Question about pendulums, gravity etc
Adrian
Could someone wise in the ways of science settle an argument for me? :)
On a message board I have a person who claims that a hypothetical pendulum
set in motion, in a pure vacuum and with a perfectly frictionless bearing
would swing forever, as the only thing that would cause it to slow down is
friction.
So, could someone explain precisely what would cause such a pendulum to
slow to an eventual halt, if friction is removed from the equation?
Cheers,
Adrian
--
Using M2, Opera's revolutionary e-mail client: http://www.opera.com/m2/
Greg Neill
news:oprsvs7u...@news.demon.co.uk...
>
> Could someone wise in the ways of science settle an argument for me? :)
>
> On a message board I have a person who claims that a hypothetical pendulum
> set in motion, in a pure vacuum and with a perfectly frictionless bearing
> would swing forever, as the only thing that would cause it to slow down is
> friction.
>
> So, could someone explain precisely what would cause such a pendulum to
> slow to an eventual halt, if friction is removed from the equation?
> Cheers,
> Adrian
Gravitational radiation. Be prepared to wait a
long, long time.
Gregory L. Hansen
Adrian <adr...@abarnett.demon.co.uk> wrote:
>
>Could someone wise in the ways of science settle an argument for me? :)
>
>On a message board I have a person who claims that a hypothetical pendulum
>set in motion, in a pure vacuum and with a perfectly frictionless bearing
>would swing forever, as the only thing that would cause it to slow down is
>friction.
>
>So, could someone explain precisely what would cause such a pendulum to
>slow to an eventual halt, if friction is removed from the equation?
If there is a magnetic field and the pendulum is conductive, magnetic
damping will slow it. Doppler shifting by light that illuminates it will
slow it. If there's a nonuniform charge distribution on it,
electromagnetic radiation will slow it. Otherwise, what the other Greg
said.
There's a lot of if's to this type of question because it depends on what
physical model you have in mind, and what magnitude effect you're looking
for. Magnetic damping can slow the pendulum quickly if the device was
designed that way, and some are. Slowing by Doppler shifting will be
faster than gravitational radiation, but that's not saying much. Usually,
to very high precision, friction is the only thing to worry about.
--
"A good plan executed right now is far better than a perfect plan
executed next week."
-Gen. George S. Patton
Bill Vajk
snip
> There's a lot of if's to this type of question because it depends on what
> physical model you have in mind, and what magnitude effect you're looking
> for. Magnetic damping can slow the pendulum quickly if the device was
> designed that way, and some are. Slowing by Doppler shifting will be
> faster than gravitational radiation, but that's not saying much. Usually,
> to very high precision, friction is the only thing to worry about.
I expect he was asking about a situation involving
an ideal pendulum without quite knowing how to
frame his question.
The question becomes real interesting in a case where
a pendulum forms a significant third member with
two orbiting bodies and one swinging one.
Steve Harris
argument for me? :)
>
> On a message board I have a person who claims that a
hypothetical pendulum
> set in motion, in a pure vacuum and with a perfectly
frictionless bearing
> would swing forever, as the only thing that would cause it
to slow down is
> friction.
>
> So, could someone explain precisely what would cause such
a pendulum to
> slow to an eventual halt, if friction is removed from the
equation?
> Cheers,
> Adrian
Such a device changes quadrupole moment, and thus radiates
gravity waves, which suck off energy and eventually will
stop it.
Two orbiting planets, like your frictionless pendulum, will
eventually spiral into each other and "stop." But a
rotating ring goes forever.
kdthrge
> > On a message board I have a person who claims that a
> hypothetical pendulum
> > set in motion, in a pure vacuum and with a perfectly
> frictionless bearing
> > would swing forever, as the only thing that would cause it
> to slow down is
> > friction.
> >
> > So, could someone explain precisely what would cause such
> a pendulum to
> > slow to an eventual halt, if friction is removed from the
> equation?
> > Cheers,
> > Adrian
>
>
> Such a device changes quadrupole moment, and thus radiates
> gravity waves, which suck off energy and eventually will
> stop it.
>
> Two orbiting planets, like your frictionless pendulum, will
> eventually spiral into each other and "stop." But a
> rotating ring goes forever.
......
The moon is increasing it's distance from the earth. This is because
of energy lost from tidal friction and because a lesser energy orbit
has a greater orbital radius. Please refer to your basic Kepler law
and gravitational law. Two orbiting planets will only spiral into
eachother if energy is added to their orbit. I know this is difficult
for the godchildren of Bohr to understand.
http://home.earthlink.net/~kdthrge
.............
Sam Wormley
>
> ......
> The moon is increasing it's distance from the earth. This is because
> of energy lost from tidal friction and because a lesser energy orbit
> has a greater orbital radius. Please refer to your basic Kepler law
> and gravitational law. Two orbiting planets will only spiral into
> eachother if energy is added to their orbit. I know this is difficult
> for the godchildren of Bohr to understand.
>
>
You mean like Neptune's moon, Triton?
Sam Wormley
Double-A
Interesting comparison. My understanding is that what it boils down
to is that since the Earth is rotating faster than the Moon is
revolving, energy is slowly being transferred from the Earth's
rotation to put the Moon into a higher orbit.
What I am curious about is whether the Sun, which rotates about once
every 27 days or so, is transferring energy to slowly put the Earth in
a higher orbit? If so, I wonder if it will be enough to compensate
for the Sun's gradual heating up over the next 500 million years?
Double-A
Sam Wormley
>
> Sam Wormley <swor...@mchsi.com> wrote in message news:<3F2301B9...@mchsi.com>...
> > Sam Wormley wrote:
> > >
> > > kdthrge wrote:
> > > >
> >
> > > > ......
> > > > The moon is increasing it's distance from the earth. This is because
> > > > of energy lost from tidal friction and because a lesser energy orbit
> > > > has a greater orbital radius. Please refer to your basic Kepler law
> > > > and gravitational law. Two orbiting planets will only spiral into
> > > > eachother if energy is added to their orbit. I know this is difficult
> > > > for the godchildren of Bohr to understand.
> > > >
> > > >
> > >
> > > You mean like Neptune's moon, Triton?
> > http://www.mhhe.com/physsci/astronomy/fix/student/chapter14/14f39.html
>
> Interesting comparison. My understanding is that what it boils down
> to is that since the Earth is rotating faster than the Moon is
> revolving, energy is slowly being transferred from the Earth's
> rotation to put the Moon into a higher orbit.
No No No.... When rotation of the planet slows, angular momentum is
conserved and the orbital radius of the satellite increases for a
prograde satellite, but decreases for a retrograde satellite.
Tidal Torque
http://scienceworld.wolfram.com/physics/TidalTorque.html
Sam Wormley
>
> Sam Wormley <swor...@mchsi.com> wrote in message news:<3F2301B9...@mchsi.com>...
> > Sam Wormley wrote:
> > >
> > > kdthrge wrote:
> > > >
> >
> > > > ......
> > > > The moon is increasing it's distance from the earth. This is because
> > > > of energy lost from tidal friction and because a lesser energy orbit
> > > > has a greater orbital radius. Please refer to your basic Kepler law
> > > > and gravitational law. Two orbiting planets will only spiral into
> > > > eachother if energy is added to their orbit. I know this is difficult
> > > > for the godchildren of Bohr to understand.
> > > >
> > > >
> > >
> > > You mean like Neptune's moon, Triton?
> > http://www.mhhe.com/physsci/astronomy/fix/student/chapter14/14f39.html
>
> Interesting comparison. My understanding is that what it boils down
> to is that since the Earth is rotating faster than the Moon is
> revolving, energy is slowly being transferred from the Earth's
> rotation to put the Moon into a higher orbit.
>
When rotation of the planet slows, angular momentum is
kdthrge
> > > > > ......
> > > > > The moon is increasing it's distance from the earth. This is because
> > > > > of energy lost from tidal friction and because a lesser energy orbit
> > > > > has a greater orbital radius. Please refer to your basic Kepler law
> > > > > and gravitational law. Two orbiting planets will only spiral into
> > > > > eachother if energy is added to their orbit. I know this is difficult
> > > > > for the godchildren of Bohr to understand.
> > > > >
> > > > >> > > >
> > to is that since the Earth is rotating faster than the Moon is
> > revolving, energy is slowly being transferred from the Earth's
> > rotation to put the Moon into a higher orbit.
> >
>
> When rotation of the planet slows, angular momentum is
> conserved and the orbital radius of the satellite increases for a
> prograde satellite, but decreases for a retrograde satellite.
>
> Tidal Torque
> http://scienceworld.wolfram.com/physics/TidalTorque.html
Perhaps there is someome on this channel that knows some astrophysics
and can look at this problem objectively.
An orbit is not a circular level of distance as pictured in Bohrs
atomic model. It is always an elipse of some degree. The energy of an
orbit can be defined by the area it encompasses. Less area =greater
energy
Imagine firing the rockets briefly on an orbiting spacecraft.
Regardless of the direction the rockets are burned (unless the burn
exactly counters the angular momentum)this will only have the effect
of increasing the eccentricity of the orbit. The resulting ellipse
will have less area than the previous orbit. When the Mir space
station was brought down, the rockets were fired to increase the
eccentricity of the orbit taking the station out and then with the
greater ellipse it returns into the atmosphere and burns up. An orbit
decays only from the friction of the atmosphere.
Energy is lost from the earth-moon system as heat,generated from the
friction of the motion of the tides.
Angular momentum is always conserved unless it is converted into
another form of energy.
http://home.earthlink.net/~kdthrge
Rob Fatland
the earth is slowed by tidal friction. That is, the moon recedes and is moving
slower (R^(-1/2)) but is further away and has more gravitational potential (~R).
By the way, the rate of recession of the moon is about 4 cm / year, and I think
every century our days are .002 seconds longer. At this rate the earth will
become tidally locked to the moon in something like 15 billions years but will
long since have been burnt to a cinder as the sun runs low on fuel in something
like 6 billions years.
As to the earth slowing down the sun's rate of rotation, part of the reason this
effect would be very small is the earth's mass compared to the sun. Another
important factor is that the sun doesn't have any continental shelves to provide
tidal drag, so the tidal bulge is free to flow across its surface unobstructed.
(In contrast the earth has a lot of nice continents in the way of the bulge which
serve to increase tidal friction.) My guess is that the effect of the earth (and
other planets) stealing energy from the sun and moving into higher orbit is
negligible over the time scales mentioned.
Bill Vajk
> As to the earth slowing down the sun's rate of rotation, part of the reason this
> effect would be very small is the earth's mass compared to the sun. Another
> important factor is that the sun doesn't have any continental shelves to provide
> tidal drag, so the tidal bulge is free to flow across its surface unobstructed.
> (In contrast the earth has a lot of nice continents in the way of the bulge which
> serve to increase tidal friction.)
snip
Consider earth tides. Here's a pedestrian description to
get you started:
http://www.almanac.com/tides/earthtide.html
As a specific area of study, earth tides are relatively
recent though there is an international commission
involved:
http://www.astro.oma.be/IAG/earthti.html
see also:
http://www.agu.org/pubs/crossref/2003/2001JB000569.shtml
http://www.kms.dk/fags/ps04icet.htm
http://www.miz.nao.ac.jp/ets2000/first.html
Sentient discussions of luna's tidal deformations are
rare to nonexistent.
Greg Neill
news:1eb8d2f6.03072...@posting.google.com...
> Perhaps there is someome on this channel that knows some astrophysics
> and can look at this problem objectively.
> An orbit is not a circular level of distance as pictured in Bohrs
> atomic model. It is always an elipse of some degree. The energy of an
> orbit can be defined by the area it encompasses. Less area =greater
> energy
The total mechanical energy of an obit is related to the size
of its semimajor axis:
E = -GM/(2a)
Consider a circular orbit with radius a and an elliptical
orbit with semimajor axis a. These orbits will have
equal energy.
Now suppose that the eccentricity of the elliptical orbit
is given by e, where 0 < e < 1. The semiminor axis of
the ellipse with eccentricity e is given by:
b = a*sqrt(1-e^2)
Now compute the areas of the two orbits. For the circle:
Ac = pi*a^2
For the ellipse:
Ae = pi*a*b = pi*a^2*sqrt(1-e^2)
Clearly, given the constraints on e, Ae < Ac. Yet the
energies of the two orbits are identical.
Steve Harris
"Greg Neill" <gnei...@OVE.netcom.ca> wrote in message
news:mpfVa.10741$vB6.3...@weber.videotron.net...
> "kdthrge" <kdt...@yahoo.com> wrote in message
> news:1eb8d2f6.03072...@posting.google.com...
>
> > Perhaps there is someome on this channel that knows some
astrophysics
> > and can look at this problem objectively.
> > An orbit is not a circular level of distance as pictured
in Bohrs
> > atomic model. It is always an elipse of some degree. The
energy of an
> > orbit can be defined by the area it encompasses. Less
area =greater
> > energy
>
> The total mechanical energy of an obit is related to the
size
> of its semimajor axis:
>
> E = -GM/(2a)
>
> Consider a circular orbit with radius a and an elliptical
> orbit with semimajor axis a. These orbits will have
> equal energy.
They'll have equal periods, so NOT equal energy. In fact it
takes energy (a rocket burn at the appropriate distance a)
to convert one orbit to the other (either circular orbit
with radius a to elliptical with semimajor a, or vice
versa). The simple reason is that a thing in orbit
elliptically with semimajor axis a isn't moving as fast when
it gets to "a" as if it were in a circular orbit with radius
"a". Thus, kinetic and mechanical energy must be added.
Greg Neill
news:bg47cl$nfg$1...@slb0.atl.mindspring.net...
The expression E = -GM/(2a) is valid for all conic orbits.
Despite the fact that two orbits have the same mechanical
energy, it takes energy to perform the manouver from one
to another. For example, in the case of changing the
elliptical orbit to a circular one, you need to raise the
perihelion and then circularize. Two manouvers which
combined, first add, then subtract energy. If you do
the sums you'll find that the net change (to the orbital
energy) is zero.
kdthrge
> > > The total mechanical energy of an obit is related to the
> size
> > > of its semimajor axis:
> > >
> > > E = -GM/(2a)
> > >
> > > Consider a circular orbit with radius a and an elliptical
> > > orbit with semimajor axis a. These orbits will have
> > > equal energy.
> >
> >
> > They'll have equal periods, so NOT equal energy. In fact it
> > takes energy (a rocket burn at the appropriate distance a)
> > to convert one orbit to the other (either circular orbit
> > with radius a to elliptical with semimajor a, or vice
> > versa). The simple reason is that a thing in orbit
> > elliptically with semimajor axis a isn't moving as fast when
> > it gets to "a" as if it were in a circular orbit with radius
> > "a". Thus, kinetic and mechanical energy must be added.
>
> The expression E = -GM/(2a) is valid for all conic orbits.
>
> Despite the fact that two orbits have the same mechanical
> energy, it takes energy to perform the manouver from one
> to another. For example, in the case of changing the
> elliptical orbit to a circular one, you need to raise the
> perihelion and then circularize. Two manouvers which
> combined, first add, then subtract energy. If you do
> the sums you'll find that the net change (to the orbital
> energy) is zero.
I think what confuses people is the concept of total mechanical energy
of the orbit. The term more energetic orbit would more clearly
describe the simple reality of Kepler's second law. In evaluating the
energy of two different orbits, it is important to evaluate the energy
of the complete orbits. But if one simply evaluates them by mechanical
energy for unit of time, it is very clear which orbit is more
energetic. Just because a slow car travels a greater distance on a
long circuit expending a large amount of mechanical energy in one
circuit, does not make it higher energy than a race car on a short
circuit. The race car's energy per hour is much greater although it's
mechanical energy per circuit is not so great on the shorter track.
To decrease angular momemtum of an orbital body increases its orbital
radius.
Keith F. Lynch
> On a message board I have a person who claims that a hypothetical
> pendulum set in motion, in a pure vacuum and with a perfectly
> frictionless bearing would swing forever, as the only thing that
> would cause it to slow down is friction.
> So, could someone explain precisely what would cause such a pendulum
> to slow to an eventual halt, if friction is removed from the equation?
Steve Harris <sbha...@ix.RETICULATEDOBJECTcom.com> wrote:
> Such a device changes quadrupole moment, and thus radiates gravity
> waves, which suck off energy and eventually will stop it.
Yes. But have you calculated how long this would take for a
reasonable sized pendulum?
I think that the pendulum's interaction with the cosmic ray background
would bring it to a halt far sooner, since it would experience
alternating blue shifts and red shifts tending to counteract its
motion. This would still be extremely slow as compared to, say,
geological ages. But it wouldn't take more eons than there have
been nanoseconds since the Big Bang. Gravitational waves would.
If there was a source of light, or if the pendulum was warmer than the
cosmic ray background, the slowdown would be faster still, for similar
reasons.
> Two orbiting planets, like your frictionless pendulum, will
> eventually spiral into each other and "stop." But a rotating
> ring goes forever.
A rotating ring won't radiation gravitational waves, but it will still
be slowed by the cosmic microwave background, which is a much stronger
effect.
--
Keith F. Lynch - k...@keithlynch.net - http://keithlynch.net/
I always welcome replies to my e-mail, postings, and web pages, but
unsolicited bulk e-mail (spam) is not acceptable. Please do not send me
HTML, "rich text," or attachments, as all such email is discarded unread.
Greg Neill
news:1eb8d2f6.03072...@posting.google.com...
>
> I think what confuses people is the concept of total mechanical energy
> of the orbit. The term more energetic orbit would more clearly
> describe the simple reality of Kepler's second law. In evaluating the
> energy of two different orbits, it is important to evaluate the energy
> of the complete orbits.
The term is unambiguous since the total energy is
an integral of the motion, that is, it is a constant.
It is the sum of the kinetic and potential energy
of the system.
> But if one simply evaluates them by mechanical
> energy for unit of time, it is very clear which orbit is more
> energetic.
Of course, since it is constant it will be the same
value for all instants of time.
> Just because a slow car travels a greater distance on a
> long circuit expending a large amount of mechanical energy in one
> circuit, does not make it higher energy than a race car on a short
> circuit. The race car's energy per hour is much greater although it's
> mechanical energy per circuit is not so great on the shorter track.
Your analogy is weak, as the cars are not performing their
antics in an energentically closed system and are deriving
their impetus from fuel.
>
> To decrease angular momemtum of an orbital body increases its orbital
> radius.
h = r x v (x denotes the cross product)
= r*v at perihelion or aphelion
In the case of a circular orbit, v = sqrt(GM/r), so we have
h = r*sqrt(GM/r)
= sqrt(rGM)
Sure looks like the angular momentum increases with an
increase in radius.
kdthrge
.....
I just want to point out that circular orbits do not exist. An orbital
body cannot maintain moving 'perpendicular' to the scource of
attraction and will always be either receding or closing.
Edward Green
> I just want to point out that circular orbits do not exist. An orbital
> body cannot maintain moving 'perpendicular' to the scource of
> attraction and will always be either receding or closing.
Why would that be?
Possibly in case two unequal masses m < M orbiting each other neither
can move in a circle; but in the special cases m = M or m << M (M =
oo) a circular orbit is as good as any other, except in the trivial
sense that any real orbit will always deviate from the precise value e
= 0.
Come to think of it, a circular orbit works for arbitrary dissimilar
masses as well. What kind of gravitation have you been smoking?
Edward Green
"Use the antonym of 'retrograde' in a sentence". :-)
> When rotation of the planet slows, angular momentum is
> conserved and the orbital radius of the satellite increases for a
> prograde satellite, but decreases for a retrograde satellite.
Let's see: rw^2 = constant x r^-2; w = constant' x r^-3/2
KE = constant'' x (rw)^2 = constant''' x (1/r), and
PE = yet some other constant x (1/r), so overall E = C/r
Well, except that the KE and PE terms have different signs, which
means either they cancel, and all orbits have the same energy, or one
dominates: I assume the potential dominates (not all orbits have
escape velocity), so C < 0.
OK, so larger orbits have greater energy, up to 0 at oo. So a
prograde satellite acquires some of the rotational energy of the host,
whereas a retrograde satellite loses some of its orbital energy in
heating the host.
I'm a little confused about the trend in angular momentum, though:
L = constant x r^2w = constant' x r^1/2
OK... that's a positive constant, and larger orbits have larger L.
kdthrge
> h = r x v (x denotes the cross product)
>
> = r*v at perihelion or aphelion
>
> In the case of a circular orbit, v = sqrt(GM/r), so we have
>
> h = r*sqrt(GM/r)
>
> = sqrt(rGM)
>
> Sure looks like the angular momentum increases with an
> increase in radius.
I hope the martians don't mind the two more pieces of scrape metal
headed to crash on their planet like the last two missions did. If the
boys at NASA are using your math or your comphrehension of what
momentum is, theres not a chance in hell of a succesful mission.
No matter how you slice it, momemtum is a product of velocity and
mass, angular or non-angular momentum.
Well according to your math, pluto has a greater orbital velocity than
mercury. Being in it's higher energy state. What goes up must come
down huh. This is the extent of your actual understanding.. The higher
you lift something the harder it comes down, huh... more energeee
If you had any analytical thought, you could even see that a
gravitational field increases its strength as an inverse square to
distance. The orbital body with less radius is being slung around by
the greater gravitational attraction; therfore greater velocity and
momentum.
Mean orbital radius squared = period cubed.
Mars mean orbital radius is 1.52369 AU. (earth orbital radius)
Its period therfore is 1.8808 years.
It travels a distance 1.5 greater than earth (C=2pir)in a time 1.9
greater but it has greater momentum????????. Maybe you should get a
job driving a truck and give up science and math. Or stick to quantum
theory which can never be tested.
Simply to diminish angular momentum increases radius. I was taught the
analyses of the earth-moon system by a German astrophysicist so
orthodox that he would never even talk to you. And he was correct and
you are a fool!!
http://home.earthlink/net/~kdthrge
Greg Neill
> >
> > h = r x v (x denotes the cross product)
> >
> > = r*v at perihelion or aphelion
> >
> > In the case of a circular orbit, v = sqrt(GM/r), so we have
> >
> > h = r*sqrt(GM/r)
> >
> > = sqrt(rGM)
> >
> > Sure looks like the angular momentum increases with an
> > increase in radius.
> ...............
> I hope the martians don't mind the two more pieces of scrape metal
> headed to crash on their planet like the last two missions did. If the
> boys at NASA are using your math or your comphrehension of what
> momentum is, theres not a chance in hell of a succesful mission.
>
> No matter how you slice it, momemtum is a product of velocity and
> mass, angular or non-angular momentum.
You know nothing. When you're dealing with angular momentum
you must use the moment of inertia, which depends upon both
the mass and its relative distribution.
> Well according to your math, pluto has a greater orbital velocity than
> mercury. Being in it's higher energy state. What goes up must come
> down huh. This is the extent of your actual understanding.. The higher
> you lift something the harder it comes down, huh... more energeee
Again you show that you know nothing. Where in the equation
h = sqrt(rGM) do you see a velocity dependence? Also, in
the *definition* of angular momentum, h = r x v, it's obvious
to anyone but a complete bozo that there are two terms that
determine the angular momentum: velocity and radius. A large
radius implies a lower velocity for the same value of angular
momentum. Guess what, Sparky? Pluto's orbital radius is
larger than that ov Mercury.
>
> If you had any analytical thought, you could even see that a
> gravitational field increases its strength as an inverse square to
> distance. The orbital body with less radius is being slung around by
> the greater gravitational attraction; therfore greater velocity and
> momentum.
My, my. You can't even state Netwton's law properly. Are you
quite sure that gravity increases its strength as the inverse square
of the distance? Your old curmudgeon of a professor would be most
displeased. You must now write out Newton's law 100 times on the
blackboard.
> Mean orbital radius squared = period cubed.
So?
> Mars mean orbital radius is 1.52369 AU. (earth orbital radius)
> Its period therfore is 1.8808 years.
So?
> It travels a distance 1.5 greater than earth (C=2pir)in a time 1.9
> greater but it has greater momentum????????.
Angular momentum. This is quite different from linear momentum.
In the case of Mars, its specific angular momentum would be
about 1.5 * 1.9 = 2.8 times that of the Earth. Note I said
specific angular momentum. Multiply by the ratio of Mars'
mass to Earth's mass to get the actual relative value.
> Maybe you should get a
> job driving a truck and give up science and math. Or stick to quantum
> theory which can never be tested.
> Simply to diminish angular momentum increases radius. I was taught the
> analyses of the earth-moon system by a German astrophysicist so
> orthodox that he would never even talk to you. And he was correct and
> you are a fool!!
Your delusions of competence must be a great comfort to you.
> http://home.earthlink/net/~kdthrge
kdthrge
>
> Again you show that you know nothing. Where in the equation
> h = sqrt(rGM) do you see a velocity dependence? Also, in
> the *definition* of angular momentum, h = r x v, it's obvious
> to anyone but a complete bozo that there are two terms that
> determine the angular momentum: velocity and radius. A large
> radius implies a lower velocity for the same value of angular
> momentum. Guess what, Sparky? Pluto's orbital radius is
> larger than that ov Mercury.
>
At some point it will become clear what is correct. To add energy to
an orbit increases area of the orbit or reduces it. This is the
question.
Of course bad science always relies on bad analogy. Yours is your
hangup with definitions of angular momentum. If one is on a merry-go
round your definitions of angular momentum hold. An orbit is a state
of free fall. Gravity is not a string tied onto the orbital body as
you imagine giving your factor of radius any value.
If one throws a ball upward at an angle it makes an ellipse. This
ellipse is described by only two vectors. The force initially applied.
And the acceleration of gravity which in this case is a direct and
continually applied vector towards the center of the earth. The
momentum of the object must be described as angular and it describes
the force of the ball at any particular point and the eventual path of
the ball. If you tied a string to the ball it may fit your formulae.
The inital vector of the ball is constantly diminishing due to the
tangental force of gravity against it until the arc is no longer
receding against gravity. Then what remains of this vector remains
until impact with the ground. What you need to remember is that
vectors of momentum in a free fall can never change direction. There
is no memory possible for vectors in a free fall like there is on your
merry go round.
Escuse my mistake on my last post since (a cubed = p squared)
http://home.earthlink.net/~kdthrge
Greg Neill
news:1eb8d2f6.03080...@posting.google.com...
> At some point it will become clear what is correct. To add energy to
> an orbit increases area of the orbit or reduces it. This is the
> question.
> Of course bad science always relies on bad analogy. Yours is your
> hangup with definitions of angular momentum.
[mercy snip]
Excuse me, but you are an empirical moron. Why do you even bother?
Consider yourself consigned to the kill-file. Yeesh! What a maroon!
The Ghost In The Machine
<kdt...@yahoo.com>
wrote
on 3 Aug 2003 18:56:32 -0700
<1eb8d2f6.03080...@posting.google.com>:
> "
>>
>> Again you show that you know nothing. Where in the equation
>> h = sqrt(rGM) do you see a velocity dependence? Also, in
>> the *definition* of angular momentum, h = r x v, it's obvious
>> to anyone but a complete bozo that there are two terms that
>> determine the angular momentum: velocity and radius. A large
>> radius implies a lower velocity for the same value of angular
>> momentum. Guess what, Sparky? Pluto's orbital radius is
>> larger than that ov Mercury.
>>
>
>
> At some point it will become clear what is correct. To add energy to
> an orbit increases area of the orbit or reduces it. This is the
> question.
That depends on the vector of the momentum. Part of the problem
is that there are only two ways we know of to add energy to
an orbit, and the second, involving electromagnetic fields, is
not all that practical. The first way is through old tried
and true rocketry, where we impart energy to exhaust gases,
losing part of the energy in the process as the gas goes
into its own orbit.
It takes two rocket burns to raise a satellite from one
circular orbit to a higher circular orbit. This ain't
rocket science (although it is part thereof). :-)
As for angular momentum -- I don't see why it can't be applied
to this problem, as we have two objects (Sun and Earth, say)
spinning around a common mass-point (deep within the Sun).
Angular momentum must always be conserved in a closed system,
much like regular momentum.
[rest snipped]
--
#191, ewi...@earthlink.net
It's still legal to go .sigless.
Greg Neill
message news:l09101-...@lexi2.athghost7038suus.net...
> In sci.physics, kdthrge <kdt...@yahoo.com> wrote
> >
> > At some point it will become clear what is correct. To add energy to
> > an orbit increases area of the orbit or reduces it. This is the
> > question.
>
> That depends on the vector of the momentum. Part of the problem
> is that there are only two ways we know of to add energy to
> an orbit, and the second, involving electromagnetic fields, is
> not all that practical. The first way is through old tried
> and true rocketry, where we impart energy to exhaust gases,
> losing part of the energy in the process as the gas goes
> into its own orbit.
>
> It takes two rocket burns to raise a satellite from one
> circular orbit to a higher circular orbit. This ain't
> rocket science (although it is part thereof). :-)
>
> As for angular momentum -- I don't see why it can't be applied
> to this problem, as we have two objects (Sun and Earth, say)
> spinning around a common mass-point (deep within the Sun).
> Angular momentum must always be conserved in a closed system,
> much like regular momentum.
It's a simple matter to compare the area encompassed
by two sample orbits. The total specific mechanical
energy is given by -GM/(2a), where M is the total mass
of the two body system, and a is the semimajor axis of
the orbit. If the primary is much more massive than
the secondary (as is the case for the Sun and the Earth,
for example), then M can be taken to be simply the mass
of the primary.
Clearly we can produce two orbits of equal energy but
different area by keeping the semimajor axes the same
and adjusting the eccentricity. The area of an ellipse
is pi*a*b, with b being the semiminor axis. b can be
had from a and the eccentricity e as:
b = a*sqrt(1-e^2)
giving for the area of the orbit:
A = pi*(a^2)*sqrt(1-e^2)
Now, recalling that the energy of the orbit is -GM/(2a),
A = -pi*GM/(2E)*sqrt(1-e^2)
Keep in mind that the energy for a bound orbit is
negative, so the area will be positive!
We can see from the above that the area can be adjusted
by the two parameters E and e. There will be combinations
of E and e that produce the same area.
How about the angular momentum? The magnitude of the
specific angular momentum is given by:
h = sqrt(a*GM)*(1-e^2)
or in terms of the energy and eccentricity:
h = GM*sqrt((e^2 -1)/2E)
After a little manipulation and substitution we find:
A = pi*h/sqrt(-2E)
Again, keep in mind that E is negative for a bound orbit.
A Williams
> Could someone wise in the ways of science settle an argument for me? :)
>
> set in motion, in a pure vacuum and with a perfectly frictionless bearing
> would swing forever, as the only thing that would cause it to slow down is
> friction.
>
> So, could someone explain precisely what would cause such a pendulum to
> slow to an eventual halt, if friction is removed from the equation?
> Adrian
in pure vacoum in the absense of other forces other than the
one whict starts the motion will not swing at all because a pendulum
neds constant force
perpendicular to its motion.
bill university of kerala
trivandrum