 73,696,438,000,000,000,000 Metric Tons
Reduced Mass 20,228,796,000,000,000 metric tons-force
So the reality is that the reduced mass is the EQUIVALENT of
20,228,796,000,000,000 metric tons-force moving at the surface, at only
1023 m/s (or 1018 m/s if you prefer).
We have demonstrated ... that your inverse square law put
together with Newton's rule balancing gravity pull and
centrifugal force would have the equivalent of a million,
trillion metric ton Moon up where your satellites ...
position themselves, moving at only ... 1/4 the speed of
your stationary satellites. ... you must explain how the
Moon could have an equivalent [mass] of a million,
trillion metric tons Earth surface weight, while only
moving at 1023 m/s.
ZetaTalk™
Greg Neill
> In Article <QrX77.26410$PA1.2...@news20.bellglobal.com> Greg Neill
> wrote:
> >>>> And where the force of gravity between two objects is another
> >>>> Newton law:
> >>>> F = G*M1*M2/r^2
> >>>> Should the velosity not be countering the force of gravity, to
> >>>> keep the Moon from plunging to Earth? But it's out there
> >>>> sedately poking along! Floating up there! Almost as though
> >>>> it were floating on .... a bed of gravity particles jammed
> >>>> between the Earth and Moon, as in the REPULSION FORCE!
> >
> >>> Who says that the velocity is not sufficient to maintain the orbit?
> >
> >> Because in your velocity formula, you EXCLUDE the mass
> >> the gravity attraction between the Earth and Moon! This is,
> >> per Newton, a factor of BOTH masses.
> >
> > Look above. F = G*M1*M2/r^2.
> > What do you think M2 is?
>
> The Moon, but this is the Inverse Square law.
>
> My complaint was that you're not putting the laws TOGETHER, as the
> velocity and acceleration laws do NOT compute the force of gravity in
> the same way. You cannot put them TOGETHER, as the Zetas have been
> saying.
Two force laws:
F = G*M1*M2/r^2 This one concerns gravity.
F = M2*v^2/r This one concerns centrifugal force (due to inertia).
One is an inward force, the other outward. In order to have a
balanced system, thus a circular orbit, the two must be equal.
Setting them equal (on the same page!) yields
G*M1*M2/r^2 = M2*v^2/r
G*M1/r = v^2
v = sqrt(G*M1/r)
Now, precicely where do you get lost?
Greg Neill
> >
> > Gravity is the force pulling inwards, and is
> > F = G*M1*M2/r^2.
> > Centrifugal force is the outward force, and is equal to
> > F = M*v^2/r
> > as shown above. Got it? Since the gravitational force and centrifugal
>
> > force on a body in free-fall are equal and opposite, the net weight
> > is zero.
>
> Equal and opposite? If Centrifugal force has to EQUAL the force of
> gravity pulling inward, it does NOT in this math. The force inward
> takes into consideration both masses. The force outward is only dealing
> with the mass of the secondary. How can they NOT both consider the same
> factors!
Please show me where they are not equal if they are written as equal:
G*M1*M2/r^2 = M2*v^2/r
The equation above says that they're equal. We know that they are
equal by observation (circular orbit ==> inward force = outward force),
so we write the two forces down and equate them. Now, where are you
getting lost?
Greg Neill
> In Article <lGX77.26412$PA1.2...@news20.bellglobal.com> Greg Neill
> wrote:
> >> Is there a Newton formula for velocity for a HIGH mass
> >> orbiting object?
> >
> > Yes. It is v = sqrt(G*(M1 + M2)/r)
>
> This is again stating that velocity must only take into consideration a
> LESSER force of gravity than the Inverse Square law pronounces. You are
> NOT putting your math together, and this is because it is WRONG and DOES
> NOT WORK. The force of gravity between the Earth and a satellite is
> trivial, because the satellite is a nit.
Quick, what are the units of force?
What are the units of velocity?
What are the units of acceleration?
Do you understand why velocity and acceleration are not the same thing?
Do you understand why force and accelerationa re not the same thing?
Do you understand that square yards of capet are not the same as
linear yards of string?
You are complaining because the formula for velocity does not
contain a mass-squared term. That's just silly. The result
would be a "velocity" specified in units of kg*m/sec. Do you
drive to the store at 60 kilogram miles per hour?
What is it in your head that makes you want to see mass squared
in the velocity expression? Do you understand that centrifugal
force depends only on the mass of the body in motion, its velocity,
and the radius f curvature of its path?
Greg Neill
> In Article <9jqvrj$5cn$3...@sevenofnine.peak.org> Bill Nelson wrote:
> > Nothing can orbit the Earth at surface level at the
> > velocity the Moon has at 200,000+ miles distance. It
> > has to move much faster.
>
> Here's the figures, again.
>
> Earth density: 5.5170 g/cm^3
> Moon density: 3.3411 g/cm^3
> Earth: 5.9763e+24 kg ( 5.9763e+21 Metric Tons)
> or 2.70789475e+21 Metric Tons
> Moon: 7.3508e+22 kg ( 7.3508e+19 Metric Tons)
> or 5.49973424e+19 Metric Tons
> or 73,696,438,000,000,000,000 Metric Tons
>
> gravitational equation F = G*m1*m2/r^2
> acceleration of the moon due to gravity toward the Earth
> F = 2.20228796E+16 metric ton force
> or = 20,228,796,000,000,000 metric tons-force
>
> C = Pi*2*R = Pi*D
> C = 2415768.45 km
> gives a circular orbital speed of: 1023.183 m/s
>
> Note that the mass of the Moon has been REDUCED by the distance to be
> only 20,228,796,000,000,000 metric tons-force.
Mass is not force. Units of mass: kg
Units of force: N (kg*m/sec^2)
> So if you bluster away
> saying I cannot put this REDUCED Moon on the surface of the Earth,
> floating by slowly at only 1023 m/s, because now it is closer to the
> Earth, you are not dealing with the fact that it has ALREADY BEEN
> REDUCED by the distance factor.
What utter nonsense. Mass does not reduce. The centrifugal force
depends only upon the mass of the Moon, the radius of curvature of
its path (the distance) and its velocity. Mass is invariant.
>
> Mass of Moon 73,696,438,000,000,000,000 Metric Tons
> Reduced Mass 20,228,796,000,000,000 metric tons-force
Mass does not reduce. Where do you think it goes?
Your logic is flawed. It is broken.
John Shakespeare
Nancy Lieder wrote:
> In Article <lGX77.26412$PA1.2...@news20.bellglobal.com> Greg Neill
> wrote:
> >> Is there a Newton formula for velocity for a HIGH mass
> >> orbiting object?
> >
> > Yes. It is v = sqrt(G*(M1 + M2)/r)
>
> This is again stating that velocity must only take into consideration a
> LESSER force of gravity than the Inverse Square law pronounces. You are
> NOT putting your math together, and this is because it is WRONG and DOES
> NOT WORK. The force of gravity between the Earth and a satellite is
> trivial, because the satellite is a nit.
The only nit in this whole thread has been the person posting as Nancy Lieder.
People have put the equations together for you, and shown you the result
clearly and concisely. The onus is on you to study it and understand it. If
you don't understand it, ask for explanations, don't just pout and whine.
[snip]
Best Regards,
John.
P.S. news:sci.astro.amateur removed from followups.
Bob May
understand what you are saying because you aren't using the right numbers.
Please redo with the correct numbers and all will come out.
And no, I'm not going to show you because you're a big woman now and you
should know the proper way.
--
Bob May
Remember that computers do exactly what you tell them to do, not what you
think you told them to do.
Bill Nelson
:> What do you think M2 is?
: The Moon, but this is the Inverse Square law.
: My complaint was that you're not putting the laws TOGETHER, as the
: velocity and acceleration laws do NOT compute the force of gravity in
: the same way. You cannot put them TOGETHER, as the Zetas have been
: saying.
Why would we want to calculate "gravitational forces" in the same way.
We are not interested in the causes of the forces, only the magnitudes
and directions of the forces.
So, we use one equation that accounts for the gravitational attraction
between the Earth and the Moon. We use a different equation to account
for the centripital force (due to momentum) of the Moon.
--
Bill Nelson (bi...@peak.org)
Bill Nelson
: Note that the mass of the Moon has been REDUCED by the distance to be
: only 20,228,796,000,000,000 metric tons-force. So if you bluster away
: saying I cannot put this REDUCED Moon on the surface of the Earth,
: floating by slowly at only 1023 m/s, because now it is closer to the
: Earth, you are not dealing with the fact that it has ALREADY BEEN
: REDUCED by the distance factor.
But it HASN'T been "reduced". Just because you perform a silly calculation
does not make the result useful.
By your own statement - a heavy mass at a given distance from the Earth
must move faster than a lighter mass at the same distance - if it is going
to remain in orbit.
If a satellite (low mass) has to move much faster than 1023 m/s to remain
in orbit at 100 miles distance from the surface - then, by your own
statements in an earlier post, the Moon must move faster.
Since your calculations do not show this, then something has to be very
wrong with your calculations.
: So the reality is that the reduced mass is the EQUIVALENT of
: 20,228,796,000,000,000 metric tons-force moving at the surface, at only
: 1023 m/s (or 1018 m/s if you prefer).
No, it isn't. See above.
--
Bill Nelson (bi...@peak.org)
John Jones
Yes it does.
> The force inward
> takes into consideration both masses. The force outward is only dealing
> with the mass of the secondary. How can they NOT both consider the same
> factors!
Because they are different forces. One balances the other. Ask your
imaginary Zeta friends.
Nancy Lieder
wrote:
>> Equal and opposite? If Centrifugal force has to EQUAL
>> the force of gravity pulling inward, it does NOT in this
>> masses. The force outward is only dealing with the mass
>> of the secondary. How can they NOT both consider the
>> same factors!
>
> as equal:
> G*M1*M2/r^2 = M2*v^2/r
> The equation above says that they're equal.
> We know that they are equal by observation
> (circular orbit ==> inward force = outward force)
The centrifugal force is to OFFSET the gravity attraction of the
primary, or the force of gravity between the two bodies, then why does
distance matter? Per Newton, these two bodies are not AWARE of each
other EXCEPT for the force of gravity. Do these bodies put out trip
wires, so they cannot come closer without an alarm going off? The
Newton trip-wire law? He deals with mass, distance, and speed. Thus he
computes that the force of gravity is OFFSET by the pull outward of
centrifugal force, a faster speed producing more centrifugal force.
In Article <QHi87.16219$Tn3.7...@wagner.videotron.net> Greg Neill
wrote:
>>> v = sqrt(G*(M1 + M2)/r)
>
>> This is again stating that velocity must only take into
>> consideration a LESSER force of gravity than the
>> Inverse Square law pronounces. You are NOT
>> putting your math together
>
>You are complaining because the formula for velocity
> does not contain a mass-squared term. That's just silly.
> The result would be a "velocity" specified in units of
> kg*m/sec. Do you drive to the store at 60 kilogram miles
> per hour?
You do if you are considering the force of IMPACT if you car crashes!
Centrifugal Force must avoid an IMPACT due to the force of gravity, no?
Nancy Lieder
Greg and Bill are saying that their math, Newton, explains why something
that is a million trillion metric tons can orbit the Earth at only 1023
m/s or so. Because:
Inverse Square F = G*M1*M2/r^2
Centrifugal Force F = M2*v^2
where M1=Earth=5.9763e+24 kg
M2=Moon=7.3508e+22 kg
r=distance=3.844E8
and G*M1*M2/r^2 = M2*v^2
giving G*M1/r = v^2
thus v = sqrt(GM1 / r)
so F1=F2
Constant as m * p^2/d^3 is constant for all orbits.
where m = mass of primary
d = distance
p = period
I'm asserting that the force of gravity has not been taken into
consideration as it is not used in THE SAME manner in all the
equations. They are saying the the math is right because it balances on
paper. I'm saying it does not jibe with reality. This produces some
interesting results, where the Moon could theoretically orbit at the
same distance as Satellites, at the same velocity (see next post).
Nancy Lieder
> If a satellite (low mass) has to move much faster than
> 1023 m/s to remain in orbit at 100 miles distance from
In Article <nLi87.16221$Tn3.7...@wagner.videotron.net> Greg Neill
wrote:
>> Note that the mass of the Moon has been REDUCED by the
>> distance:
>> Mass of Moon 73,696,438,000,000,000,000 Metric Tons
>> Reduced Mass 20,228,796,000,000,000 metric tons-force
>
> The centrifugal force depends only upon the mass of the Moon,
> the radius of curvature of its path (the distance) and its velocity.
The IMPLICATION:
So if your math RULES, then moving the Moon closer to the Earth only
requires that the Moon move as fast as the satellites, to stay aloft.
So we could move the Moon in from 200,000 miles to be as close as the
satellites at 100 miles, and all would be well. Right? And we could
move the Moon to orbit at the ground or surface level and all would be
well, as long as it's going at the right speed. Right? Or am I missing
something? Or are YOU missing something, like the obvious!
In Article <8u_67.10984$e5.16...@newsb.telia.net> Magnus Nyborg wrote:
> Orbital speed for ideal circular motion of a low-mass object
> circling a high-mass object M (which refers to it's mass) is
> determined by the formula
>
> v = sqrt( G*M / r )
>
> Ground orbit (if possible) -
> v = sqrt( 6.67E-11 * 5.976E24 / 6.378E6 ) = 7905 m/s
> Satellite orbit -
> v = sqrt( 6.67E-11 * 5.976E24 / 6.478E6 ) = 7844 m/s
> Moon orbit -
> v = sqrt( 6.67E-11 * 5.976E24 / 3.844E8 ) = 1018 m/s
>
Imagine the Moon orbiting there alongside the satellites! This is what
you're saying, that all would be well, per Newton! The elephant and the
mosquito, side by side! NO problem whatsoever! If this is the rule of
the Universe, per your flawless math, then we should indeed have these
types of arrangments out there. DO we? Give me an example of this type
of near-touching orbits of large mass objects! Why are they staying
apart, if Newton explains ALL. What is missing is the Repulsion Force
Nancy Lieder
Since the orbiting satellites or space stations are infinitesimal in
mass compared to the Earth,
v = sqrt( G*M / r )
works for them, consistently. M2, the orbiter, being relatively
inconsequential. It also works on paper for the Moon, except that per
this math you could have the moon at near ground level, moving no faster
than the satellites. Intuitively, this is wrong, somewhere. What is
wrong is that the force of gravity is STRONGER between large mass bodies
so the mass of the primary is not sufficient for Newton's math. It
DOES NOT WORK. Let's do a Newton (being exploratory with math) and see
if we can come up with a Repulsion Force R factor that would allow the
Moon to float by at the level the satellites orbit, without being an
absurd and intuitively wrong concept.
Inverse Square F = G*M1*M2/r^2
Centrifugal Force F = M2*v^2
Becomes
Inverse Square F = (G*M1*M2/r^2) - R
Centrifugal Force F = (M2 - R)* v^2)
So that
Inverse Square F = 0 at the point of contact
Centrifugal Force v = 0 and an object can hover at ground level
Per the Zetas:
Why would the planets not drift into the Sun? Are the orbits
all that swift so that centrifugal force is extreme? ... The reason
Mankind is Unaware of a repulsive force, also inherent in
gravity, is that for this to become evident there must be a
semblance of equality in size and weight, i.e. the mass of the
objects, and freedom of movement such as exists in space,
and lack of undue influence from other nearby objects. ... The
repulsion force is generated as a result of two bodies exerting a
gravitational force on each other. ... Where the repulsion force
comes to equal the force of gravity by the time the objects in
play would make contact, it builds at a rate that differs from
gravity. ... The repulsion force is infinitesimally smaller than
the force of gravity, but has a sharper curve so that it equals
the force of gravity at the point of contact.
ZetaTalk™, Repulsion Force
(http://www.zetatalk.com/science/s34.htm)
So, what is the R factor?
And after we have this factored, we can plug it into the perturbations
of planets also under discussion, and see if THIS explains why the
planets RETURN to their orbits after having been perturbed in CLOSER to
the sun - what I've described for lack of a better term as the missing
push-away law.
John Shakespeare
Nancy Lieder wrote:
[snip: prelude to erroneous ideas]
> absurd and intuitively wrong concept.
Now you've summed up your whole fallacy in just five words.
It's obvious you have not yet grasped Kepler's laws, Newtonian gravitation, or
any other facet of basic basic physics relevant to celestial mechanics.
Try learning about angular momentum first; it's a simple concept, and very
useful, in that it is conserved. You might then fantasize a bit less less
about your imaginary "pushing" forces of gravity.
[snip: Nancy's latest funnies]
Greg Neill
> In Article <Hvi87.16200$Tn3.7...@wagner.videotron.net> Greg Neill
> wrote:
> >> Equal and opposite? If Centrifugal force has to EQUAL
> >> the force of gravity pulling inward, it does NOT in this
> >> math. The force inward takes into consideration both
> >> masses. The force outward is only dealing with the mass
> >> of the secondary. How can they NOT both consider the
> >> same factors!
> >
> > Please show me where they are not equal if they are written
> > as equal:
> > G*M1*M2/r^2 = M2*v^2/r
> > The equation above says that they're equal.
> > We know that they are equal by observation
> > (circular orbit ==> inward force = outward force)
>
> The centrifugal force is to OFFSET the gravity attraction of the
> primary, or the force of gravity between the two bodies, then why does
> distance matter?
Not "is to", "does". In order to have a circular orbit, the
centrifugal force must equal the gravitational force. Why
does distance matter? Because the gravitational force drops
off with the square of the distance, and the centrifugal
force drops off with the distance too, only not as the square
of the distance.
> Per Newton, these two bodies are not AWARE of each
> other EXCEPT for the force of gravity. Do these bodies put out trip
> wires, so they cannot come closer without an alarm going off?
No need. The magnitude and direction of the gravitational force
is all the information that the bodies need. It tells them how
to accelerate (f = m*a). The acceleration leads to the
centrifugal force.
> The
> Newton trip-wire law? He deals with mass, distance, and speed. Thus he
> computes that the force of gravity is OFFSET by the pull outward of
> centrifugal force, a faster speed producing more centrifugal force.
Yes, you're catching on. Except the bit obout trip wires, of course.
>
> In Article <QHi87.16219$Tn3.7...@wagner.videotron.net> Greg Neill
> wrote:
> >>> v = sqrt(G*(M1 + M2)/r)
> >
> >> This is again stating that velocity must only take into
> >> consideration a LESSER force of gravity than the
> >> Inverse Square law pronounces. You are NOT
> >> putting your math together
> >
> >You are complaining because the formula for velocity
> > does not contain a mass-squared term. That's just silly.
> > The result would be a "velocity" specified in units of
> > kg*m/sec. Do you drive to the store at 60 kilogram miles
> > per hour?
>
> You do if you are considering the force of IMPACT if you car crashes!
> Centrifugal Force must avoid an IMPACT due to the force of gravity, no?
You must be a terrible driver to require a momentum meter in your
vehicle rather than a spedometer.
Greg Neill
> The GIVENS:
>
> Greg and Bill are saying that their math, Newton, explains why something
> that is a million trillion metric tons can orbit the Earth at only 1023
> m/s or so. Because:
>
> Inverse Square F = G*M1*M2/r^2
> Centrifugal Force F = M2*v^2
>
> where M1=Earth=5.9763e+24 kg
> M2=Moon=7.3508e+22 kg
> r=distance=3.844E8
>
> and G*M1*M2/r^2 = M2*v^2
> giving G*M1/r = v^2
> thus v = sqrt(GM1 / r)
>
> so F1=F2
>
> Constant as m * p^2/d^3 is constant for all orbits.
> where m = mass of primary
> d = distance
> p = period
>
> I'm asserting that the force of gravity has not been taken into
> consideration as it is not used in THE SAME manner in all the
> equations. They are saying the the math is right because it balances on
> paper. I'm saying it does not jibe with reality. This produces some
> interesting results, where the Moon could theoretically orbit at the
> same distance as Satellites, at the same velocity (see next post).
The force of gravity is used once. Centrifugal force is used once.
They are stated to be equal (F1 = F2). The results follow without
introducing any other form for the forces, it is simply algebra after
that.
It is true that the velocity for circular orbit of a relatively small
mass around a large mass depends only upon the distance from the
large mass. So the Moon could indeed orbit at practically the same
velocity of a man-made satellite at the same distance as such a
satellite.
Greg Neill
> In Article <9jt3d3$rcn$4...@sevenofnine.peak.org> Bill Nelson wrote:
> > If a satellite (low mass) has to move much faster than
> > 1023 m/s to remain in orbit at 100 miles distance from
> > the surface - then ... the Moon must move faster.
>
> In Article <nLi87.16221$Tn3.7...@wagner.videotron.net> Greg Neill
> wrote:
> >> Note that the mass of the Moon has been REDUCED by the
> >> distance:
> >> Mass of Moon 73,696,438,000,000,000,000 Metric Tons
> >> Reduced Mass 20,228,796,000,000,000 metric tons-force
> >
> > The centrifugal force depends only upon the mass of the Moon,
> > the radius of curvature of its path (the distance) and its velocity.
>
> The IMPLICATION:
>
> So if your math RULES, then moving the Moon closer to the Earth only
> requires that the Moon move as fast as the satellites, to stay aloft.
Yes.
> So we could move the Moon in from 200,000 miles to be as close as the
> satellites at 100 miles, and all would be well. Right?
Sure, if you could ignore the radius of the Moon. You'll have a
problem with the Moon intersecting the Earth... remember, the
distance must be calculated from the centers of the masses.
> And we could
> move the Moon to orbit at the ground or surface level and all would be
> well, as long as it's going at the right speed. Right? Or am I missing
> something? Or are YOU missing something, like the obvious!
I fear that I am not missing anything obvious. Perhaps you are
avoiding it?
>
> In Article <8u_67.10984$e5.16...@newsb.telia.net> Magnus Nyborg wrote:
>
> > Orbital speed for ideal circular motion of a low-mass object
> > circling a high-mass object M (which refers to it's mass) is
> > determined by the formula
> >
> > v = sqrt( G*M / r )
> >
> > Ground orbit (if possible) -
> > v = sqrt( 6.67E-11 * 5.976E24 / 6.378E6 ) = 7905 m/s
> > Satellite orbit -
> > v = sqrt( 6.67E-11 * 5.976E24 / 6.478E6 ) = 7844 m/s
> > Moon orbit -
> > v = sqrt( 6.67E-11 * 5.976E24 / 3.844E8 ) = 1018 m/s
> >
>
> Imagine the Moon orbiting there alongside the satellites! This is what
> you're saying, that all would be well, per Newton! The elephant and the
> mosquito, side by side! NO problem whatsoever! If this is the rule of
> the Universe, per your flawless math, then we should indeed have these
> types of arrangments out there. DO we? Give me an example of this type
> of near-touching orbits of large mass objects! Why are they staying
> apart, if Newton explains ALL. What is missing is the Repulsion Force
> (see next post).
There are examples of very closely orbiting binary stars, where
indeed they are nearly touching. In fact, they are so close that
they exchange gases one to the other. One form of periodic nova
is the result of gas spiralling down from one star onto the
surface of a neutron star. When the resulting gas layer reaches
a critical mass, it ignites in a fusion reaction producing a
Nova.
One thing that can prevent the very close co-orbiting of solid
bodies in practice is what is known as the Roche Limit. It is
a tidal effect. A body in a gravitational field is stretched
by the difference in the strength of gravity across its
diameter (stronger pull on parts nearer the attracting body than
on parts further away). Earth's tides are an example of this, with
a bulge showing on each side of the planet, one under the Moon
and one on the side opposite.
Bodies that are held together by self gravitation find themselves
broken apart when they get to the Roche Limit, that is, the
proximity where the tidal force exceeds their self gravitational
force.
Greg Neill
> The MISSING:
>
> Since the orbiting satellites or space stations are infinitesimal in
> mass compared to the Earth,
> v = sqrt( G*M / r )
> works for them, consistently. M2, the orbiter, being relatively
> inconsequential. It also works on paper for the Moon, except that per
> this math you could have the moon at near ground level, moving no faster
> than the satellites. Intuitively, this is wrong, somewhere.
That is a statement about your intuition, not anyone else's.
> What is
> wrong is that the force of gravity is STRONGER between large mass bodies
> so the mass of the primary is not sufficient for Newton's math.
You were shown that in the case where the secondary mass cannot be
considered to be negligible with respect to the primary, then
the full precision expression for the velocity is:
v = sqrt(G*(M1+M2)/r)
> It
> DOES NOT WORK.
Please cite a physical example (in real life, not in your
imagination) of a case where it does not work.
> Let's do a Newton (being exploratory with math) and see
> if we can come up with a Repulsion Force R factor that would allow the
> Moon to float by at the level the satellites orbit, without being an
> absurd and intuitively wrong concept.
>
> Inverse Square F = G*M1*M2/r^2
> Centrifugal Force F = M2*v^2
>
> Becomes
> Inverse Square F = (G*M1*M2/r^2) - R
> Centrifugal Force F = (M2 - R)* v^2)
Except that you have created obviously wrong equations.
The units do not work. How can you subtract distance
from force, or distance from mass? Please show me
100 yards of kilogram.
Remainder of sillyness snipped.
Bill Nelson
: The centrifugal force is to OFFSET the gravity attraction of the
: primary, or the force of gravity between the two bodies, then why does
: distance matter? Per Newton, these two bodies are not AWARE of each
: other EXCEPT for the force of gravity. Do these bodies put out trip
: wires, so they cannot come closer without an alarm going off? The
: Newton trip-wire law? He deals with mass, distance, and speed. Thus he
: computes that the force of gravity is OFFSET by the pull outward of
: centrifugal force, a faster speed producing more centrifugal force.
Bingo. So, if the velocity is higher, then the object can orbit closer to
the primary, as the centrifugal force is higher.
In the subject under discussion - that means the Moon, if orbiting at
a faster velocity, would orbit closer to the Earth.
In other words, G, M1 and M2 do not change. This means that if you change
v in the centrifugal force formula, you must change r as well - as they
are the only other two variables.
: In Article <QHi87.16219$Tn3.7...@wagner.videotron.net> Greg Neill
: wrote:
:>
:>You are complaining because the formula for velocity
:> does not contain a mass-squared term. That's just silly.
:> The result would be a "velocity" specified in units of
:> kg*m/sec. Do you drive to the store at 60 kilogram miles
:> per hour?
: You do if you are considering the force of IMPACT if you car crashes!
We are not talking about impact here - so your comment is immaterial.
: Centrifugal Force must avoid an IMPACT due to the force of gravity, no?
That is exactly the case. The centrifugal force must equal the force of
gravity - or the Moon will either collide with the Earth or escape from
Orbit.
But figuring energy transfer of an impact has nothing to do with orbital
dynamics.
--
Bill Nelson (bi...@peak.org)
Bill Nelson
: I'm asserting that the force of gravity has not been taken into
: consideration as it is not used in THE SAME manner in all the
: equations. They are saying the the math is right because it balances on
: paper. I'm saying it does not jibe with reality. This produces some
: interesting results, where the Moon could theoretically orbit at the
: same distance as Satellites, at the same velocity (see next post).
Unfortunately, you have shown that you do not understand orbital
dynamics - which can be demonstrated right here on earth with a
weight, a piece of string, a scale and a stopwatch.
Nor have you provided any mathematical support for what is really a
very simple physical system.
--
Bill Nelson (bi...@peak.org)
Bill Nelson
:> the radius of curvature of its path (the distance) and its velocity.
: The IMPLICATION:
: So if your math RULES, then moving the Moon closer to the Earth only
: requires that the Moon move as fast as the satellites, to stay aloft.
No, it does not say that. While the velocity of the Moon and the satellite
would be similar, they would be different - with the Moon having a higher
velocity. Why? Because the gravitational attraction of the Earth/Moon
system is higher than the gravitational attraction of the Earth/satellite
system.
You can plug the constants that have been given into the various formulas
and figure out what the velocity would have to be.
: So we could move the Moon in from 200,000 miles to be as close as the
: satellites at 100 miles, and all would be well. Right? And we could
: move the Moon to orbit at the ground or surface level and all would be
: well, as long as it's going at the right speed. Right? Or am I missing
: something? Or are YOU missing something, like the obvious!
If the Moon would not be torn apart by the increased gravitational
attraction, then it could orbit at the distances you mentioned.
: In Article <8u_67.10984$e5.16...@newsb.telia.net> Magnus Nyborg wrote:
:> Orbital speed for ideal circular motion of a low-mass object
:> circling a high-mass object M (which refers to it's mass) is
:> determined by the formula
:>
:> v = sqrt( G*M / r )
:>
:> Ground orbit (if possible) -
:> v = sqrt( 6.67E-11 * 5.976E24 / 6.378E6 ) = 7905 m/s
:> Satellite orbit -
:> v = sqrt( 6.67E-11 * 5.976E24 / 6.478E6 ) = 7844 m/s
:> Moon orbit -
:> v = sqrt( 6.67E-11 * 5.976E24 / 3.844E8 ) = 1018 m/s
:>
: Imagine the Moon orbiting there alongside the satellites! This is what
: you're saying, that all would be well, per Newton! The elephant and the
: mosquito, side by side! NO problem whatsoever! If this is the rule of
: the Universe, per your flawless math, then we should indeed have these
: types of arrangments out there. DO we? Give me an example of this type
: of near-touching orbits of large mass objects! Why are they staying
: apart, if Newton explains ALL. What is missing is the Repulsion Force
: (see next post).
No problem at all is correct.
There are near touching orbits in space. There are various binary suns
that have such orbits. They are so close together that mass from one sun
is stripped off and pulled to the surface of the other sun.
--
Bill Nelson (bi...@peak.org)
tho...@antispam.ham
> Greg and Bill are saying that their math, Newton, explains why something
> that is a million trillion metric tons can orbit the Earth at only 1023
> m/s or so.
What does the mass have to do with it? You could put a penny at the
same distance and it would orbit the Earth with essentially the same
speed.
Still waiting for the answers to these questions:
] Nancy Lieder writes:
]
] > I was told by an astronomer who works for a large California observatory
] > the following "We were looking for it, then we found it, and now we're
] > tracking it.
]
] Which "astronomer"? Which "large California observatory"?
]
] Recall that you have repeatedly and erroneously accused me of evading
] answers. Will you act hypocritically and evade the answers to these
] questions?
tho...@antispam.ham
> Why would the planets not drift into the Sun?
Conservation of angular momentum. Basic physics. Very basic.
tho...@antispam.ham
> So if your math RULES, then moving the Moon closer to the Earth only
> requires that the Moon move as fast as the satellites, to stay aloft.
You have a problem with that?
> So we could move the Moon in from 200,000 miles to be as close as the
> satellites at 100 miles, and all would be well. Right?
Well, I wouldn't say that. The tides would be considerably stronger,
perhaps to the point of flooding coastal cities.
> And we could move the Moon to orbit at the ground or surface level
> and all would be well, as long as it's going at the right speed.
> Right? Or am I missing something?
You're missing the atmosphere. But hey, that's par for your course.
> Or are YOU missing something, like the obvious!
What I'm obviously missing are the answers to these questions, which
you hypocritically continue to evade:
Bob May
First is the centrifigal force is just that, a force which is generated by
something else. That something else is the gravitational attraction between
the two objects. You have to remember that the orbiting objects are merely
falling in the gravaitational well, unfortunately, the two objects are
constantly missing each other because the velocity that they have relative
to each other is putting them in the same orientation as they were in,
another way of putting it is that the Moon is going fast enough that when
it gets the 90deg. of orbit from where it was, it's already traveled the
radius of difference between them and thus is still in the same relationship
as they were.
Second is that these are theories that we are dealing with. There's a whole
lot of little modifiers that we aren't talking about, including the
time-space dialation, atmosphere effects and so forth.
Go do some reading and exploring asking the questions of the books (they
make books so that you can recover the knowledge of others too, you know!)
and you will get a lot of interesting answers. Start with a high school
level physics book and then go on to the different levels of University
physics. You will understand then why you are being so repeatedly nailed to
the wall.
Bob May
your nitwit brain says:
> Why would the planets not drift into the Sun? Are the orbits all that
>swift so that centrifugal force is extreme?
the high school physics book.
You're so dumb that you probably put your hand into the flame trying to pick
it up.
Nancy Lieder
wrote:
> In Article <3B62DAC0...@zetatalk.com> Nancy Lieder wrote:
>> The GIVENS:
>> Constant as m * p^2/d^3 is constant for all orbits.
>> where m = mass of primary
>> d = distance
>> p = period
>> could theoretically orbit at the same distance as Satellites,
>
> It is true that the velocity for circular orbit of a relatively
> small mass around a large mass depends only upon the
> distance from the large mass. So the Moon could indeed
> orbit at practically the same velocity of a man-made
Greg Neill, meet Greg Neill. You guys seem to be contradicting each
other. In the same sit-down-and-respond-to-sci..astro-posts session,
yet. Newton's centrifugal force law only takes into account the mass of
the Primary. You are saying, below, what I've been asserting - that
this does not fit with the Inverse Square law. However, up until now,
you've not given that argument of mine any credence. Plus, you're
contradicting yourself. Want to have a second go at that?
In Article <9jvdv8$s04$4...@sevenofnine.peak.org> Greg Neill wrote:
> In sci.astro Nancy Lieder <zeta...@zetatalk.com> wrote:
>> The IMPLICATION:
>> So if your math RULES, then moving the Moon closer
>> to the Earth only requires that the Moon move as fast as
>> the satellites, to stay aloft.
>
> No, it does not say that. While the velocity of the Moon and
> the satellite would be similar, they would be different - with
> the Moon having a higher velocity. Why? Because the
> gravitational attraction of the Earth/Moon system is higher
> than the gravitational attraction of the Earth/satellite
> system.
Nancy Lieder
> gravitational attraction, then it could orbit at the
> distances you mentioned.
>
>> whatsoever! If this is the rule of the Universe, per your
>> flawless math, then we should indeed have these types
>> of arrangments out there. DO we? Give me an example
>> of this type of near-touching orbits of large mass objects!
>
> binary suns that have such orbits. They are so close together
> that mass from one sun is stripped off and pulled to the
> surface of the other sun.
And how far are they apart? How far are they KEPT apart, by the
Repulsion Force.
If the mass is pulled from one to the other, then the gravity attraction
is strong enough to do this. So why doesn't the second star just merge
with the first? One part of the star is honoring Newton and abiding by
his laws, while the other is not? Why is the mass moving from one star
to the other, if the centrifugal force is strong enough to keep it in
its orbit?
Nancy Lieder
> In sci.astro Nancy Lieder <zeta...@zetatalk.com> wrote:
>> This produces some interesting results, where the Moon
>> could theoretically orbit at the same distance as Satellites,
>
> Unfortunately, you have shown that you do not understand
> support for what is really a very simple physical system.
Bill Nelson, meed Magnus Nyborg. I was using his math.
This law takes into consideration ONLY the mass of the primary. So my
statement above is correct, as confirmed by (one of the) Greg Neills
personas and David Tholen. (Greg is still arguing with himself, having
been momentarily discombobulated by cracks in Newton appearing before
him. Where is that contact cement!)
In Article <8u_67.10984$e5.16...@newsb.telia.net> Magnus Nyborg wrote:
> Orbital speed for ideal circular motion of a low-mass object
> circling a high-mass object M (which refers to it's mass) is
> determined by the formula
>
> v = sqrt( G*M / r )
>
> Ground orbit (if possible) -
> v = sqrt( 6.67E-11 * 5.976E24 / 6.378E6 ) = 7905 m/s
> Satellite orbit -
> v = sqrt( 6.67E-11 * 5.976E24 / 6.478E6 ) = 7844 m/s
> Moon orbit -
> v = sqrt( 6.67E-11 * 5.976E24 / 3.844E8 ) = 1018 m/s
>
Nancy Lieder
wrote:
>> Greg and Bill are saying that their math, Newton,
>> explains why something that is a million trillion metric
>> tons can orbit the Earth at only 1023 m/s or so.
>
> What does the mass have to do with it? You could put a
> penny at the same distance and it would orbit the Earth with
> essentially the same speed.
David Tholen, meet Greg Neill and Bill Nelson. Are we on the same page
here, guys? Yes or no, are the implications of Newton's math as I've
described them? If the mass of the primary is the ONLY thing an
orbiting object needs to be concerned with, in how close it can come and
how fast it orbits about the primary (these being the only factors in
the law), then what seems to be the problem? The elephant and the
mosquito could zoon around the Earth, at near surface level, side by
side, same speed, no problem whatsoever. So says Newton.
In Article <9jvdv8$s04$4...@sevenofnine.peak.org> Greg Neill wrote:
> In sci.astro Nancy Lieder <zeta...@zetatalk.com> wrote:
>> The IMPLICATION:
>> So if your math RULES, then moving the Moon closer
>> to the Earth only requires that the Moon move as fast as
>> the satellites, to stay aloft.
>
> No, it does not say that. While the velocity of the Moon and
> the satellite would be similar, they would be different - with
> the Moon having a higher velocity. Why? Because the
> gravitational attraction of the Earth/Moon system is higher
> than the gravitational attraction of the Earth/satellite
> system.
In Article <9jvdd4$s04$3...@sevenofnine.peak.org> Bill Nelson wrote:
> In sci.astro Nancy Lieder <zeta...@zetatalk.com> wrote:
>> This produces some interesting results, where the Moon
>> could theoretically orbit at the same distance as Satellites,
>> at the same velocity (see next post).
>
> Unfortunately, you have shown that you do not understand
> orbital dynamics ... Nor have you provided any mathematical
> support for what is really a very simple physical system.
-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
Nancy Lieder
wrote:
>> So we could move the Moon in from 200,000 miles to be
>> as close as the satellites at 100 miles, and all would be well.
>> Right?
>
> Well, I wouldn't say that. The tides would be considerably
> stronger, perhaps to the point of flooding coastal cities.
>
>> And we could move the Moon to orbit at the ground or
>> surface level and all would be well, as long as it's going
>> at the right speed. Right?
>
So we'd have a flaming ball of rock, 1/4 the size of the Earth, orbiting
at surface level at a speed just a tick up from the speed required to
orbit the satellites. We could all get a 6-pack and watch. "Duck!
Here she comes again!" A new ball game, Dodge the Moon.
David Tholen, meet Greg Neill, who seems to thing the ball of rock would
fragment. Due to the Moon's molten core and fragile crust, no doubt
(just kiding here, Dave, so don't go off on a tangent). Perhaps, being
the astronomer, Dave, you could give us examples of where Newton proved
correct, and we have such massive objects orbiting each other, at the
literal touch point.
In Article <9jvdv8$s04$4...@sevenofnine.peak.org> Greg Neill wrote:
> If the Moon would not be torn apart by the increased
> gravitational attraction, then it could orbit at the
> distances you mentioned.
tho...@antispam.ham
> Bill Nelson wrote:
>> Nancy Lieder wrote:
>>> This produces some interesting results, where the Moon
>>> could theoretically orbit at the same distance as Satellites,
>>> at the same velocity (see next post).
>> Unfortunately, you have shown that you do not understand
>> orbital dynamics ... Nor have you provided any mathematical
>> support for what is really a very simple physical system.
> This law takes into consideration ONLY the mass of the primary. So my
> statement above is correct, as confirmed by (one of the) Greg Neills
> personas and David Tholen.
Bill Nelson's statement above is correct, as confirmed by me, namely
that you have shown that you do not understand orbital dynamics.
Still waiting for the answers to these questions:
] Nancy Lieder writes:
Greg Neill
> In Article <sgG87.41806$PA1.4...@news20.bellglobal.com> Greg Neill
> wrote:
> > In Article <3B62DAC0...@zetatalk.com> Nancy Lieder wrote:
> >> The GIVENS:
> >> Constant as m * p^2/d^3 is constant for all orbits.
> >> where m = mass of primary
> >> d = distance
> >> p = period
> >> This produces some interesting results, where the Moon
> >> could theoretically orbit at the same distance as Satellites,
> >> at the same velocity.
> >
> > It is true that the velocity for circular orbit of a relatively
> > small mass around a large mass depends only upon the
> > distance from the large mass. So the Moon could indeed
> > orbit at practically the same velocity of a man-made
> > satellite at the same distance as such a satellite.
>
> Greg Neill, meet Greg Neill. You guys seem to be contradicting each
> other. In the same sit-down-and-respond-to-sci..astro-posts session,
> yet. Newton's centrifugal force law only takes into account the mass of
> the Primary. You are saying, below, what I've been asserting - that
> this does not fit with the Inverse Square law. However, up until now,
> you've not given that argument of mine any credence. Plus, you're
> contradicting yourself. Want to have a second go at that?
You have not been reading, just ranting.
The centrifugal force does not depend at all upon the mass of
the primary, although you keep insisting that it should.
The centrifugal force depends only upon the mass of the
secondary and its path, i.e., acceleration. It just so
happens that this acceleration is due to the gravitational
force, so that the centrifugal force and gravitational force
are equal and opposite in all cases.
Please show where I (or my evil twin) stated that the
centrifugal force depends upon the mass of the primary
in any other way than as the result of the secondary's
acceleration due to gravity.
Greg Neill
> In Article <9jvdv8$s04$4...@sevenofnine.peak.org> Greg Neill wrote:
> > If the Moon would not be torn apart by the increased
> > gravitational attraction, then it could orbit at the
> > distances you mentioned.
> >
> >> The elephant and the mosquito, side by side! NO problem
> >> whatsoever! If this is the rule of the Universe, per your
> >> flawless math, then we should indeed have these types
> >> of arrangments out there. DO we? Give me an example
> >> of this type of near-touching orbits of large mass objects!
> >
> > There are near touching orbits in space. There are various
> > binary suns that have such orbits. They are so close together
> > that mass from one sun is stripped off and pulled to the
> > surface of the other sun.
>
> And how far are they apart? How far are they KEPT apart, by the
> Repulsion Force.
non sequitur.
>
> If the mass is pulled from one to the other, then the gravity attraction
> is strong enough to do this. So why doesn't the second star just merge
> with the first? One part of the star is honoring Newton and abiding by
> his laws, while the other is not? Why is the mass moving from one star
> to the other, if the centrifugal force is strong enough to keep it in
> its orbit?
Eventually these co-orbiting stars do coalesce. Such closely orbiting
bodies raise terrific tides on eachother, and orbital energy is lost
to tidal friction -- they spiral into eachother.
As for whay mass is moving from one to the other, see my post on the
Roche Limit. As one star enters the Roche Limit, it can lose surface
material.
tho...@antispam.ham
>>> Greg and Bill are saying that their math, Newton,
>>> explains why something that is a million trillion metric
>>> tons can orbit the Earth at only 1023 m/s or so.
>> What does the mass have to do with it? You could put a
>> penny at the same distance and it would orbit the Earth with
>> essentially the same speed.
> David Tholen, meet Greg Neill and Bill Nelson.
Unnecessary.
> Are we on the same page here, guys?
Certainly not on your page.
> Yes or no, are the implications of Newton's math as I've
> described them?
You've certainly not described them properly.
> If the mass of the primary is the ONLY thing an
> orbiting object needs to be concerned with, in how close it can come and
> how fast it orbits about the primary (these being the only factors in
> the law), then what seems to be the problem?
The problem is that you don't understand what happens when the atmosphere
comes into play.
> The elephant and the mosquito could zoon around the Earth, at near
> surface level, side by side, same speed, no problem whatsoever.
> So says Newton.
No, that's not what Newton said. Newton talked about forces, and
you're now ignoring the frictional force introduced by the
atmosphere. Newton did not ignore it.
Greg Neill
> In Article <9jvdd4$s04$3...@sevenofnine.peak.org> Bill Nelson wrote:
> > In sci.astro Nancy Lieder <zeta...@zetatalk.com> wrote:
> >> This produces some interesting results, where the Moon
> >> could theoretically orbit at the same distance as Satellites,
> >> at the same velocity (see next post).
> >
> > Unfortunately, you have shown that you do not understand
> > orbital dynamics ... Nor have you provided any mathematical
> > support for what is really a very simple physical system.
>
> Bill Nelson, meed Magnus Nyborg. I was using his math.
>
> This law takes into consideration ONLY the mass of the primary. So my
> statement above is correct, as confirmed by (one of the) Greg Neills
> personas and David Tholen. (Greg is still arguing with himself, having
> been momentarily discombobulated by cracks in Newton appearing before
> him. Where is that contact cement!)
Your statement is false. You have merely chosen to fail to
understand or to purposely misinterpret what has been said.
This belies low character.
The formula stated, namely
v = sqrt(G*M/r)
is valid within the stated conditions.
tho...@antispam.ham
>>> So we could move the Moon in from 200,000 miles to be
>>> as close as the satellites at 100 miles, and all would be well.
>>> Right?
>> Well, I wouldn't say that. The tides would be considerably
>> stronger, perhaps to the point of flooding coastal cities.
>>> And we could move the Moon to orbit at the ground or
>>> surface level and all would be well, as long as it's going
>>> at the right speed. Right?
>> You're missing the atmosphere.
> So we'd have a flaming ball of rock, 1/4 the size of the Earth, orbiting
> at surface level at a speed just a tick up from the speed required to
> orbit the satellites. We could all get a 6-pack and watch. "Duck!
> Here she comes again!" A new ball game, Dodge the Moon.
>
> David Tholen, meet Greg Neill, who seems to thing the ball of rock would
> fragment.
Irrelevant to the issue of orbital mechanics.
> Due to the Moon's molten core and fragile crust, no doubt
> (just kiding here, Dave, so don't go off on a tangent).
On what basis do you claim that the Moon has a molten core?
> Perhaps, being the astronomer, Dave, you could give us examples of
> where Newton proved correct, and we have such massive objects
> orbiting each other, at the literal touch point.
What is the "literal touch point"?
Nancy Lieder
wrote:
>> If the mass is pulled from one to the other, then the
>> gravity attraction is strong enough to do this. So why
>> doesn't the second star just merge with the first? One
>> part of the star is honoring Newton and abiding by his
>> laws, while the other is not? Why is the mass moving
>> from one star to the other, if the centrifugal force is
>> strong enough to keep it in its orbit?
>
> Eventually these co-orbiting stars do coalesce. Such
> closely orbiting bodies raise terrific tides on each other,
> and orbital energy is lost to tidal friction -- they spiral
> into each other.
Here we have the mathematician getting sloppy, injecting, when faced
with the failure of Newton to explain a phenomena, a vague "energy
trading". This is happening behind the back of Newton, presumably, when
Newton is not looking.
In Article <Hvi87.16200$Tn3.7...@wagner.videotron.net> Greg Neill
wrote:
>> Equal and opposite? If Centrifugal force has to EQUAL
>> the force of gravity pulling inward, it does NOT in this
>> math. The force inward takes into consideration both
>> masses. The force outward is only dealing with the mass
>> of the secondary. How can they NOT both consider the
>> same factors!
>
> Please show me where they are not equal if they are written
> as equal:
> G*M1*M2/r^2 = M2*v^2/r
> The equation above says that they're equal.
> We know that they are equal by observation
> (circular orbit ==> inward force = outward force)
Nice and tight. The mass, which presumably includes ALL of that
co-orbiting star that starts to coalesce, should NOT, per Newton, want
to do anything but orbit about nicely, even if torn apart. Does the
Asteroid Belt not do so? Why should PART of that sun deviate? Do
Newton's neat little equations not balance, perfectly, as you stated?
Nancy Lieder
wrote:
>>>> The GIVENS:
>>>> Constant as m * p^2/d^3 is constant for all orbits.
>>>> where m = mass of primary
>>>> d = distance
>>>> p = period
>>>> could theoretically orbit at the same distance as Satellites,
>
>>> It is true that the velocity for circular orbit of a relatively
>>> small mass around a large mass depends only upon the
>>> distance from the large mass.
>
>> mass of the Primary. You are saying, below, what I've been
>> asserting - that this does not fit with the Inverse Square law.
>
> mass of the primary ... Please show where I (or my evil twin)
> stated that the centrifugal force depends upon the mass of the
> primary in any other way than as the result of the secondary's
> acceleration due to gravity.
Sure. Greg Neill meet Greg Neill, and have both of you met Magnus
Nyborg?
In Article <Hvi87.16200$Tn3.7...@wagner.videotron.net> Greg Neill
wrote:
> Please show me where they are not equal if they are written
> as equal:
> G*M1*M2/r^2 = M2*v^2/r
> The equation above says that they're equal.
> We know that they are equal by observation
> (circular orbit ==> inward force = outward force)
And this reduces to the Velocity equation, by factoring OUT the mass of
the secondary, M2, so that ONLY the mass of the primary is a concern.
In Article <8u_67.10984$e5.16...@newsb.telia.net> Magnus Nyborg wrote:
> v = sqrt( G*M / r )
>
> Ground orbit (if possible) -
> v = sqrt( 6.67E-11 * 5.976E24 / 6.378E6 ) = 7905 m/s
> Satellite orbit -
> v = sqrt( 6.67E-11 * 5.976E24 / 6.478E6 ) = 7844 m/s
> Moon orbit -
> v = sqrt( 6.67E-11 * 5.976E24 / 3.844E8 ) = 1018 m/s
Greg Neill, meet Greg Neill, etc.
In Article <B_U87.4131$uH4.1...@news20.bellglobal.com> Greg Neill
wrote:
> The formula stated, namely
> v = sqrt(G*M/r)
> is valid within the stated conditions.
-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
Greg Neill
> In Article <VUU87.4032$uH4.1...@news20.bellglobal.com> Greg Neill
> wrote:
> >> If the mass is pulled from one to the other, then the
> >> gravity attraction is strong enough to do this. So why
> >> doesn't the second star just merge with the first? One
> >> part of the star is honoring Newton and abiding by his
> >> laws, while the other is not? Why is the mass moving
> >> from one star to the other, if the centrifugal force is
> >> strong enough to keep it in its orbit?
> >
> > Eventually these co-orbiting stars do coalesce. Such
> > closely orbiting bodies raise terrific tides on each other,
> > and orbital energy is lost to tidal friction -- they spiral
> > into each other.
>
> Here we have the mathematician getting sloppy, injecting, when faced
> with the failure of Newton to explain a phenomena, a vague "energy
> trading". This is happening behind the back of Newton, presumably, when
> Newton is not looking.
Show me perfectly rigid, perfectly spherical planets and I'll
show you a case where you can apply Newton's gravitation and
inertial laws perfectly. You seem to think that all of existence
should be ideal cases without secondary affects like friction.
If you had an education in the sciences, you might also be aware
that Newton explains tides, too.
>
> In Article <Hvi87.16200$Tn3.7...@wagner.videotron.net> Greg Neill
> wrote:
> >> Equal and opposite? If Centrifugal force has to EQUAL
> >> the force of gravity pulling inward, it does NOT in this
> >> math. The force inward takes into consideration both
> >> masses. The force outward is only dealing with the mass
> >> of the secondary. How can they NOT both consider the
> >> same factors!
> >
> > Please show me where they are not equal if they are written
> > as equal:
> > G*M1*M2/r^2 = M2*v^2/r
> > The equation above says that they're equal.
> > We know that they are equal by observation
> > (circular orbit ==> inward force = outward force)
>
> Nice and tight. The mass, which presumably includes ALL of that
> co-orbiting star that starts to coalesce, should NOT, per Newton, want
> to do anything but orbit about nicely, even if torn apart. Does the
> Asteroid Belt not do so? Why should PART of that sun deviate? Do
> Newton's neat little equations not balance, perfectly, as you stated?
If a body is not held together by internal forces, then it does
not behave as a single, rigid body. Each part will individually
follow its own course per the force laws. As I explained, and you
clearly chose to ignore, the surface of at least one of the Suns
in this case is within the Roche Limit. Gas at the surface is not
firmly held to the Sun because the gravity of the companion Sun
is lifting it away via tidal stretching. Do you deny the Earth's
tides?
The Asteroid belt consists of myriad individual bodies, each
with its own orbit obeying Newton's laws. It does not behave
as a single entity. Your observation, in this context, is
therefore specious.
Before you go making silly statements about the failure of
Newton's laws, you should consider whether you are not
trying to impose conditions which clearly necessitate the
consideration of other factors, like the rigidity and
consistency of the bodies in question.
Greg Neill
I can see the centrifugal force on the right side of the
equation:
M2*v^2/r
It does not involve the mass of the primary, M1. It does
involve the mass of the secondary M2. What's your problem?
>
> And this reduces to the Velocity equation, by factoring OUT the mass of
> the secondary, M2, so that ONLY the mass of the primary is a concern.
You seem to think that algebraic cancelling of the terms for M2
on either side constitutes some grievous physical offence. Can
you be that naive about mathematics? If so, how can you think
to be taken seriously about *anything* you try to support in
the realm of astrodynamics?
Okay, I'll assume that you're a mischievos dolt incapable of
simple algebra. Here's the same formula without cancelling
balancing terms:
G*M1*M2/r^2 = M2*v^2/r
v^2 = G*M1*M2*r/(M2*r)
v = sqrt(G*M1*M2*r/(M2*r))
You may now plug in your values for the variables and arrive at
the same result as before.
john Latala
> If the mass of the primary is the ONLY thing an orbiting object needs
> to be concerned with, in how close it can come and how fast it orbits
> about the primary (these being the only factors in the law), then what
> seems to be the problem?
As long as the mass of the primary is significantly larger than the mass
of the orbiting body then the mass of the orbiting body can be ignored.
I think you're missing one small point. For a given primary if you pick a
velocity you want your orbiting body to move at that sets the radius of
the orbit. Conversely if you pick a radius then that sets the velocity the
body will orbit at.
--
john R. Latala
jrla...@golden.net
Bob May
wrong. The concepts are clear and easy to understand but you don't seem to
have the necessary intelligence to understand that.
Bob May
tho...@antispam.ham
> Here we have the mathematician getting sloppy, injecting, when faced
> with the failure of Newton to explain a phenomena, a vague "energy
> trading".
What alleged "failure of Newton"?
Meanwhile, we have your failure to answer the questions:
Bill Nelson
:> Unfortunately, you have shown that you do not understand
:> orbital dynamics ... Nor have you provided any mathematical
:> support for what is really a very simple physical system.
: Bill Nelson, meed Magnus Nyborg. I was using his math.
No you weren't. You were only vaguely using what you interpreted to
be his math.
: This law takes into consideration ONLY the mass of the primary. So my
Gravitational attraction requires the mass of both objects, although if
the mass of the satellite is small compared to the primary, then the mass
of the secondary can be ignored unless you are trying to figure out the
answer to a whole bunch of decimal places.
In other words, there is not much difference between 100,0001 and 100,000.
For most calculation, the 100,000 value is precise enough.
The Moon could possibly - and note that I say possibly - a borderline
case. Since it has a mass of only a bit more than 1% of that of the Earth,
the Moon's mass may also be ignored for most calculation. There is not a
lot of difference between 100 and 101 - unless you need high precision.
When figuring out orbital velocities etc, such precision is not needed
for most purposes.
: statement above is correct, as confirmed by (one of the) Greg Neills
: personas and David Tholen. (Greg is still arguing with himself, having
: been momentarily discombobulated by cracks in Newton appearing before
: him. Where is that contact cement!)
: In Article <8u_67.10984$e5.16...@newsb.telia.net> Magnus Nyborg wrote:
:> Orbital speed for ideal circular motion of a low-mass object
:> circling a high-mass object M (which refers to it's mass) is
:> determined by the formula
Note the "low-mass" requirement. Your claims are not being supported by
either Greg or David.
--
Bill Nelson (bi...@peak.org)
Bill Nelson
:> as equal:
:> G*M1*M2/r^2 = M2*v^2/r
:> The equation above says that they're equal.
:> We know that they are equal by observation
:> (circular orbit ==> inward force = outward force)
: Nice and tight. The mass, which presumably includes ALL of that
: co-orbiting star that starts to coalesce, should NOT, per Newton, want
: to do anything but orbit about nicely, even if torn apart. Does the
: Asteroid Belt not do so? Why should PART of that sun deviate? Do
: Newton's neat little equations not balance, perfectly, as you stated?
As has been explained before - the distances are measured to the center
of masses of the two objects. The center of mass is NOT at the surface.
Further, the orbital velocity is the velocity of the center of mass, so
some areas of that mass or moving at somewhat higher velocities than the
average. If you cannot figure out how this can be true, then you have
no possibility of every understanding orbital dynamics.
So, since the part of star-1 that is closer to star-2 is moving at a
slower velocity than necessary to remain in orbit (remember, the closer
the object the faster it must go to remain in orbit) - the gravitational
attraction of star-2 may be sufficiently higher than both the gravitational
attraction of star-1 and the centrifugal force of the gases of star-1
that are closest to star-2. The result is that gases are then stripped
off star-1 and attracted to star-2.
If you can understand this, and agree with it, then we may discuss to which
star the gasses are stripped.
--
Bill Nelson (bi...@peak.org)
The Small Kahuna
>
> Several things that you don't understand.
Actually, the next sentence shows that there is a lot that you don't
understand.
[Like compute the "centrifugal force"...]
Watching this thread has been really interesting so I decided to jump
into the fun.
Newton was sitting under an apple tree one day eating an apple. Another
apple fell on his head. It occurred to him that what must have happened
is that the earth itself had attracted the apple. When he began to get
a bump on his head it occurred to him that the problem was symmetric,
the earth attracted the apple, and the apple attracted the earth, but
since the earth was so much bigger, it had the larger influence.
Besides it really didn't matter what attracted what because his head
still hurt.
[Actually none of that happened. Its JUST A STORY.]
After thinking about it some more and looking at the data for the
observation of objects in the solar system, he decided that a reasonable
fit was that the force of attraction was proportional to the product of
the masses and inversely proportional to the square of the distance
between the two masses. All that was needed was a constant of
proportionality to get the units right. The result was:
F = (G * M1 * M2) / r^2
In addition, it was also discovered that a constant application of a
force to a fixed mass will produce a fixed acceleration:
F = M * A
All this is fine, but useless because you cannot tell where things came
from or where they are going. In order to do this, you have to project
the objects into a coordinate system, say for example, a cartesian
coordinate system (X,Y,Z).
Now you have:
F(X,Y,Z) = (G * M1 * M2 ) * ((x1-x2)^2,(y1-y2)^2,(z1-z2)^2)
and
F(X,Y,Z) = M * A(X,Y,Z)
In other words, you have two vector equations. Now you have something
you can work with and compute trajectories. However, the problem is
that the fundamental equation is only defined for *point* masses, not
large masses such as the earth. This must be true because a little bit
of stuff closer to the moon has a larger gravitational attraction than a
little bit of stuff on the other side.
So what you really need is two triple integrals, one to sum every dV
(delta volume) of earth against every dV of the moon. I won't write
this out because text is a really bad way to enter these equations, but
stick with me.
If you have any doubt that the integrals are actually needed, go sit
with your toes in the surf at low tide. After a while, you will be in
water up to your eyebrows and you will see the light, so to speak.
Now you have a force on one side and two triple integrals on the other,
but this does not help determine where the moon is. So now you need to
plug the force equations into the acceleration equation (and do it
twice, because the earth pulls on the moon and the moon pulls on the
earth).
The MA equation is a differential velocity, so you need to solve the
differential equation to determine what happens to the position. In
order to solve the differential equation, you need initial conditions
for the constants, (so you might as well look out the window and see
where the moon is and where it is going and use those values).
Now, you have something you can use to determine where the moon is going
to go next. (You also know why Newton had to invent calculus).
If I wrote this equation out, it would get lost in all the text, but it
would be general and accurate.
Needless to say, all this math is a pain in the ass, and not necessarily
helpful for someone trying to figure it all out. So normally, some
simplifying assumptions are made.
The first assumption is that there are no special locations so the
location of the (0,0,0) point is not important, so why not set it
equivalent to the center of the earth? Any place is as good as any
other. This means that the motion of the earth has been "nulled out" by
allowing the coordinate system to move. This assumption is reasonable
and supported by data that shows that we cannot seem to measure absolute
motion. There is no special place or preferred direction, and we can
only measure our velocity in reference to something else. So we might
as well drag the coordinate system with us, it simplifies the math.
The second assumption is that since the earth is symmetric and
homogeneous (except at the smallest level) it is OK to assume the earth
is a point mass. The same argument can be made for the moon. This
conveniently removes the triple integrals.
The third assumption is that it is OK to have the coordinate system
ROTATE in conjunction with the moon's orbit. This simplifies the
equations to:
F = G*M1*M2/r^2
F = M*A
Now, we are sitting here on a rotating and moving frame of reference.
It is like we are sitting on a rotating stool with a weight at the end
of a string. We can measure the force we need to apply to the string to
keep the weight a constant distance away, as measured from our eyes,
while sitting on a moving and rotating stool. This is comparable to the
gravitational force between the earth and moon.
Suddenly we let go of the string and notice that the weight moves away
from us AS IF a force had acted on it. We call this force the
"centrifugal" force when we are in elementary school. (When we are in
college, our physics professors are quite clear in stating that there is
no such thing as the "centrifugal force" as it is only an illusion based
on our (moving) frame of reference.) (Um, all you posters did go to
college, no?)
Now all of this makes the math simple, but it is wrong. The only way to
really do it is to setup and solve the differential velocity equation in
three space, and do the full integral.
To do all this correctly all we need is the constant of proportionality,
G. Fortunately, Cavendish measured this and subsequent repetitions of
(essentially) the same experiment have made it more accurate. Once we
have the constant of proportionality, all we need is a detailed
microscopic map of the interior of both the earth and the moon.
Oops. Houston, we have a problem.
Since we cannot measure the earth and the moon without grinding both up
into little pieces, we go back to our equations and SOLVE FOR THE
MASSES. It is a nice trick, actually, we "measure" the mass of the
earth indirectly. But we still do not really have any idea *exactly*
what the earth weighs, we just know that, to within experimental error,
our equations balance with the objects we can see and measure in the
solar system.
But the bottom line is we cheat. And there is no escaping this fact.
Your *only* point of contention can only be not whether we cheat, but
only if it matters.
Since nobody has shown me a differential equation with two triple
integrals, I can only assume that *none* of you know what you are doing,
not just Nancy.
The simplifying assumptions are reasonable and help clarify, and
thinking about a mysterious "centrifugal force" may keep our brains from
exploding, but it is really all wrong. So sorry, guys, Nancy is right,
all your math is wrong, even if hers is too.
Of course none of this takes into account relativistic corrections, so
it is all wrong twice.
Lately it has been released from "respected" researchers in "the
establishment" that the latest measurements of the expansion of the
universe indicate that this expansion is accelerating, not
decelerating. This can only be true if there is a repulsion force
operating.
So the bottom line is the latest thinking among the cosmological and
astronomical community is that, yes, Virginia, there really is a
repulsive force. Now the current explanation is that it only operates
at intergalactic distances, but this is, of course, bull. How can a
force know where to operate and where not to? This is like the other
bit of conventional wisdom about the "expansion of the universe", the
expansion only occurs in space where nobody is looking so we do not have
to confront the fact that the "expansion of the universe" violates the
conservation of energy. (Um, either conservation is a law or it isn't -
you can't have it both ways).
The bottom line is that *all* of the math being thrown around in this
thread is wrong. The other bottom line is that the current thinking
includes a repulsive force. The question is not *if* the repulsive
force is present in the universe, but only what the characteristics of
this force are. These questions include:
- Is this force static ala Einstein's cosmological constant?
- Is this force (directly) related to any other forces such as the
normal attractive gravity? Does this mean that gravity is inherently
bipolar similar to electrostatics and magnetics? Does this bipolar
nature explain *why* gravitation is 10^43 times smaller in magnitude
than the other forces? Is it because the forces are really both
powerful, but ever so slightly out of balance?
- Why do we not observe this force in everyday life? Is it simply
because it is very much smaller than gravity (which is incredibly much
smaller than the other known forces)? How much smaller is it? Why is
it smaller?
- What is this force a function of? Mass? Surface Area? Time? Density?
Composition?
- What is the propagation effect of this force, inverse square like
gravity and the electromagnetic force or some other function (like the
strong force)?
- Can this force be engineered (i.e. manipulated by us, somehow) like
the strong force or electromagnetics?
So you can stop picking on Nancy for her bad math until yours gets MUCH
better and you can also stop discounting the repulsive force because
some very smart people claim to have found it.
Greg Neill
"The Small Kahuna" <per...@company.com> wrote in message
news:3B66099A...@company.com...
>
> The first assumption is that there are no special locations so the
> location of the (0,0,0) point is not important, so why not set it
> equivalent to the center of the earth? Any place is as good as any
> other. This means that the motion of the earth has been "nulled out" by
> allowing the coordinate system to move. This assumption is reasonable
> and supported by data that shows that we cannot seem to measure absolute
> motion. There is no special place or preferred direction, and we can
> only measure our velocity in reference to something else. So we might
> as well drag the coordinate system with us, it simplifies the math.
Except that, if you choose to place your (0,0,0) point at the
center of the Earth you'd better take stock of what your assumption
implies. It implies that you've chosen a non-inertial refererence
frame and all the baggage that goes along with it. You can deal
with this in the vector equations all right, but not without some
pain. As far as doing math in coordinate systems goes, inertial
frames are preferable. In a non-inertial reference frame you
*can* determine that you are moving by acceleration effects.
>
> The second assumption is that since the earth is symmetric and
> homogeneous (except at the smallest level) it is OK to assume the earth
> is a point mass. The same argument can be made for the moon. This
> conveniently removes the triple integrals.
>
> The third assumption is that it is OK to have the coordinate system
> ROTATE in conjunction with the moon's orbit. This simplifies the
> equations to:
>
> F = G*M1*M2/r^2
>
> F = M*A
By now choosing a rotating frame, you'd better take into account
the coriolis terms, too, for general calculations.
>
> Now, we are sitting here on a rotating and moving frame of reference.
> It is like we are sitting on a rotating stool with a weight at the end
> of a string. We can measure the force we need to apply to the string to
> keep the weight a constant distance away, as measured from our eyes,
> while sitting on a moving and rotating stool. This is comparable to the
> gravitational force between the earth and moon.
>
> Suddenly we let go of the string and notice that the weight moves away
> from us AS IF a force had acted on it. We call this force the
> "centrifugal" force when we are in elementary school. (When we are in
> college, our physics professors are quite clear in stating that there is
> no such thing as the "centrifugal force" as it is only an illusion based
> on our (moving) frame of reference.) (Um, all you posters did go to
> college, no?)
Right. There is no separate entity that is the centrifugal force.
It is in fact an inertial effect due to acceleration of a mass
performing describing a curving trajectory. It is, however, a
convenient and perfectly valid analysis technique.
>
> Now all of this makes the math simple, but it is wrong. The only way to
> really do it is to setup and solve the differential velocity equation in
> three space, and do the full integral.
In this case wrong is a matter of degree. If you want answers that
reflect reality to several decimal places, the simplifying assumptions
are perfectly valid. If you want more decimal places, you choose
a better model. Like including topographic anomalies and mascons
and so forth in a slightly more complex geoid. It does not invalidate
the general results of the simpler model, especially in terms of the
gross effects which are being discussed.
>
> To do all this correctly all we need is the constant of proportionality,
> G. Fortunately, Cavendish measured this and subsequent repetitions of
> (essentially) the same experiment have made it more accurate. Once we
> have the constant of proportionality, all we need is a detailed
> microscopic map of the interior of both the earth and the moon.
>
> Oops. Houston, we have a problem.
>
> Since we cannot measure the earth and the moon without grinding both up
> into little pieces, we go back to our equations and SOLVE FOR THE
> MASSES. It is a nice trick, actually, we "measure" the mass of the
> earth indirectly. But we still do not really have any idea *exactly*
> what the earth weighs, we just know that, to within experimental error,
> our equations balance with the objects we can see and measure in the
> solar system.
Not quite. The Cavendish and related experiments give us G, and we
define our unit of mass with a standard. The Earth can be "weighed"
in terms of that standard, and its gravity mapped to great precision
with laser ranging satellites. In effect, reather than grinding it
up and measuring it mote by mote, we can perform a gravitational
tomographic scan to determine its construction to sufficient
precision for any desired orbit calculations.
>
> But the bottom line is we cheat. And there is no escaping this fact.
> Your *only* point of contention can only be not whether we cheat, but
> only if it matters.
Right.
>
> Since nobody has shown me a differential equation with two triple
> integrals, I can only assume that *none* of you know what you are doing,
> not just Nancy.
How many decimal places of accuracy are required to make decent
predictions of orbital characteristics in the models under
discussion? What would be gained by doing the integrals
every time you came across a model with homogenious spherical
masses?
The points under discussion are not about arguing the value of
the seventh or eighth decimal place in a prediction of the
orbital velocity of the Moon.
>
> The simplifying assumptions are reasonable and help clarify, and
> thinking about a mysterious "centrifugal force" may keep our brains from
> exploding, but it is really all wrong. So sorry, guys, Nancy is right,
> all your math is wrong, even if hers is too.
An approximation is not wrong if it is applied with full
knowledge of the limits of its applicability. Just as it's not
wrong to use Newtonian physics to do (most) solar system
calculations, even though General Relativity is known to be
a better model.
>
> Of course none of this takes into account relativistic corrections, so
> it is all wrong twice.
See above. Not knowing the position of your car in your
driveway to within a tenth of an angstrom does not prevent you
from finding it and driving to work. It is enough to know that
you can find it in your driveway. Just because you aren't
making full relativistic corrections in calculation for an
ephemeris of an asteroid does not mean you can't locate it
to within a fraction of an arcsecond using "wrong" Newtonian
physics.
>
> Lately it has been released from "respected" researchers in "the
> establishment" that the latest measurements of the expansion of the
> universe indicate that this expansion is accelerating, not
> decelerating. This can only be true if there is a repulsion force
> operating.
>
> So the bottom line is the latest thinking among the cosmological and
> astronomical community is that, yes, Virginia, there really is a
> repulsive force. Now the current explanation is that it only operates
> at intergalactic distances, but this is, of course, bull. How can a
> force know where to operate and where not to? This is like the other
> bit of conventional wisdom about the "expansion of the universe", the
> expansion only occurs in space where nobody is looking so we do not have
> to confront the fact that the "expansion of the universe" violates the
> conservation of energy. (Um, either conservation is a law or it isn't -
> you can't have it both ways).
In General Relativity, conservation of energy is a rather vague
concept for non-local physics. So this is really nothing new.
It is also broadening the topic into areas where other threads
have been hashing things out. I don't think it would serve any
purpose but obfuscation to get sidetracked down this path.
>
> The bottom line is that *all* of the math being thrown around in this
> thread is wrong.
The math is not wrong. Perhaps the models are too crude for your
liking, but the math for the models is not wrong.
> The other bottom line is that the current thinking
> includes a repulsive force. The question is not *if* the repulsive
> force is present in the universe, but only what the characteristics of
> this force are. These questions include:
>
> - Is this force static ala Einstein's cosmological constant?
>
> - Is this force (directly) related to any other forces such as the
> normal attractive gravity? Does this mean that gravity is inherently
> bipolar similar to electrostatics and magnetics? Does this bipolar
> nature explain *why* gravitation is 10^43 times smaller in magnitude
> than the other forces? Is it because the forces are really both
> powerful, but ever so slightly out of balance?
>
> - Why do we not observe this force in everyday life? Is it simply
> because it is very much smaller than gravity (which is incredibly much
> smaller than the other known forces)? How much smaller is it? Why is
> it smaller?
>
> - What is this force a function of? Mass? Surface Area? Time? Density?
> Composition?
>
> - What is the propagation effect of this force, inverse square like
> gravity and the electromagnetic force or some other function (like the
> strong force)?
>
> - Can this force be engineered (i.e. manipulated by us, somehow) like
> the strong force or electromagnetics?
All fine questions for other threads.
>
> So you can stop picking on Nancy for her bad math until yours gets MUCH
> better and you can also stop discounting the repulsive force because
> some very smart people claim to have found it.
No. This is nonsense. First of all, Nancy is claiming a repulsive
force that rises to overwhelm gravity at close surface distances.
This is demonstrably silly as evidenced by the fact that you and
I can both stand on the Earth without being tied down. It is
also empirically false as demonstrated by recent close proximity
measurements of G.
Second, her math is self inconsistent in that it does not even
respect units.
Third, the math that is being presented by others describes
adequately a model capable of distinguishing the relevant claims.
The Small Kahuna
>
> "The Small Kahuna" <per...@company.com> wrote in message
> news:3B66099A...@company.com...
>
> >
> > The first assumption is that there are no special locations so the
> > location of the (0,0,0) point is not important, so why not set it
> > equivalent to the center of the earth? Any place is as good as any
> > other. This means that the motion of the earth has been "nulled out" by
> > allowing the coordinate system to move. This assumption is reasonable
> > and supported by data that shows that we cannot seem to measure absolute
> > motion. There is no special place or preferred direction, and we can
> > only measure our velocity in reference to something else. So we might
> > as well drag the coordinate system with us, it simplifies the math.
>
> Except that, if you choose to place your (0,0,0) point at the
> center of the Earth you'd better take stock of what your assumption
> implies. It implies that you've chosen a non-inertial refererence
> frame and all the baggage that goes along with it. You can deal
> with this in the vector equations all right, but not without some
> pain. As far as doing math in coordinate systems goes, inertial
> frames are preferable. In a non-inertial reference frame you
> *can* determine that you are moving by acceleration effects.
I'm glad you basically agree that the "real" math is much more complex.
Including non-intertial reference frames is part of what I was referring
to when I mentioned the relativistic effects. Newton's theories are
first presented to kids in High School or Junior High School (or what
ever else it is called locally), and presenting everything all at once
would probably blow most people's minds.
Like you went on to point out about how finding your car in the morning
does not require angstrom accuracy, one has to differentiate between
what is useful for performing engineering calculations and what
"explains reality". If you want to put a man on the moon, there are
enough other unknown variables and an overwhelming desire to ensure a
safety margin that taking into account second order relativistic effects
is probably pointless. A mid course correction burn will take care of
it. But if you want to *explain* why the moon has been where it has for
an extended period, you just cannot hand wave, not really. A four
billion year integral is a terrible task master. Small effects really
do matter. Small effects have been used to explain why the moon always
shows us the same side and tidal forces (and I mean the literal tide of
the oceans) has been shown to have an effect on the rotation of the
earth. Neither effect is obvious at first glance and a simple equation
would not predict either effect.
> No. This is nonsense. First of all, Nancy is claiming a repulsive
> force that rises to overwhelm gravity at close surface distances.
> This is demonstrably silly as evidenced by the fact that you and
> I can both stand on the Earth without being tied down. It is
Well, not exactly because what has been stated is that the repulsive
force requires an approximate balance between objects. Neither you or I
are comparable to the earth in mass, so this is not philosophically
inconsistent (although I agree that the relevant equations have been
lacking). Since we know so little about the repulsive force, it is
impossible to say with any certainty that it should behave one way or
another, we will need further experiments and data to develop a theory.
> also empirically false as demonstrated by recent close proximity
> measurements of G.
Can you provide a pointer for this? I'd like to read up on it.
Greg Neill
Absolutely. Physics is about making predictions and describing
nature as accurately as possible, or at the engineering stage,
accurately enough. Physics does not make claims to answer "why"
questions, but rather "how" questions.
> If you want to put a man on the moon, there are
> enough other unknown variables and an overwhelming desire to ensure a
> safety margin that taking into account second order relativistic effects
> is probably pointless. A mid course correction burn will take care of
> it. But if you want to *explain* why the moon has been where it has for
> an extended period, you just cannot hand wave, not really. A four
> billion year integral is a terrible task master. Small effects really
> do matter. Small effects have been used to explain why the moon always
> shows us the same side and tidal forces (and I mean the literal tide of
> the oceans) has been shown to have an effect on the rotation of the
> earth. Neither effect is obvious at first glance and a simple equation
> would not predict either effect.
>
> > No. This is nonsense. First of all, Nancy is claiming a repulsive
> > force that rises to overwhelm gravity at close surface distances.
> > This is demonstrably silly as evidenced by the fact that you and
> > I can both stand on the Earth without being tied down. It is
>
> Well, not exactly because what has been stated is that the repulsive
> force requires an approximate balance between objects. Neither you or I
> are comparable to the earth in mass, so this is not philosophically
> inconsistent (although I agree that the relevant equations have been
> lacking). Since we know so little about the repulsive force, it is
> impossible to say with any certainty that it should behave one way or
> another, we will need further experiments and data to develop a theory.
The Moon is also not comparable to the Earth in mass. It makes for
sloppy physics to have to impose ad-hoc rules for every case.
>
> > also empirically false as demonstrated by recent close proximity
> > measurements of G.
>
> Can you provide a pointer for this? I'd like to read up on it.
Absolutely. A great place to start is here:
http://www.npl.washington.edu/eotwash/index.html
Enjoy.
Quantum Certainty
I find The Small Kahuna's arguments generally lucid and enlightening
but..
> Lately it has been released from "respected" researchers in "the
> establishment" that the latest measurements of the expansion of the
> universe indicate that this expansion is accelerating, not
> decelerating. This can only be true if there is a repulsion force
> operating.
This observation, however, follows from the observed red shift of
light over long distance. In another of your posts you stated, and I
agree, that light losses energy when traveling over long distances due
to the influence of gravity and particles in-between the source and
us--space is not, at least on the quantum level, a vacuum. With this
assumption, the red shift does not imply a repulsion force.
> So the bottom line is the latest thinking among the cosmological and
> Astronomical community is that, yes, Virginia, there really is a
> Repulsive force. Now the current explanation is that it only operates
> at intergalactic distances, but this is, of course, bull. How can a
> force know where to operate and where not to? This is like the other
> bit of conventional wisdom about the "expansion of the universe", the
> Expansion only occurs in space where nobody is looking so we do not have
> to confront the fact that the "expansion of the universe" violates the
> Conservation of energy. (Um, either conservation is a law or it isn't -
> you can't have it both ways).
I do agree that there is a repulsion force. No, I am not contradicting
myself because if gravity is both repulsive and attractive the
universe would be in some sort of stable state balanced between these
two forces (no expansion). Also the idea that gravity is a quantum
event has been around for some time. We even have a name for a gravity
particle--the graviton and a theory called quantum gravity. Seeing
gravity as a particle phenomenon allows us to explain failures of GR
like singularities. I am not saying that GR is wrong, just that it is
incomplete.
So 1) if gravity is composed of particles that interact with bodies by
pressing through them and 2) the repulsion force moves with a greater
velocity than the regular, constant speed, attractive gravity, then we
can do some thought experiments to determine the nature and
consequences of this new theory. Suppose following Einstein's example
that we are in an elevator sitting STILL on the surface of the earth.
Gravity particles flow through us pressing us down toward the floor.
This is true gravity but not the gravity we feel. We feel the floor
pushing back up on us. No problems here. Now if we take the elevator
and make move at a constant velocity upwards we run into problems. By
experience we know that we feel NO more gravity AT A COPNSTANT UPWARD
VELOCITY in the elevator than simply sitting still on the earth. If we
did it would contradict relativity. The gravity particles, now, are
pressing through us with a greater velocity and should exert more
force (addition of velocities according to SR). This however cannot be
the case.
In the previous example I supposed that the downward particles moved
with a speed less than the speed of light (because supposedly the
repulsion particles move faster). Not only does this scenario present
the addition of velocities problem, but also another problem. Suppose
that we are floating towards a black hole. If the downward press of
particles moves with less than the speed of light we will never
accelerate to the speed of light, and thus there is no such thing as
black holes. This violates GR and must also be incorrect.
We have to assume then that the downward particles move at the speed
of light. Now, due to relativistic addition of velocities, when moving
uniformly upward in our elevator we still see the gravity particles
moving at the same speed and experience the same force. Also, we can
now predict black holes. The strength of gravity would now not be
related so much to the velocity of gravity particle as the quantity
per given unit of mass. But this view is also inconsistent with
relativity since one of our premises is that the repulsive gravity
travels faster than the attractive gravity, and now the attractive
force is traveling at the speed of light.
Thus, it seems, we have to retreat to some odd quantum explanation to
purge the inconsistencies
Quantum Certainty
P.S. Small Kahuna please email me your email address so I can discuss
this more with you, as you seem open minded.