Re: Will the Moon Crash Into Earth?
Jo Schaper
> I asked a science prof if the moon will eventually move away or into
> the earth.
>
> He said neither, it will remain revolving the earth for eternity.
>
> Obviously after this insulting response I stopped taking him too
> serious.
Well, barring a collision with an asteroid, the moon probably will
continue moving away (as it is currently by a few cm a year.) It likely
will remain in orbit at some distance, until both it and earth are
crisped to cinders by the expanding sun, a couple of billion years in
the future. A couple of billion years is close enough to an eternity for
me.
Frogwatch
wrote:
If earth and the moon were truly solid (elastic) bodies, then the moon
would stay at the same nominal distance until disturbed by something.
This is due to Conservation of Angular Momentum and Conservation of
Energy. HOWEVER, neither the earth or moon are truly elastic. The
earth has tides due to the moon and the moon has "tides" due to
earth. Thus some of the energy associated with the moons orbiting
earth turns into heat in the tides. This means it loses energy.
Kinetic energy = 1/2 I w^2 where I is the moment of inertia and w is
angular frequency of rotation ( I know, this is really simplified).
So if kinetic energy goes down, so does w. If w goes down, the radius
of the orbit INCREASES. Due to tidal losses, the moons orbit
increases about an inch each year. Apparently, a billion years ago,
it was closer, roughly a billion inches or 100 million feet or about
16000 miles closer.
WillE1
<effa...@hotmail.com> wrote in message
news:1170989273.1...@m58g2000cwm.googlegroups.com...
>I asked a science prof if the moon will eventually move away or into
> the earth.
>
> He said neither, it will remain revolving the earth for eternity.
>
> Obviously after this insulting response I stopped taking him too
> serious.
>
>
> Barring some unforeseen circumstance, my belief leans to toward. This
> stems from several factors one being due to the fact that the duration
> the moon has chummed with the earth ... billions of years supposedly.
> I'm convinced that the escape possibilities are near zero by now.
>
>>From another thread, I'm outnumbered about four to one, but this is
> meaningless because as my science teacher, they were out to lunch. I'm
> sure some of these trolls will turn up here too.
>
> Obviously this is a good subject for (great) debate.
>
> Again this is ONLY a BELIEF. I'm not SURE. I don't profess to KNOW. So
> please keep the insults to a minimum. Bear in mind, I leave few
> insults unanswered. If this gets too ugly I will just restart another
> (clean) thread.
>
> Let's have a healthy debate.
>
> So what eventually happens to the moon-earth system? Try to convince
> me either way.
>
> Thanks,
> Steve
>
You have to start with the question "What forces are acting on this three
body problem that keeps the Moon from crashing into us tonight?" Then, we
can answer your question properly. Time to start thinking about parking the
Moon in a Lagrangian.
http://en.wikipedia.org/wiki/Lagrangian_point Will E.
sir.jp...@neuf.fr
> I asked a science prof if the moon will eventually move away or into
> the earth.
>
> He said neither, it will remain revolving the earth for eternity.
>
> Obviously after this insulting response I stopped taking him too
> serious.
>
> What say you, toward? away?
>
> Barring some unforeseen circumstance, my belief leans to toward. This
> stems from several factors one being due to the fact that the duration
> the moon has chummed with the earth ... billions of years supposedly.
> I'm convinced that the escape possibilities are near zero by now.
>
> >From another thread, I'm outnumbered about four to one, but this is
>
> meaningless because as my science teacher, they were out to lunch. I'm
> sure some of these trolls will turn up here too.
>
> Obviously this is a good subject for (great) debate.
>
> Again this is ONLY a BELIEF. I'm not SURE. I don't profess to KNOW. So
> please keep the insults to a minimum. Bear in mind, I leave few
> insults unanswered. If this gets too ugly I will just restart another
> (clean) thread.
>
> Let's have a healthy debate.
>
> So what eventually happens to the moon-earth system? Try to convince
> me either way.
>
> Thanks,
> Steve
It has already happened 3 times and all Orogenesis are consecutive to
Moon crash on this Earth !
( Caledonian, Hercynian & Alpine orogenesis 120 centuries ago )
The last time it was 11700 years ago indeed right in the middle of
the Pacific at the geometric centre of what is known now as the
Pacific Ring of Fire.
( when the Sun was in the Lion hence the Sphinx symbol made by the
Antediluvian meaning that anticipation of that event )
The present fraud masquerading as Geology is unable to understand
first the real environment of pressure into which we are immersed
... then where from we come and where we are heading to ...
Nor of course the origin of forces putting us in motion every minutes
since birth of Planet from the Sun a mere 170 millions years ago and
its consecutive drift on the plane of Ecliptics towards Pluto ever
since !
I have a simple suggestion to you so as to bring a permanent answer to
your well commendable questioning, next time don't ask one of those
Universilities brainwashed idiots masquerading as Geologist or Science
Teacher, ask instead the rocks in the fields, the strata's of
alternate Carbonate & Clay bedding, the folded bedding seen in the
apparent Relief !!! Ask a toad or a frog for a better answer as to why
they are such ... ask fossils why they are imbedded in vertical
positions in the 1000 m thick Red Sandstone of Scotland with all their
colours of scales & freshness of eyes maintained intact !
... of course for economic reasons learn the idiocies foraged down
your throat by the sycophants of that fraudulous alleged Geology
science , but don't trust them at all with their narrow views and
short minded answers LEARN BY YOURSELF AND NEVER TRUST ANY OF THAT
UNIVERSILITIES RABBLE
Indeed they are unable on some other fields to cure their very own
diseases ... the ones they pretend to cure in others !
With kindest regards
Sir Jean-Paul Turcaud
Exploration Geologist & Offshore Consultant
Mobile +33 650 171 464
Australia Mining Pioneer
Founder of the True Geology
http://www.tnet.com.au/~warrigal/grule.html
http://users.indigo.net.au/don/tel/index.html
http://members.iimetro.com.au/~hubbca/turcaud.htm
http://www.abc.net.au/rn/talks/bbing/stories/s28534.htm
ray
>
>
Come ON, Turd, you're obviously getting desperate.
Icy Wiener
soup
> ( when the Sun was in the Lion hence the Sphinx symbol made by the
> Antediluvian meaning that anticipation of that event )
WTF?
sir.jp...@neuf.fr
> Gezus, Turdy you really need to switch brands
I was pleased to read the different answers from the gutless
australian turds ...it can be noted indeed that the shithead
Antipodean beasts remain carefully anonymous, expound their ignorance
shamelessly, repeat like the toads they are what has been foraged in
their thick degenerate morons 's skulls by the official bodies of
deceit & official Geology teaching institutions such as Curtin,
Monash, Sydney Universities & the ANU in Canberra. In brief they
demonstrate to the world at large what a congregation of hopeless
froggish australian Toads they are, and what Hell it is to go to such
debased country & what shame it is to have achieved some exceptional
feats of geological intelligence, courage, abnegation, sacrifice
indeed for such australian manure to benefit in centuries to
come !!!
. Never any of those fatly paid Universilities parasites, by the
way, has ever questioned a single one of their infinitively hopeless
theories whatever ! Indeed according to their Sci000nce the Earth was
'Created' by accretion of banging meteorites following that famous Big
Bang
NOTE
( ou Gros Boum ! ... oh la la, Quelle Grande Intelligence Révélée en
cela par les Bètes Intellectuelles ! )
( of what great & amazing discoveriezzz are able all those
Intellectual Beasts rooming the campus & paddocks of Oxford, Caltech,
ze MITH or the ANU ... I am flabbergasted !!!)
Any child of 4 knowns that explosion following firing is not going to
put together material blasted away in the process ... to bring it back
together again compression factors SUPERIOR to the original
environment are required ... but the Universilities brainwashed beasts
making profession indeed of high learning of course superbly ignore
this !
All through my personal scrutiny of official Geology theories* , I
have discovered that not only all were not only false without
exception & moreover misleading, but further infantile attempts by
primitive & crude minds to adapt what the blinded fools were seeing to
reach interpretation of past events having modelled Relief ... What I
have discovered in fact are not so much the overall fraud related to
that ridiculous circus of Intellectual Geology Clowns indeed, but the
fact that the same childish inductive or deductive mechanical logic or
say dishonest sophisms were the basis of all Certitudes of that
present alleged knowledge on our History & the History of our Solar
System ... further still I have found the same flaws in awareness &
reasoning in all the other Intellectual beasts making profession of
knowledge in other fields like chemistry, palaeontology, archaeology,
medecine etc
The noted beliefs of all religious sects is a good indicator of the
state of degeneracy & easy mind programming of the Human Race ....
just to realise that such Human Animal 's immediate response to any of
his fears & questioning is bleating prayers is in this a strong
confirmation that the COUNTRY OF THE BLIND well described by H.G Wells
in that short novel of that name, is a perfect analogy of a state of
affairs extending to the whole world indeed !
What are the reasons ? ...what are the causes ? ... what are the
remedies ?
It is not opportune to discuss this now .... indeed some good &
appropriate chastiment are required before that .
* Information or Geological teaching is made of 3 elements :
1) Sur-information of redundant material comforting official
beuuliefs ,
2) Dis-information especially by play on word like by ex: Erosion. Of
course there is erosion of carbonates, oxides by water, and as well
secondary erosion of unconsolidated sedimentations ... but the
Original erosion of intrusive or extrusive rocks structures by water
since Water does not dissolve silicates ....& certainly not under
current pressure & temperature systems, and the millionzzz of yearzzz
are no operative factors non obstant the affirmation of the
Universilities Intellectual beasts TIME DOES NOT ERODE.
3) Non-information on any things infirming the official blind fools '
established theories, like by ex. the Glaciations theories where the
mere PRESSURE POINT OF FUSION OF ICE prevents indeed any pretension of
having such alleged erosion of rock formation by such weak tools ! Why
not use wrenches make of butter & ice scream .... it would lead any
fitter or tradesman to laugh his head off at the jokes ... but nothing
of the kind would move the Oxford, Curtin or ANU sci000ntific beasts
from their certitudes indeed !
Nom de Dieu... mais c'est pas possible !!!
pete
This is a great explanation except for not being right. Yes, the
moon is expanding its orbit. Yes, tidal interactions are lossy
and generate heat; yes, the energy of a spinning object
is given by 1/2Iw^2. However, I is computed as <sum>[m(i)r^2(i)]
for all the component bits m(i) of the spinning mass. For a point
mass (good enough approx for a 3km object at a 250km radius), this
reduces to mr^2, so r is in the computation, so you can't treat
I as a constant. And in fact, the total energy for an orbitting
object is traditionally regarded as negative, being zero at infinite
distance, ie no gravitational interaction, and getting lower as
it gets closer. Thus the energy becomes greater with larger orbit,
which is what you would expect, as you have to continually add
energy to an object to get it to move farther away from its primary.
Otherwise, things would immediately fly away, as objects seek
a state of minimum energy.
Ok, so if tidal interactions are frittering away energy, how can
the moon be acquiring energy and moving further away? Simple: these
two conditions are not incompatible. The oceans on the earth can
tidally couple to the moon, basically spinning it up. The energy
comes from the vast available energy from the earth's spin, which
is slightly declining as it gives spin away to the moon and also
heat away in tidal strain.
--
==========================================================================
vincent@triumf[munge].ca Pete Vincent
Disclaimer: all I know I learned from reading Usenet.
Timberwoof
wrote:
So in short, as the tides slow the Earth's rotation, that increases the
moon's orbital period.
--
Timberwoof <me at timberwoof dot com> http://www.timberwoof.com
"Like this cup," the master daid, "you are full of your own opinions and
speculations. How can I show you anything unless you first empty your cup?"
Alan Johnson
He means he doesn't know!
Jo Schaper
> On Feb 8, 7:31 pm, Jo Schaper <jospamnotschaper34@5socket78dot9net>
> wrote:
>> effac...@hotmail.com wrote:
>
>> Well, barring a collision with an asteroid, themoonprobably will
>> continue moving away (as it is currently by a few cm a year.)
>
> supporting that the moon will continue moving away. I mean forty years
> of data is completely useless. Even 1000 years of data could well be
> inside a cyclical eliptical movement.
>
> I'd love to have one billion years of distance data. Then were
> talking. I think that data would show the moon is just in the outgoing
> part of a cycle inside a cycle inside a cycle .. just a guess of
> course.
>
People in hell want icewater, too.
>> It likely
>> will remain in orbit at some distance, until both it and earth are
>> crisped to cinders by the expanding sun, a couple of billion years in
>> the future. A couple of billion years is close enough to an eternity for
>> me.
>
> So the sun will expand first and swallow up both? Maybe. I think by
> that time the moon and earth will already be one clump .. perhaps
> millions of years in advance or perhaps only days or hours before that
> happens.
Scientists have 40 years of data that the moon is moving away.
Steve has no years of data that the moon is coming closer.
Scientists are wrong and Steve is right,
only because he says he thinks so.
There is no debate in that argument. 40 weighed against 0 tips the scale
in favor of the 40.
Phil Hays
> On Feb 8, 7:31 pm, Jo Schaper <jospamnotschaper34@5socket78dot9net>
> wrote:
>> effac...@hotmail.com wrote:
>
>> Well, barring a collision with an asteroid, themoonprobably will
>> continue moving away (as it is currently by a few cm a year.)
>
> Problem with that theory is that there's no substantial data supporting
> that the moon will continue moving away. I mean forty years of data is
> completely useless. Even 1000 years of data could well be inside a
> cyclical eliptical movement. I'd love to have one billion years of
> distance data.
Oh? Tidal sediments from a billion years ago called "tidal rhythmites"
cover that range. Quick google search finds me this:
http://www.sciencenews.org/pages/sn_arch/7_6_96/fob1.htm
> Then were talking. I think that data would show the moon is just in the
> outgoing part of a cycle inside a cycle inside a cycle .. just a guess
> of course.
Why do you suspect a cycle? Tidal friction is lossy, not lossless.
--
Phil Hays
Ian Smith
a planet without being broken up by the planet's gravitational field, i.e.
tidal forces. Look it up on Google.
Peter Webb
"Ian Smith" <ianin...@btinternet.naespam.com> wrote in message
news:11711695...@proxy00.news.clara.net...
> The Roche Limit. The closest distance at which a natural satellite can
> orbit a planet without being broken up by the planet's gravitational
> field, i.e. tidal forces. Look it up on Google.
>
Nothing to do with it.
The moon is moving further away from the earth, as the earth transfers
angular momentum (and energy) to the moon via tidal forces. The earth-moon
distance increases, as does the length of a day.
Its pretyy easy to work out exactly where it will end up - the final
position is where the earth and moon keep the same face to each other (ie
where a day equals a month). This is easy to calculate through conservation
of angular momentum, this allows the final distance from the earth to the
moon to be calculated, though working out how long it takes is way harder.
Androcles
"Peter Webb" <webbfamily...@optusnet.com.au> wrote in message news:45cea46c$0$16555$afc3...@news.optusnet.com.au...
It's pretty easy to guess where it will end up, and your guess isn't
a good one.
http://antwrp.gsfc.nasa.gov/apod/ap051113.html
This is easy to calculate through conservation of angular momentum,
this allows the final distance from the earth to the moon to be
calculated, though working out the eccentricity of the orbit is
beyond your guess.
Enkidu
>
> What keeps the moon away from the earth other than centripital force?
>
A centripetal force acts *towards* the centre. A centrifugal force acts
away from the centre.
Actually, the force that stops the moon from flying away is gravity. If
gravity was to be suddenly abolished the moon would travel away from the
earth along a tangent to its orbit. It is only gravity pulling it
towards the earth that causes it to swing around in an orbit. There is
no force keeping the moon away from the earth.
Cheers,
Cliff
--
Have you ever noticed that if something is advertised as 'amusing' or
'hilarious', it usually isn't?
Androcles
"Steve" <philh...@aol.com> wrote in message news:1171185075.0...@j27g2000cwj.googlegroups.com...
> On Feb 10, 11:11 pm, "Androcles" <Engin...@hogwarts.physics.co.uk>
> wrote:
>> "Peter Webb" <webbfamily-diespam...@optusnet.com.au> wrote in messagenews:45cea46c$0$16555$afc3...@news.optusnet.com.au...
>
>> It's pretty easy to guess where it will end up, and your guess isn't
>> a good one.
>> http://antwrp.gsfc.nasa.gov/apod/ap051113.html
>
> moon swayed like that. It's not as static as I thought. Interesting.
> Thanks!
You can even measure the major/minor axis ratio and compute
eccentricity from that if you notice the change in apparent size.
>> This is easy to calculate through conservation of angular momentum,
>> this allows the final distance from the earth to the moon to be
>> calculated, though working out the eccentricity of the orbit is
>> beyond your guess.- Hide quoted text -
>>
>> - Show quoted text -
>
> And your guess is?
Sorry, I never guess. Anyway, who is going to be around long
enough to verify it?
But don't you find it curious that all planetary orbits are nearly
circular rather than highly elliptical as a comet's?
(Except Pluto/Charon of course, but that is probably a captured
comet anyway.)
The inference is the solar system as a whole has had enough time
to settle down, the most eccentric planet being Mercury and
that is advancing its longitude of perihelion as Jupiter, Venus
and the Earth drag it along. Mars is too small and too far away
to have as much effect as the other three.
http://faculty.ifmo.ru/butikov/Projects/Collection1.html
Example 5 gives a clear indication of the result, example 3
what could happen with a spacecraft in Lunar orbit, Example 9
gives you Shoemaker Levy.
Example 10 is fascinating because it appears at first to be stable.
Collection3 is mind boggling.
Peter Webb
"Steve" <philh...@aol.com> wrote in message
news:1171185075.0...@j27g2000cwj.googlegroups.com...
> On Feb 10, 11:11 pm, "Androcles" <Engin...@hogwarts.physics.co.uk>
> wrote:
>> "Peter Webb" <webbfamily-diespam...@optusnet.com.au> wrote in
>> messagenews:45cea46c$0$16555$afc3...@news.optusnet.com.au...
>
>> a good one.
>> http://antwrp.gsfc.nasa.gov/apod/ap051113.html
>
> moon swayed like that. It's not as static as I thought. Interesting.
> Thanks!
>
Ha ha. The wobbling you think you see is a result of the inclination of the
earth's axis affecting the time lapse photography. The effect is due to the
camera moving, not the earth.
>> This is easy to calculate through conservation of angular momentum,
>> this allows the final distance from the earth to the moon to be
>> calculated, though working out the eccentricity of the orbit is
>> beyond your guess.- Hide quoted text -
It won't be particularly eccentric. Its current very slight eccentricity
will be smoothed out as the earth-moon distance increases.
>>
>> - Show quoted text -
>
> And your guess is?
>
> Thanks,
> Steve
>
The moon is retreating from the earth at about 3.8 cms per year (see
http://en.wikipedia.org/wiki/Lunar_distance_%28astronomy%29
As I said, its easy to work out the distance it will eventually reach
through conservation of angular momentum. With the benefit of the LIDAR
experiments, the time frames can be estimated but not precisiely defined
(unlike the terminal distance, which can be easily calculated).
Peter Webb
> On Feb 10, 9:06 pm, "Peter Webb" <webbfamily-
> diespam...@optusnet.com.au> wrote:
>> "Ian Smith" <ianinho...@btinternet.naespam.com> wrote in message
>
>>> The Roche Limit. The closest distance at which a natural satellite can
>>> orbit a planet without being broken up by the planet's gravitational
>>> field, i.e. tidal forces. Look it up on Google.
>
>> Nothing to do with it.
>
>
>> The moon is moving further away from the earth, as the earth transfers
>> angular momentum (and energy) to the moon via tidal forces.
>
>
>>The earth-moon distance increases, as does the length of a day.
>>
>> Its pretyy easy to work out exactly where it will end up - the final
>> position is where the earth and moon keep the same face to each other (ie
>> where a day equals a month). This is easy to calculate through
>> conservation
>> of angular momentum, this allows the final distance from the earth to the
>> moon to be calculated, though working out how long it takes is way
>> harder.
>
> face the moon permanently. How does this lessen earth's gravitational
> force? Gravity is a function of rotation? A gyro effect won't play a
> role since the moon is not part of the earth's gyro equation.
>
> So by this theory, if a huge star spins fast enough it loses most all
> of its gravitational force acting on neighboring bodies. That doesn't
> compute.
>
> Thanks,
> Steve
>
You have badly mangled some physics concepts.
Why don't you first satisfy yourself that the moon is actually retreating
from the earth, for example by Googling Earth-Moon distance or LIDAR?
Once we have established that it is an astronomical fact that the moon's
orbital distance is increasing (and has done so since it was formed), we can
discuss the physics of why it happens, which are reasonably straightforward.
Ian Smith
"Peter Webb" (webbfamily...@optusnet.com.au) shared this wisdom:
I cited the Roche Limit in the sense that the moon can't notionally collide
with the Earth as a result of orbit decay; at least in 1 piece it can't. I'm
aware that it is, in reality, moving away.
Androcles
"Peter Webb" <webbfamily...@optusnet.com.au> wrote in message news:45ceee21$0$16557$afc3...@news.optusnet.com.au...
>
> "Steve" <philh...@aol.com> wrote in message
> news:1171185075.0...@j27g2000cwj.googlegroups.com...
>> On Feb 10, 11:11 pm, "Androcles" <Engin...@hogwarts.physics.co.uk>
>> wrote:
>>> "Peter Webb" <webbfamily-diespam...@optusnet.com.au> wrote in
>>> messagenews:45cea46c$0$16555$afc3...@news.optusnet.com.au...
>>
>>> It's pretty easy to guess where it will end up, and your guess isn't
>>> a good one.
>>> http://antwrp.gsfc.nasa.gov/apod/ap051113.html
>>
>> Nice time lapse shots of the entire moon day cycle. Didn't know the
>> moon swayed like that. It's not as static as I thought. Interesting.
>> Thanks!
>>
>
> Ha ha. The wobbling you think you see is a result of the inclination of the
> earth's axis affecting the time lapse photography. The effect is due to the
> camera moving, not the earth.
Seeing more than 180 degrees of the Moon's surface is caused by
the Earth's axis?
>
>>> This is easy to calculate through conservation of angular momentum,
>>> this allows the final distance from the earth to the moon to be
>>> calculated, though working out the eccentricity of the orbit is
>>> beyond your guess.- Hide quoted text -
>
> It won't be particularly eccentric. Its current very slight eccentricity
> will be smoothed out as the earth-moon distance increases.
What is its current slight eccentricity?
>
>
>>>
>>> - Show quoted text -
>>
>> And your guess is?
>>
>> Thanks,
>> Steve
>>
>
> The moon is retreating from the earth at about 3.8 cms per year (see
> http://en.wikipedia.org/wiki/Lunar_distance_%28astronomy%29
>
> As I said, its easy to work out the distance it will eventually reach
> through conservation of angular momentum.
Well, it is so easy, what is it?
> With the benefit of the LIDAR
> experiments, the time frames can be estimated but not precisiely defined
> (unlike the terminal distance, which can be easily calculated).
>
Yeah yeah... what is this easy calculation you can do through conservation
of angular momentum, as you said you can?
John Kepler
> 40 years is completely meaningless.
Based on what laws of orbital mechanics? Let me guess....you had a hard
time with improper fractions back in 4th grade?
And as I also said, not even 1000
> years of distance data is to solve where the moon will be billions of
> years from now.
Scooter....how long do you feel you need to observe a trajectory before you
can mathmatically calculate and predict said trajectory?
We'd need much! much! much! more movement data before
> anyone could sanely use as an argument one way or the other.
Again....based on WHAT flavor of physics and mathmatics?
>
> Jo what would you say if the moon was currently in an inward portion
> of an unseen eliptical cycle?
Huh....."unseen eliptical cycle"????? Scooter, we've LANDED on the
sucker....unless you can slide a chunk of Black Hole by it to perturb it's
orbit....it's movement is as predictable as, quite literally, a falling
rock! I'd say you spent too much of your apparently limited educational
time in the "Fuzzy Subjects"....(poli-sci...communications....english lit?).
FWIW, based on my ancient ancestors work....you only need TWO (three's
better) accurate observations to accurately predict the orbital mechanics of
a given body. "40 years" of observation is MORE than enough to precisely
predict the Moon's orbit 40 years from now, 40,000 years from now, 40
million years from now to long beyond the potential life-span of the inner
solar system....or the reverse! The math not only predicts the future....it
recapitulates the past. But since you already sound like one of those
"flat-earth" non-mathmatic, "Let's 'discuss' this!" types.....you aren't
going to accept anything that can't be observed via Prof Peabody's Way-Back
Machine, so I've just wasted my time, and will predicatbly have to deal with
your adolescent ranting in response via your attempts to "debate" rather
than "calculate"! My BEST answer to you is to study Kepler's three
laws....and DO THE MATH!
>
> But thanks for the debate. Love it.
Figures you'd bring up the term "debate"....like yammering and "concensus"
is a substitute for calculation! Scooter....if it can't be expressed in
numbers, it's an opinion, not a fact...and all you've brought to this is
opinion to refute physics.
I await your "righteously indignant", and probably profanity-filled attempt
at flaming! To quote another scientist confronting a "debater", "It still
moves!"
John
"The principal difference between genius and stupidity is that genius has
limits!" A. Einstein
Odysseus
"Steve" <philh...@aol.com> wrote:
<snip>
>
> So the earth is rotating slower and slower and may eventually stop and
> face the moon permanently. How does this lessen earth's gravitational
> force? Gravity is a function of rotation?
Not at all. According to Newton's theory, gravity is a function of mass
and distance only; the configuration and motions of a body have no
effect on its gravity. The only way to generally "lessen earth's
gravitational force" would be to reduce its mass.
--
Odysseus
Peter Webb
Yeah yeah... what is this easy calculation you can do through conservation
of angular momentum, as you said you can?
**********************
Do you first agree that the moon is actually moving away from the earth?
And that there will be an orbit it asymptotically approaches? (Equivalently,
whatever it is that is causing its orbital distance to increase will not
impart enough energy for it to escape the earth entirely?).
If the answers to both of these are yes, then:
The Earth has some angular momentum that could be calculated. I won't, as
the density of the earth varies with depth, so its a tedious arithmetic
calculation. Lets call it E.
So has the moon; this is easier as the major contributor to its angular
mamoentum is its movement around the earth. Lets call it M. Again, easily
calculated.
So the current total angular momentum is M + E.
When the two are in lock step and showing the same face towards each other
at all times, the total angular momentum will still be M + E (as it is
conserved).
As they are showing the same faces to each other, the period of rotation of
each will be the same. Let the common point they rotate about be "d_e" from
the earth and "d_m" from the moon. Let m_e be the mass of the earth and m_m
the mass of the moon.
Then:
(d_e)^2 * m_e + (d_m)^2 * m_m + some components related to the individual
rotational speeds of the earth and the moon= E+M
You get a second equation linking the rotational speeds to the distance
apart using conservation of energy and Newtons law of gravity.
Two equations and two unknowns (d_e and d_m). The earth-moon distance will
simply be d_e + d_m.
You should be able to find a link where somebody has dones the calculation
... its absolutely straightforward, but tedious to compute accurately
without knowing the approximate density at different depths of the earth.
For the theory, see http://en.wikipedia.org/wiki/Angular_momentum or Google
"angular momentum".
Like many uses of Conservation theories, this tells you what will happen,
but not how long it will take - the whole point about conserved quantities
is that they don't vary over time ...
Androcles
"Odysseus" <odysseu...@yahoo-dot.com> wrote in message news:odysseus1479-at-53...@news.telus.net...
> In article <1171184419....@l53g2000cwa.googlegroups.com>,
> "Steve" <philh...@aol.com> wrote:
>
> <snip>
>>
>> So the earth is rotating slower and slower and may eventually stop and
>> face the moon permanently. How does this lessen earth's gravitational
>> force? Gravity is a function of rotation?
>
> Not at all.
The rise and fall of the tides are a function of rotation (according to
Newton's theory).
Androcles
"Peter Webb" <webbfamily...@optusnet.com.au> wrote in message news:45cf2eed$0$16554$afc3...@news.optusnet.com.au...
>
>
> Yeah yeah... what is this easy calculation you can do through conservation
> of angular momentum, as you said you can?
>
> **********************
>
> Do you first agree that the moon is actually moving away from the earth?
Hey fuckhead! I neither agree nor disagree! Just prove you can do it.
Show us this easy calculation you can do through conservation of angular
momentum, as you said you can.
Jo Schaper
John Kepler wrote:
>
> Again....based on WHAT flavor of physics and mathmatics?
>> Jo what would you say if the moon was currently in an inward portion
>> of an unseen eliptical cycle?
I think Steve needs to put down his 1950s Amazing Science Stories
comics, and heck--even read some decent science fiction, if he can't
handle the real thing.
>
> Huh....."unseen eliptical cycle"????? Scooter, we've LANDED on the
> sucker....unless you can slide a chunk of Black Hole by it to perturb it's
> orbit....it's movement is as predictable as, quite literally, a falling
> rock! I'd say you spent too much of your apparently limited educational
> time in the "Fuzzy Subjects"....(poli-sci...communications....english lit?).
No, John, Steve hasn't spent his time in communications nor English
lit-- if he had, he'd understand the art of persuasion by logical
statement, and the rhetorical varieties of supporting an argument, even
if numerical arguments are beyond him.
Instead, he careens about like a piece of space debris, occasionally
colliding with objects in reasonable orbits, making sparks and going
"Whee!" I haven't quite understood why anyone would seek attention by
making nonsensical propositions, and then poorly defending them. It's
like the poor soul jumping up and down and screaming, "I'm not nuts!"
for hours on end. In this case, the medium (and the format)is truly the
message, not what is putatively being said.
Timberwoof
> On Feb 10, 9:06 pm, "Peter Webb" <webbfamily-
> diespam...@optusnet.com.au> wrote:
> > "Ian Smith" <ianinho...@btinternet.naespam.com> wrote in message
>
> >> The Roche Limit. The closest distance at which a natural satellite can
> >> orbit a planet without being broken up by the planet's gravitational
> >> field, i.e. tidal forces. Look it up on Google.
>
> > Nothing to do with it.
>
> Agree.
>
> > The moon is moving further away from the earth, as the earth transfers
> > angular momentum (and energy) to the moon via tidal forces.
>
> We're back to tides again?
>
> >The earth-moon distance increases, as does the length of a day.
> >
> > Its pretyy easy to work out exactly where it will end up - the final
> > position is where the earth and moon keep the same face to each other (ie
> > where a day equals a month). This is easy to calculate through conservation
> > of angular momentum, this allows the final distance from the earth to the
> > moon to be calculated, though working out how long it takes is way harder.
>
> So the earth is rotating slower and slower and may eventually stop and
> face the moon permanently. How does this lessen earth's gravitational
> force? Gravity is a function of rotation? A gyro effect won't play a
> role since the moon is not part of the earth's gyro equation.
>
> So by this theory, if a huge star spins fast enough it loses most all
> of its gravitational force acting on neighboring bodies. That doesn't
> compute.
Mostly because your model of orbital mechanics is too simple, and
although the mechanisms have been explained to you, you ignored it
because you don't want to understand.
Ignorance can be cured. Stupidity is another matter.
--
Timberwoof <me at timberwoof dot com> http://www.timberwoof.com
"Like this cup," the master daid, "you are full of your own opinions and
speculations. How can I show you anything unless you first empty your cup?"
Timberwoof
Odysseus <odysseu...@yahoo-dot.com> wrote:
What you're both missing is that the moon's orbit gets larger not
through a lessening of the Earth's mass but a different mechanism. Pay
attention now!
As the moon orbits the Earth, it drags water around with it, causing
tides and slowing the Earth's rotation. So what happens to the energy of
that rotation? Some of it is converted to heat through viscous friction.
But most of it is transferred to the moon's orbit, which gets bigger as
a result.
Timberwoof
"Steve" <philh...@aol.com> wrote:
> On Feb 10, 8:51 pm, "Ian Smith" <ianinho...@btinternet.naespam.com>
> Ok let's say the moon breaks up first before it hits the earth. It
> still hits.
> What keeps the moon away from the earth other than centripital force?
That's what orbits do. Do you understand how orbits work?
Timberwoof
Enkidu <enkid...@com.cliffp.com> wrote:
> Steve wrote:
> >
> > What keeps the moon away from the earth other than centripital force?
> >
> A centripetal force acts *towards* the centre. A centrifugal force acts
> away from the centre.
>
> Actually, the force that stops the moon from flying away is gravity. If
> gravity was to be suddenly abolished the moon would travel away from the
> earth along a tangent to its orbit. It is only gravity pulling it
> towards the earth that causes it to swing around in an orbit. There is
> no force keeping the moon away from the earth.
Oh, boy, I can already see it coming: a failure to distinguish between
force and acceleration.
Timberwoof
"Steve" <philh...@aol.com> wrote:
> On Feb 10, 6:56 pm, Jo Schaper <jospamnotschaper34@5socket78dot9net>
> wrote:
> > Scientists have 40 years of data that the moon is moving away.
> > Steve has no years of data that the moon is coming closer.
> >
> > Scientists are wrong and Steve is right,
> > only because he says he thinks so.
> >
> > There is no debate in that argument. 40 weighed against 0 tips the scale
> > in favor of the 40.- Hide quoted text -
>
> What percent of 4 Billion years (the time needed to observe all the
> movement cycles) is 40 years?
> I rest my case (again).
That's not much of an argument. You claim there's some kind of cycle
going on. People who know more than you say it's a simple monotonic
process. Maybe they're onto something you're not?
> 40 years is completely meaningless. And as I also said, not even 1000
> years of distance data is to solve where the moon will be billions of
> years from now. We'd need much! much! much! more movement data before
> anyone could sanely use as an argument one way or the other.
And where did you get your PhD in astronomy? Your MS? Your BS? Did you
even take a freshman honors astronomy course?
> Jo what would you say if the moon was currently in an inward portion
> of an unseen eliptical cycle?
> And where have I even hinted that scientists were wrong with distance
> data?
>
> But thanks for the debate. Love it.
It's not much of a debate, Steve. Most of your arguments are from
incredulity and ignorance. You put up this wall of misunderstanding and
refuse to pay attention to what people are telling you who are more
knowledgeable than you. If you'd stop to ask questions now and then
instead of screaming at the top of your lungs that it can't be that way,
then it might be a real discussion.
Timberwoof
Jo Schaper <jospamnotschaper34@5socket78dot9net> wrote:
There was a documentary on PBS a while back titled something like "The
Biography of E=MC^2". The first chapter was on the M part of the
equation; it focused on the work of Lavoisier. The whole tedious
business about conservation of matter came from a French tax collector.
Lavoisier had pissed off some revolutionary who had made some nutsy
proposal about the nature of matter. Lavoisier told him that if he
didn't have numbers, he had only opinion, not science. The guy reported
him to the revolutionary authorities for doing undemocratic science and
Lavoisier got the chop.
Timberwoof
"Peter Webb" <webbfamily...@optusnet.com.au> wrote:
> "Steve" <philh...@aol.com> wrote in message
> news:1171185075.0...@j27g2000cwj.googlegroups.com...
> > On Feb 10, 11:11 pm, "Androcles" <Engin...@hogwarts.physics.co.uk>
> > wrote:
> >> "Peter Webb" <webbfamily-diespam...@optusnet.com.au> wrote in
> >> messagenews:45cea46c$0$16555$afc3...@news.optusnet.com.au...
> >
> >> It's pretty easy to guess where it will end up, and your guess isn't
> >> a good one.
> >> http://antwrp.gsfc.nasa.gov/apod/ap051113.html
> >
> > Nice time lapse shots of the entire moon day cycle. Didn't know the
> > moon swayed like that. It's not as static as I thought. Interesting.
> > Thanks!
> >
>
> Ha ha. The wobbling you think you see is a result of the inclination of the
> earth's axis affecting the time lapse photography. The effect is due to the
> camera moving, not the earth.
No, I don't think so. The wobble happens with lunar cycles, not with an
Earth annual cycle. It has more to do with the inclination of the moon's
orbit around the earth and, as the text says, with the moon's elliptical
orbit.
> >> This is easy to calculate through conservation of angular momentum,
> >> this allows the final distance from the earth to the moon to be
> >> calculated, though working out the eccentricity of the orbit is
> >> beyond your guess.- Hide quoted text -
>
> It won't be particularly eccentric. Its current very slight eccentricity
> will be smoothed out as the earth-moon distance increases.
>
>
> >>
> >> - Show quoted text -
> >
> > And your guess is?
> >
> > Thanks,
> > Steve
> >
>
> The moon is retreating from the earth at about 3.8 cms per year (see
> http://en.wikipedia.org/wiki/Lunar_distance_%28astronomy%29
>
> As I said, its easy to work out the distance it will eventually reach
> through conservation of angular momentum. With the benefit of the LIDAR
> experiments, the time frames can be estimated but not precisiely defined
> (unlike the terminal distance, which can be easily calculated).
--
Landy
> So far, people seem fixated on water. What percent of the earth is
> water? 5%?
> (Not wrt the surface only). What percent of the earth is molten lava?
> 90%?
Where on earth did you get that idea?
The mantle is NOT molten.
cheers
Bill
Peter Webb
"Androcles" <Engi...@hogwarts.physics.co.uk> wrote in message
news:lqHzh.302909$MO2.1...@fe3.news.blueyonder.co.uk...
******************
I showed how in the bit that was snipped.
What part don't you understand?
The Conservation of angular momentum is explained here:
http://www.astronomynotes.com/angmom/s1.htm
(It uses the earth-moon system as an example).
The calculation has been done at:
http://users.zoominternet.net/~matto/M.C.A.S/notes_tides.htm
They provide an estimate of the stable, terminal orbit as 50% greater than
today, or a bit under 600,000 kms.
At http://arxiv.org/pdf/astro-ph/0107024
They use the equations for conservation of angular momentum to calculate:
"the resonant condition between the Earth and Moon is such that the length
of the month is 50.7 days, a distance of 581098 km"
http://www.astronomynotes.com/gravappl/s10.htm
Calculates much the same thing - a resonant orbit with a period of 47 days
(this can be used to calculate the distnce using Newton's Law of Gravity).
There are plenty of others.
The interesting (for me) part of this is that the rate of recession is very
hard to calculate (but easy to measure with lasers) because it depends upon
the degree of tidal coupling of the earth-moon system, which in turn depends
upon the ocean depth and continental topography. For example, if the earth
had no water, then the rate of recession would be less than half the present
value (a few centimetres a year), as the major component of the tidal lock
is ocean tides - deformation of the solid structure of the earth is less
than half the tidal effect. With lots more water, or re-arranged continents
such that tides were not disrupted by large land masses, the rate would
increase. The law of conservation of angular momentum can't be used to
directly calculate the time until resonance (a stable orbit) occurs, but it
can directly calculate the distance at which this occurs (581,098 kms) and
the final shared earth/moon day length (47 days). This is independent of any
topographic considerations, which only affect the time it takes for the
synchronisation to occur.
The behaviour of the earth-moon system is somewhat paradoxical until tidal
forces are introduced. You would expect the moon to "fall towards" the
earth; as the concept of tidal locking demonstrates (and the LIDAR
experiment measured) the exact opposite is occuring
I haven't googled for it, but doubtless tightly coupled binary star systems
would have similar tidal effects which would increase their period.
http://en.wikipedia.org/wiki/Tidal_acceleration
has a pretty good non-mathematical treatment; they state that the earth-moon
system will not achive the stable 580,000 kms orbit before the Sun turns
into a red giant. This is partially predicated on the oceans boiling off as
the Sun expands, decreasing tidal locking.
If you actually want to understand this a bit better - and my explanations
and links aren't enough - feel free to ask specific questions. I studied
this in either my Uni physics or astronomy classes (or maybe both), but as
long as you understand the Law of Conservation of Angular Momentum it is
really just high school maths (apart from calculating the current angular
momentum of the earth, which uses a tiny bit of calculus, but you can skip
that).
Peter Webb
>> >> What keeps the moon away from the earth other than centripital force?
>
>
> Steve
>
Doesn't help. Its the concept that's flawed, not your terminology.
The fact that the earth-moon distance is increasing (and will continue to
increase until a "phase-locked" or "resonant" orbit is achieved) has been
known theoretically for 200 years and has been measured for the last 35
years.
What astounds me is that the majority of the posters (except timberwolf)
seem unaware of basic physics that has been known for 200 years. A simple
google of "earth moon distance" shows dozens or hundreds of sites that use
the conservation of angular momentum and tidal coupling to explain the
increasing earth-moon distance, and calcluate the terminal orbit (I provided
some in another post). Why people would bother posting their own bullshit
theories about the moon falling into the earth when the physics behind this
is relatively simple and easily found on the net is a mystery to me. If you
don't know much about physics - and I suspect from your use of the term
"centrifugal force" that you don't - wouldn't it make a little sense to
Google the question before posting your own hypothesis?
Peter Webb
> On Feb 11, 2:27 am, "Peter Webb" <webbfamily-> Cite. Funny how you say this without one reference! Credibility hit or
> what?
>
You want a web reference that discusses the terminology in your post?
You said:
"So the earth is rotating slower and slower and may eventually stop and
face the moon permanently. How does this lessen earth's gravitational
force? Gravity is a function of rotation? A gyro effect won't play a
role since the moon is not part of the earth's gyro equation.
So by this theory, if a huge star spins fast enough it loses most all
of its gravitational force acting on neighboring bodies. That doesn't
compute."
What do you mean by "gravitational force"? You mean the earth's
gravitational force on the moon? Yes, this decreases because of Newton's law
of gravitation.
F = G*M*E/r^2
(where M and E are the masses of the earth and moon).
Yes, this decreases because r increases.
>>
>> Why don't you first satisfy yourself that the moon is actually retreating
>> from the earth, for example by Googling Earth-Moon distance or LIDAR?
>
> Cite where I said this was a lie? If this is the case fine. I said
> this proves nothing because of the short time frame or did you
> conveniently leave that out of your mind too?
>
>>
>> Once we have established that it is an astronomical fact that the moon's
>> orbital distance is increasing (and has done so since it was formed), we
>> can
>> discuss the physics of why it happens, which are reasonably
>> straightforward.- Hide quoted text -
>
> I thought your entire premise was that indeed we've proven the moon is
> currently moving away?!
Yes.
> Wow maybe your just really tired (been there done that myself).
Huh?
>
> Steve
Peter Webb
> On Feb 11, 2:21 am, "Peter Webb" <webbfamily-
> diespam...@optusnet.com.au> wrote:
>> Ha ha. The wobbling you think you see is a result of the inclination of
>> the
>> earth's axis affecting the time lapse photography. The effect is due to
>> the
>> camera moving, not the earth.
>
> "twisting" around it's center of gravity. It's not perfectly static as
> I've read. Watch particularily the edges. The camera would have to
> move tens of thousands of miles to make the moon move that much it
> were completely frozen.
Which conveniently also happens to be distance through which a camera would
have to move to photograph the moon when it is on opposite sides of the
earth.
>>
>> >> This is easy to calculate through conservation of angular momentum,
>> >> this allows the final distance from the earth to the moon to be
>> >> calculated, though working out the eccentricity of the orbit is
>> >> beyond your guess.- Hide quoted text -
>>
>> It won't be particularly eccentric. Its current very slight eccentricity
>> will be smoothed out as the earth-moon distance increases.
>>
>>
>>
>> >> - Show quoted text -
>>
>> > And your guess is?
>>
>> > Thanks,
>> > Steve
>>
>> The moon is retreating from the earth at about 3.8 cms per year
>> (seehttp://en.wikipedia.org/wiki/Lunar_distance_%28astronomy%29
>
> The moon is CURRENTLY retreating. You've left out an extremely key
> word. You can't prove it will continue "retreating" anymore than I can
> prove it is in some long term elliptical sway (which is my theory) so
> this movement could be a very stale factor and not worthy of this
> discussion, but at least you're on topic. Thanks.
>
And has done so for billions of years, and will continue to do so for
billions of years to come. As a simple google of "earth-moon distance" will
very quickly verify. Its very basic physics.
>>
>> As I said, its easy to work out the distance it will eventually reach
>> through conservation of angular momentum. With the benefit of the LIDAR
>> experiments, the time frames can be estimated but not precisiely defined
>> (unlike the terminal distance, which can be easily calculated).
>
> Again I've asked what force(s) move the moon away other than
> centrifugal?
Tidal coupling.
> If there are none
There are. Tidal coupling.
> and the moon revolution is slowing,
If its revoution aroud the earth is slowing, then its orbit must be
increasing. Low earth sattelites go round the earth in a couple of hours,
the moon (much farther away) takes 28 days or so. Is this a typo, or is it
what you actually believe?
> this PROVES without a shadow of doubt my theory that the moon will
> slam into the earth, barring of course something unforeseen.
It just proves that:
1. You know nothing about physics, and
2. You are too stupid or lazy to google for a basic astronomical explanation
of what will happen, and why, which is the exact opposite of what you claim.
> Steve
>
>
Peter Webb
> On Feb 11, 3:32 am, "Ian Smith" <ianinho...@btinternet.naespam.com>
> wrote:
>
>> I cited the Roche Limit in the sense that the moon can't notionally
>> collide
>> with the Earth as a result of orbit decay; at least in 1 piece it can't.
>> I'm
>> aware that it is, in reality, moving away.- Hide quoted text -
>
> The Roche Limit is not a force acting outwardly on the ENTIRE moon
> right? It is only a force acting on objects as multiple gravity points
> collide and shift.
>
> This phonomenon exists, BUT (HUGE BUT) wonder how much of a factor it
> is on the moon (and earth since the moon is so massive in comparison)
> if/when the moon slams into the earth. Will the moon break up before
> it hits?
>
> Interesting as hell. Great chat. Thanks,
> Steve
>
One last time:
1. The moon is moving farther from the earth, not closer. This is predicted
by theory and confirmed by experiment.
2. Therefore the moon will never come within the Roche limit or fall into
the earth (barring some huge astronomical event such as another planet
hitting the moon).
Peter Webb
"Odysseus" <odysseu...@yahoo-dot.com> wrote in message
news:odysseus1479-at-53...@news.telus.net...
> In article <1171184419....@l53g2000cwa.googlegroups.com>,
> "Steve" <philh...@aol.com> wrote:
>
> <snip>
>>
>> face the moon permanently. How does this lessen earth's gravitational
>> force? Gravity is a function of rotation?
>
> and distance only; the configuration and motions of a body have no
> effect on its gravity. The only way to generally "lessen earth's
> gravitational force" would be to reduce its mass.
>
Or to increase the distance between the earth and moon, which is what has
been happening for billions of years.
Timberwoof
"Steve" <philh...@aol.com> wrote:
> On Feb 11, 3:32 am, "Ian Smith" <ianinho...@btinternet.naespam.com>
> wrote:
>
> > I cited the Roche Limit in the sense that the moon can't notionally collide
> > with the Earth as a result of orbit decay; at least in 1 piece it can't. I'm
> > aware that it is, in reality, moving away.- Hide quoted text -
>
> The Roche Limit is not a force acting outwardly on the ENTIRE moon
> right? It is only a force acting on objects as multiple gravity points
> collide and shift.
>
> This phonomenon exists, BUT (HUGE BUT) wonder how much of a factor it
> is on the moon (and earth since the moon is so massive in comparison)
> if/when the moon slams into the earth. Will the moon break up before
> it hits?
Someone who knows more about it than you could claculate it. What
happens when they get slowly close together depends on the mass,
density, and tensional strength of the Earth and Moon. Two marbles, for
instance, have a lot of tensional strength, and would not break up as
they approached each other. But binary stars can do that sort of thing,
and their breakup as they get too close has been obvserved and the
models have been refined to match the data. So it's not all that hard to
apply the model to the Earth and Moon.
> Interesting as hell. Great chat. Thanks,
> Steve
--
Peter Webb
> On Feb 11, 2:12 pm, "Peter Webb" <webbfamily-
> diespam...@optusnet.com.au> wrote:
>> "Androcles" <Engin...@hogwarts.physics.co.uk> wrote in message
>
>>Hey fuckhead! bla bla bla
>
> Petie here earns his position as troll. Fine accomplishment in a
> usenet group.
>
> Let me pass on a bit of advice to trolls like Pete, if you want others
> to think highly of you and make a serious valid convincing point
> stick, that's not a great start.
>
>
>
Interesting. This reply does not contain a single word that I posted, and
doesn't address any of the issues I raised.
Have you any problems with the explanantion I gave of the dynamics of the
earth-moon system, or any of the links I provided that explain it further?
Do you still believe that all astronomers and physicists who (universally)
accept that the moon is moving further away from the earth, has done so for
billions of years, and will continue to do so until the Sun goes red giant,
are all wrong, and you are right?
And you call me a troll?
Red Smoke
> On Feb 11, 1:28 pm, Timberwoof <timberwoof.s...@infernosoft.com>
> wrote:
> > bla bla bla
>
> Sorry to sound trollish myself, but I've ran into this TW dude before
> and frankly he ain't worth my time especially when it involves
> science. So sorry TW I didn't read past your first post .. and that
> rule stands.
>
> However, if someone else here finds TW's comments "on topic
> stimulating" please point them out for me. Thanks.
>
> Steve
>
>
>
Gravity is like a roulette without friction.
Just so long as the moon continues to go round that invisable boundary of
that track at a constant velocity, the earth is safe.
Peter Webb
"Timberwoof" <timberw...@infernosoft.com> wrote in message
news:timberwoof.spam-F2...@nnrp-virt.nntp.sonic.net...
> In article <1171232865.6...@a75g2000cwd.googlegroups.com>,
> "Steve" <philh...@aol.com> wrote:
>
>> On Feb 11, 3:32 am, "Ian Smith" <ianinho...@btinternet.naespam.com>
>> wrote:
>>
>> > I cited the Roche Limit in the sense that the moon can't notionally
>> > collide
>> > with the Earth as a result of orbit decay; at least in 1 piece it
>> > can't. I'm
>> > aware that it is, in reality, moving away.- Hide quoted text -
>>
>> The Roche Limit is not a force acting outwardly on the ENTIRE moon
>> right? It is only a force acting on objects as multiple gravity points
>> collide and shift.
>>
>> This phonomenon exists, BUT (HUGE BUT) wonder how much of a factor it
>> is on the moon (and earth since the moon is so massive in comparison)
>> if/when the moon slams into the earth. Will the moon break up before
>> it hits?
>
> Someone who knows more about it than you could claculate it. What
> happens when they get slowly close together depends on the mass,
> density, and tensional strength of the Earth and Moon. Two marbles, for
> instance, have a lot of tensional strength, and would not break up as
> they approached each other. But binary stars can do that sort of thing,
> and their breakup as they get too close has been obvserved and the
> models have been refined to match the data. So it's not all that hard to
> apply the model to the Earth and Moon.
Its easy to apply. Tensional strength is orders of magnitude too small to
hold together a body more than a few tens of kilometers across.
Its also bullshit, because tidal forces have moved the moon into higher and
higher orbits over billions of years, and will continue to do so.
Peter Webb
>I see this thread has turned into just another troll haven (as if I
> expected anything less). Actually I was shocked to see as many
> authentic posts. Luckily, thanks to the rest of you who provided some
> solid scientific points to ponder over it has helped me mull over my
> existing theory.
>
> THANKS !!
> STEVE
>
No problem. Glad the links and explanations helped.
> PS now you trolls have the floor. Go on ripping each other new holes.
> One guy you can count on that won't be here to read any of it out of
> sheer boredom will be moi. So piss off. Touche! I get the last laugh.
>
Phil Hays
> For one thing, cycles occur in nature all over the place. "For everything
> there is a season."
Yet there are also beginnings and ends. Not everything is a cycle.
> I'm not persuaded by tides playing much of a role (at least yet) in
> determining where the moon will be billions of years from now, but maybe
> after a thorough lecture.
You can bring a horse to water, but you can't make him drink.
--
Phil Hays
Timberwoof
"Steve" <philh...@aol.com> wrote:
> On Feb 11, 2:03 am, "Androcles" <Engin...@hogwarts.physics.co.uk>
> wrote:
> > "Steve" <philhend...@aol.com> wrote in
> > messagenews:1171185075.0...@j27g2000cwj.googlegroups.com...
> > > On Feb 10, 11:11 pm, "Androcles" <Engin...@hogwarts.physics.co.uk>
> > > wrote:
> > >> "Peter Webb" <webbfamily-diespam...@optusnet.com.au> wrote in
> > >> messagenews:45cea46c$0$16555$afc3...@news.optusnet.com.au...
> >
> > >> It's pretty easy to guess where it will end up, and your guess isn't
> > >> a good one.
> > >> http://antwrp.gsfc.nasa.gov/apod/ap051113.html
> >
> > > Nice time lapse shots of the entire moon day cycle. Didn't know the
> > > moon swayed like that. It's not as static as I thought. Interesting.
> > > Thanks!
> >
> > You can even measure the major/minor axis ratio and compute
> > eccentricity from that if you notice the change in apparent size.
> >
> > >> This is easy to calculate through conservation of angular momentum,
> > >> this allows the final distance from the earth to the moon to be
> > >> calculated, though working out the eccentricity of the orbit is
> > >> beyond your guess.- Hide quoted text -
> >
> > >> - Show quoted text -
> >
> > > And your guess is?
> >
> > Sorry, I never guess.
>
> Not even a theory?
>
> >Anyway, who is going to be around long
> > enough to verify it?
>
> No one, but we can still hypothesize with existing data .. at the very
> least establishing the odds to some degree. And so far factors in this
> discussion strongly favor my tendency to believe of an eventual single
> lump .. namely the slowing of the moon's revolution around the earth.
The problem with your hypothesis is that there's neither evidence nor
rigorous math to back it up.
> > But don't you find it curious that all planetary orbits are nearly
> > circular rather than highly elliptical as a comet's?
>
> Depends. Has anyone determined the amount deviation from pure
> circular?
Yes. Eccentricity is one of the fundamental defining properties of an
ellipse.
> The ellipses will "wear off" in time. A near circular path
> *suggests* that a great deal of time has ellapsed.
>
> The moon demonstrates is definite "wobble" in its day cycle alone. I
> bet there's more than one wobble If someone could go back hundreds of
> thousands of years, they could point out more wobbles. I'm suggesting
> the LATEST outward movement is merely one of these elliptical wobbles.
> Nothing far fetched.
Indeed, people have. The orbital parameters of the Earth and Moon are
very well known, both from current measurements and from ancient ones.
It may surprise you to learn that this whole "Age of Aquarius" thing is
a real astronomical event.
> > (Except Pluto/Charon of course, but that is probably a captured
> > comet anyway.)
>
> Is Pluto Charon?
No.
> > The inference is the solar system as a whole has had enough time
> > to settle down, the most eccentric planet being Mercury and
> > that is advancing its longitude of perihelion as Jupiter, Venus
> > and the Earth drag it along. Mars is too small and too far away
> > to have as much effect as the other three.
> > http://faculty.ifmo.ru/butikov/Projects/Collection1.html
> > Example 5 gives a clear indication of the result, example 3
> > what could happen with a spacecraft in Lunar orbit, Example 9
> > gives you Shoemaker Levy.
> > Example 10 is fascinating because it appears at first to be stable.
> >
> > Collection3 is mind boggling.
>
> Interesting.
Timberwoof
"Peter Webb" <webbfamily...@optusnet.com.au> wrote:
There's a common flaw in people which is to underestimate the difficulty
of a task one has little familiarity with. I'm most familiar with this
in motorcycling, where many a newbie thinks, what's so hard about this,
and ends up steering straight for a tree.
And so, someone who knows a little bit about Newton and marbles, but not
what has actually been observed about the Earth and Moon (and Sun and so
forth) does not know what all may be involved and is astounded that
anyone might come up with answers different from his own.
People hate to be told they don't know something and they like to
believe that their ideas are on par with those of degreed scientists.
(Remember Lavoisier...)
Androcles
"Peter Webb" <webbfamily...@optusnet.com.au> wrote in message news:45cf94d2$0$4753$afc3...@news.optusnet.com.au...
>
> "Androcles" <Engi...@hogwarts.physics.co.uk> wrote in message
> news:lqHzh.302909$MO2.1...@fe3.news.blueyonder.co.uk...
>
> "Peter Webb" <webbfamily...@optusnet.com.au> wrote in message
> news:45cf2eed$0$16554$afc3...@news.optusnet.com.au...
>>
>>
>> Yeah yeah... what is this easy calculation you can do through conservation
>> of angular momentum, as you said you can?
>>
>> **********************
>>
>> Do you first agree that the moon is actually moving away from the earth?
>
>
> Hey fuckhead! I neither agree nor disagree! Just prove you can do it.
> Show us this easy calculation you can do through conservation of angular
> momentum, as you said you can.
>
> ******************
>
> I showed how in the bit that was snipped.
>
> What part don't you understand?
The bit where you showed YOUR calculations and YOUR result.
I couldn't find it.
> They use
Never mind "they". Where are YOUR calculations?
What is this easy calculation you can do through conservation
of angular momentum, as you said you can?
[snip links]
> There are plenty of others.
Never mind "others". Where are YOUR calculations?
What is this easy calculation you can do through conservation
of angular momentum, as you said you can?
> The interesting (for me) part of this is that the rate of recession is very
> hard to calculate
You said it was easy. Make up your fuckheaded mind!
What is this easy calculation you can do through conservation
of angular momentum, as you said you can?
> If you actually want to understand this a bit better
I understand it very well, what I want to see is your easy calculations.
What is this easy calculation you can do through conservation
Androcles
"Steve" <philh...@aol.com> wrote in message news:1171235814....@v45g2000cwv.googlegroups.com...
> On Feb 11, 2:12 pm, "Peter Webb" <webbfamily-
> diespam...@optusnet.com.au> wrote:
>> "Androcles" <Engin...@hogwarts.physics.co.uk> wrote in message
>
>>Hey fuckhead! bla bla bla
>
> Petie here earns his position as troll. Fine accomplishment in a
> usenet group.
>
> Let me pass on a bit of advice to trolls like Pete, if you want others
> to think highly of you and make a serious valid convincing point
> stick, that's not a great start.
>
for my personal amusement. I noticed Webb relies heavily on wackypedia,
the encyclopaedia any troll can write and no sane person can correct.
Robert Peffers.
"Steve" <philh...@aol.com> wrote in message
news:1171235814....@v45g2000cwv.googlegroups.com...
> On Feb 11, 2:12 pm, "Peter Webb" <webbfamily-
> diespam...@optusnet.com.au> wrote:
>> "Androcles" <Engin...@hogwarts.physics.co.uk> wrote in message
>
>>Hey fuckhead! bla bla bla
snip
the encyclopaedia any troll can write and no sane person can correct.
Now there is a thought! Aye! Richt!
Timberwoof
"Peter Webb" <webbfamily...@optusnet.com.au> wrote:
I should have added cohesive strength from self-gravity, for that is
what holds stars together until they get inside some other star's Roche
limit.
It turns out that all asteroids are clumps of rocks, not solid. If they
were solid, they would be able to rotate much faster than they do. There
appears to be an upper limit on the rotational speed of asteroids, and
this matches the speed (at each mass) for how fast one would fly apart.
> Its also bullshit, because tidal forces have moved the moon into higher and
> higher orbits over billions of years, and will continue to do so.
Well, there you go.
Timberwoof
"Steve" <philh...@aol.com> wrote:
> On Feb 11, 1:28 pm, Timberwoof <timberwoof.s...@infernosoft.com>
> wrote:
> > bla bla bla
>
> Sorry to sound trollish myself, but I've ran into this TW dude before
> and frankly he ain't worth my time especially when it involves
> science. So sorry TW I didn't read past your first post .. and that
> rule stands.
>
> However, if someone else here finds TW's comments "on topic
> stimulating" please point them out for me. Thanks.
>
> Steve
Oh. I've been discredited by a dufus. I can't tell you how much that
hurts.
Timberwoof
"Steve" <philh...@aol.com> wrote:
> On Feb 11, 1:02 pm, Timberwoof <timberwoof.s...@infernosoft.com>
> wrote:
> > Mostly because your model of orbital mechanics is too simple, and
> > although the mechanisms have been explained to you, you ignored it
> > because you don't want to understand.
>
> This is totally insane. On contrary, where have you been?! I'VE BEEN
> THE ONE saying there's likely all kinds of orbital movements involved,
> not just the one observed over the last 40 years that people here have
> resorted to as a "PROJECTILE MOVMENT." Sheeeeeeeeesh!
You are, of course, perfectly free to list all those other movements,
and to explain in what relevant ways projectile movement is different
from orbital movement.
> Ignorance can be cured. Stupidity is another matter. And blindness a
> third.
>
> Steve
Androcles
"Timberwoof" <timberw...@infernosoft.com> wrote in message news:timberwoof.spam-1A...@nnrp-virt.nntp.sonic.net...
> In article <1171235384.0...@h3g2000cwc.googlegroups.com>,
> "Steve" <philh...@aol.com> wrote:
>
>> On Feb 11, 1:28 pm, Timberwoof <timberwoof.s...@infernosoft.com>
>> wrote:
>> > bla bla bla
>>
>> Sorry to sound trollish myself, but I've ran into this TW dude before
>> and frankly he ain't worth my time especially when it involves
>> science. So sorry TW I didn't read past your first post .. and that
>> rule stands.
>>
>> However, if someone else here finds TW's comments "on topic
>> stimulating" please point them out for me. Thanks.
>>
>> Steve
>
> Oh. I've been discredited by a dufus. I can't tell you how much that
> hurts.
I won't bother to discredit you, fuckhead. You did that all by yourself.
*plonk*
J. Taylor
<webbfamily...@optusnet.com.au> wrote:
>
>"Steve" <philh...@aol.com> wrote in message
>news:1171185075.0...@j27g2000cwj.googlegroups.com...
>> On Feb 10, 11:11 pm, "Androcles" <Engin...@hogwarts.physics.co.uk>
>> wrote:
>>> "Peter Webb" <webbfamily-diespam...@optusnet.com.au> wrote in
>>> messagenews:45cea46c$0$16555$afc3...@news.optusnet.com.au...
>>
>>> It's pretty easy to guess where it will end up, and your guess isn't
>>> a good one.
>>> http://antwrp.gsfc.nasa.gov/apod/ap051113.html
>>
>> Nice time lapse shots of the entire moon day cycle. Didn't know the
>> moon swayed like that. It's not as static as I thought. Interesting.
>> Thanks!
>>
>
>Ha ha. The wobbling you think you see is a result of the inclination of the
>earth's axis affecting the time lapse photography. The effect is due to the
>camera moving, not the earth.
Maybe, some of it is due to cropping the images at different sizes,
but that could be due to different distances between Earth and moon.
Hard to say
JT
Timberwoof
"Androcles" <Engi...@hogwarts.physics.co.uk> wrote:
Oh, ow. I'm going to cry now.
::rolleyes::
Timberwoof
J. Taylor <nchi...@earthlink.NOSPAM.net> wrote:
> On Sun, 11 Feb 2007 21:21:18 +1100, "Peter Webb"
> <webbfamily...@optusnet.com.au> wrote:
>
> >
> >"Steve" <philh...@aol.com> wrote in message
> >news:1171185075.0...@j27g2000cwj.googlegroups.com...
> >> On Feb 10, 11:11 pm, "Androcles" <Engin...@hogwarts.physics.co.uk>
> >> wrote:
> >>> "Peter Webb" <webbfamily-diespam...@optusnet.com.au> wrote in
> >>> messagenews:45cea46c$0$16555$afc3...@news.optusnet.com.au...
> >>
> >>> It's pretty easy to guess where it will end up, and your guess isn't
> >>> a good one.
> >>> http://antwrp.gsfc.nasa.gov/apod/ap051113.html
> >>
> >> Nice time lapse shots of the entire moon day cycle. Didn't know the
> >> moon swayed like that. It's not as static as I thought. Interesting.
> >> Thanks!
> >>
> >
> >Ha ha. The wobbling you think you see is a result of the inclination of the
> >earth's axis affecting the time lapse photography. The effect is due to the
> >camera moving, not the earth.
You mean the pictures were taken from different points on the Earth's
surface?
> Maybe, some of it is due to cropping the images at different sizes,
> but that could be due to different distances between Earth and moon.
>
> Hard to say
Easy to say! Trivial to calculate.
It is reasonable to assume that the photographs were taken with the moon
always at the same right ascension and to assume that the images were
always at the same magnification. Varying either parameter woudl defeat
the purpose of the animation. So I think the changes in the moon's size
in the animation result from the changing distance to the moon. This can
be easily verified:
Perigee/apogee distances to the moon:
363,258 km
405,542 km
Moon's diameter: 3,476 km
So at perigee, the moon subtends an angle of .548 degrees. At apogee,
.491 degrees. That's a difference of more than ten percent. And that's
what we see in the animation.
The Earth's diameter is 12,756.3 km.
This means that the biggest difference in the earth-moon distance during
those photographs can be 6,378 km. The difference between the earth-moon
distances at perigee and apogee is 42,284 km, about 6.6x as much. That
would cause an apparent difference in the size of the moon of at most
~1.7% between moon at the horizon and moon overhead.
Here's the web site of the author:
http://www.astrosurf.com/cidadao/moon_obs.htm
You could write him and ask.
J. Taylor
There is little reason, in this case, to assume anything. It is
possible to find how it was done.
Any way, what you are doing is considered speculation.
JT
George Dishman
"J. Taylor" <nchi...@earthlink.NOSPAM.net> wrote in message
news:qd00t29flbfk1n0sa...@4ax.com...
Why don't you guys just follow the link on the above
page which tells you the cause:
http://en.wikipedia.org/wiki/Libration
HTH
George
Peter Webb
Never mind "others". Where are YOUR calculations?
What is this easy calculation you can do through conservation
of angular momentum, as you said you can?
*****
I gave you links, which you snipped. I also gave you the equations, and a
sketch of how they would be solved. Calculating this out for myself
independently and posting it would take a couple of hours. Why bother when
there are plenty of web sites that already have done the calculation?
BTW, my exact words were:
"Its pretyy easy to work out exactly where it will end up - the final
position is where the earth and moon keep the same face to each other (ie
where a day equals a month). This is easy to calculate through conservation
of angular momentum, this allows the final distance from the earth to the
moon to be calculated, though working out how long it takes is way harder."
I have given web sites that use the conservation of angular momentum to work
out the final earth-moon distance, and you can see its a pretty simple
calculation - somebody with a reasonably good understanding of physics
should be able to do it in a couple of hours work, as first occurred over
100 years ago.
Exactly as I said, the final earth moon distance can be calculated using
conservation of angular momentum; the stable state is where a month equals a
day; and its a pretty easy to do (assuming you know the physical parameters
such as the density of the earth at different depths).
Its a bit like me saying that the speed of light is easy to calculate using
the timing of occultation of Jupiter's moons - it is - and then you
demanding that I redo a tedious calculation to calculate c is approximately
3x10^8 m/s that was first done over 100 years ago and has never been
challenged, except perhaps by idiots in newsgroups.
Here is my claim, again:
"Its pretyy easy to work out exactly where it will end up - the final
position is where the earth and moon keep the same face to each other (ie
where a day equals a month). This is easy to calculate through conservation
of angular momentum, this allows the final distance from the earth to the
moon to be calculated, though working out how long it takes is way harder."
What part of this do you think is wrong?
> The interesting (for me) part of this is that the rate of recession is
> very
> hard to calculate
You said it was easy. Make up your fuckheaded mind!
*****
No, calculating the stable orbit is straightforward. Calculating how long it
takes to achieve (the "rate of recession") is very hard. I gave plenty of
websites that calculated the stable orbit; only one of these (from memory)
hazarded a guess as to how long it would take to achieve - for which you
need to know the rate.
> If you actually want to understand this a bit better
I understand it very well.
**********
Really? Do you understand what a vector cross-product is? (This is how
angular momentum is defined). Do you have enough calculus to calculate the
moment of inertial rotation of a spherical shell? (The calculation of the
earths angular momentum requires this). To do this rigourously, you would
also need enough perturbation theory to prove this orbit is stable and not
pseudo-stable, as is done with Lagrangian points (although there is so much
experimental verification that this is an academic exercise only; I did this
at Uni as part of course in perturbation theory literally as an academic
exercise - but that was 30 years ago).
If you do understand the orbital mechanics of the earth-moon system "very
well", you would know all this stuff already. But then, if you did
understand this stuff very well, I can't see what on earth (or on the moon)
you would object to what I have written.
So now I am confused. Do you understand basic orbital dynamics "very well" -
eg conservation of angular momentum (ie vector cross-products), integration
over the interior of a sphere, and perturbation theory?
If so, we can definitely kick this conversation up a level or two. If not,
perhaps you should learn some, and get back to this newsgroup when you know
what you are talking about.
As I said before, if there is anything in any of the many links I provided
that you don't understand, feel free to post any questions you may have.
Peter Webb
> Perigee/apogee distances to the moon:
> 363,258 km
> 405,542 km
>
> Moon's diameter: 3,476 km
>
> So at perigee, the moon subtends an angle of .548 degrees. At apogee,
> .491 degrees. That's a difference of more than ten percent. And that's
> what we see in the animation.
>
> The Earth's diameter is 12,756.3 km.
>
> This means that the biggest difference in the earth-moon distance during
> those photographs can be 6,378 km. The difference between the earth-moon
> distances at perigee and apogee is 42,284 km, about 6.6x as much. That
> would cause an apparent difference in the size of the moon of at most
> ~1.7% between moon at the horizon and moon overhead.
*** cough *** - that's 17%, not 1.7%.
But your argument is correct (well, 83% correct) and I accept that
apogee-perigee distance is a much bigger effect than the changing position
of the camera.
Enkidu
>> >> What keeps the moon away from the earth other than centripital force?
>
> I meant centrifugal force.
>
Cheers,
Cliff
--
Have you ever noticed that if something is advertised as 'amusing' or
'hilarious', it usually isn't?
Enkidu
> In article <45ce...@news2.actrix.gen.nz>, Enkidu
> <enkid...@com.cliffp.com> wrote:
>
>> Steve wrote:
>>>
>>> What keeps the moon away from the earth other than centripital
>>> force?
>>>
>> acts away from the centre.
>>
>> Actually, the force that stops the moon from flying away is
>> gravity. If gravity was to be suddenly abolished the moon would
>> travel away from the earth along a tangent to its orbit. It is only
>> gravity pulling it towards the earth that causes it to swing around
>> in an orbit. There is no force keeping the moon away from the
>> earth.
>
> Oh, boy, I can already see it coming: a failure to distinguish
> between force and acceleration.
>
Eh?
Blinky Bill
> >
>> Perigee/apogee distances to the moon:
>> 363,258 km
>> 405,542 km
>>
>> Moon's diameter: 3,476 km
>>
>> So at perigee, the moon subtends an angle of .548 degrees. At apogee,
>> .491 degrees. That's a difference of more than ten percent. And that's
>> what we see in the animation.
>>
>> The Earth's diameter is 12,756.3 km.
>>
>> This means that the biggest difference in the earth-moon distance during
>> those photographs can be 6,378 km. The difference between the earth-moon
>> distances at perigee and apogee is 42,284 km, about 6.6x as much. That
>> would cause an apparent difference in the size of the moon of at most
>> ~1.7% between moon at the horizon and moon overhead.
>
> *** cough *** - that's 17%, not 1.7%.
No - it's ~1.7%. The difference in apparent size due to perigee-apogee was
given above as .548 - .491 = .057 degrees. Expressed as a percentage of
.491 degrees it is 11.6%. This has been shown above to be ~ 6.6 times the
difference due to observing overhead compared to on the horizon . Hence the
% difference in apparent size due to overhead-horizon is 11.6% divided 6.6
or ~1.7%
Peter Webb
"Blinky Bill" <nos...@anytime.com> wrote in message
news:WuVzh.6635$sd2...@news-server.bigpond.net.au...
I accept that the difference in earth-moon distance arising from orbit
eccentricity is 405,542 - 363,258 or about 40,000 kms.
The difference due to the camera being in a diiferent position as the earth
rotates is a maximum of 12,000 kms and a minimum of 6,000 kms (6,000 kms if
you want to be able to see the moon at all, and 12,000 kms if you photograph
it when it close to overhead at all times), assuming an equatorial camera.
Therefore the MINIMUM percentage contribution from the earth's rotation is
6,000/40,000 kms, or about 15% of the observed change is size; the MAXIMUM
is 12,000/40,000, or about 30%.
Either way, not the dominant component.
Whilst I can't find the original film clip, I suspect that is also the cause
of the perceived libration of its rotational axis. As the earth's rotation
is tilted from the plane of rotation of the moon, the time of day of
observation has an effect on the perceived orientation of the moons axis -
it appears to be wobbling, but this is artefact of the camera being at
different distances from the plane of rotation (this is known as "diurnal
libration" and it really only affects our observations of the moon, as
everthing else is too far away for it to matter).
Blinky Bill
The previous poster wasn't calculating the proportion of the observed change
that the overhead-horizon difference contributed. He was calculating the
apparent difference in size that the overhead-horizon difference caused.
Androcles
"Peter Webb" <webbfamily...@optusnet.com.au> wrote in message news:45d01b52$0$4753$afc3...@news.optusnet.com.au...
>
> Never mind "others". Where are YOUR calculations?
> What is this easy calculation you can do through conservation
> of angular momentum, as you said you can?
>
>
> *****
>
> I gave you links, which you snipped.
Links are not your calculations.
What is this easy calculation you can do through conservation
of angular momentum, as you said you can, FUCKHEAD troll?
Peter Webb
> It turns out that all asteroids are clumps of rocks, not solid. If they
> were solid, they would be able to rotate much faster than they do. There
> appears to be an upper limit on the rotational speed of asteroids, and
> this matches the speed (at each mass) for how fast one would fly apart.
>
That's very interesting, I didn't know that. I found only one reference (the
original paper, http://keith.aa.washington.edu/papers/spinlimits.pdf ). It
makes an excellent argument - as you summarised it - but the implications
are huge, which makes me wonder if it is now accepted fact. It makes the
"demolished planet" theory of the asteroids formation look far more likley
than the "failed to accrete into a planet because of Jupiter's gravity"
theory of asteroid formation.
Peter Webb
OK. Do you know how to calculate a vector cross product, or should we start
with that?