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Moon By Spinoff
- AgoraBasta
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22 years 2 months ago #2962
by AgoraBasta
Replied by AgoraBasta on topic Reply from
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
1270 km straight down from the equator (still at even spin condition) material is only moving at 90% of orbital velocity, and gravity exceeds centrifugal force. Pressure is not zero here (but it is less than if there were no spin) even though we are still in the equatorial plane. The closer we get to the center the higher the pressure. Solid, liquid or gas.
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That's a simple mistake. Inside a body of constant density the gravity rises from the centre linearly to the surface and then falls as inverse square. The centrifugal force rises linearly. So the "evenspin" balance is present all through the equator plane inside the body.
Earth is not exactly a body of constant density, but not to the extent as to disrupt the overall scenario described.
Overspin=disintegration, basta!
1270 km straight down from the equator (still at even spin condition) material is only moving at 90% of orbital velocity, and gravity exceeds centrifugal force. Pressure is not zero here (but it is less than if there were no spin) even though we are still in the equatorial plane. The closer we get to the center the higher the pressure. Solid, liquid or gas.
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
That's a simple mistake. Inside a body of constant density the gravity rises from the centre linearly to the surface and then falls as inverse square. The centrifugal force rises linearly. So the "evenspin" balance is present all through the equator plane inside the body.
Earth is not exactly a body of constant density, but not to the extent as to disrupt the overall scenario described.
Overspin=disintegration, basta!
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- Larry Burford
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22 years 2 months ago #2745
by Larry Burford
Replied by Larry Burford on topic Reply from Larry Burford
[lb]
1270 km straight down from the equator (still at even spin condition) material is only moving at 90% of orbital velocity, and gravity exceeds centrifugal force. Pressure is not zero here (but it is less than if there were no spin) even though we are still in the equatorial plane. The closer we get to the center the higher the pressure. Solid, liquid or gas.
[ab]
That's a simple mistake. Inside a body of constant density the gravity rises from the centre linearly to the surface and
then falls as inverse square. The centrifugal force rises linearly. So the "evenspin" balance is present all through the equator plane inside the body.
======
It is easy to forget that we are no longer talking about a spherically symmetric situation. (I did, just above.) An object approaching even spin has a shape called oblate spheroid. In cross section it could be described as a circle with two rounded triangles attached on opposite sides. Rotate this figure to get the solid - a spherical "core" (which contains the actual core of the planet) with a triangular ring around its equator.
As even spin is approached the planet will flatten, causing the core or polar radius to shrink and the equatorial radius to increase. The circle gets smaller and the triangles get taller. The equatorial surface can be 10s, 100s or even 1000s of kilometers further from the center than the polar surface. Depending on a *lot* of variables, of course.
The amount of material in this "triangular" bulge region is much less than in the circular core. For the purpose of estimating gravitational force at any given depth the planet can now be modeled as a rather high density sphere with a radius equal to the polar radius surrounded by a rather low density spherical shell with inside radius equal to the polar radius and outside radius equal to the equatorial radius.
As you go beneath the surface at the equator the force of gravity from this low-density shell drops linearly with dereasing distance from then center. But the force of gravity from the high-density core increases quadratically as you get closer to it (you are still outside it). Depending again on numerous variables, the total of the two can actually increase with depth until you reach the "core". Once inside the core, of course, increasing depth causes a linear drop in total force.
[ab]
Earth is not exactly a body of constant density, but not to the extent as to disrupt the overall scenario described.
======
Constant density and near liquid consistency seem like reasonable approximations. If this process occurs in nature it will occur very early in the life of a planet, while it is still hot enough to shrink significantly as it cools (scenarios involving the over spin of present-day Earth are fun but not very realistic). A crust may have formed/be forming as even spin is approached. No biosphere - probably no liquid water - and lots of gas.
No one has actually seen a planet do this, but we do have some pretty good computer simulations. Perhaps it is time for someone with some experience with these simulations to say a few words.
Regards,
LB
1270 km straight down from the equator (still at even spin condition) material is only moving at 90% of orbital velocity, and gravity exceeds centrifugal force. Pressure is not zero here (but it is less than if there were no spin) even though we are still in the equatorial plane. The closer we get to the center the higher the pressure. Solid, liquid or gas.
[ab]
That's a simple mistake. Inside a body of constant density the gravity rises from the centre linearly to the surface and
then falls as inverse square. The centrifugal force rises linearly. So the "evenspin" balance is present all through the equator plane inside the body.
======
It is easy to forget that we are no longer talking about a spherically symmetric situation. (I did, just above.) An object approaching even spin has a shape called oblate spheroid. In cross section it could be described as a circle with two rounded triangles attached on opposite sides. Rotate this figure to get the solid - a spherical "core" (which contains the actual core of the planet) with a triangular ring around its equator.
As even spin is approached the planet will flatten, causing the core or polar radius to shrink and the equatorial radius to increase. The circle gets smaller and the triangles get taller. The equatorial surface can be 10s, 100s or even 1000s of kilometers further from the center than the polar surface. Depending on a *lot* of variables, of course.
The amount of material in this "triangular" bulge region is much less than in the circular core. For the purpose of estimating gravitational force at any given depth the planet can now be modeled as a rather high density sphere with a radius equal to the polar radius surrounded by a rather low density spherical shell with inside radius equal to the polar radius and outside radius equal to the equatorial radius.
As you go beneath the surface at the equator the force of gravity from this low-density shell drops linearly with dereasing distance from then center. But the force of gravity from the high-density core increases quadratically as you get closer to it (you are still outside it). Depending again on numerous variables, the total of the two can actually increase with depth until you reach the "core". Once inside the core, of course, increasing depth causes a linear drop in total force.
[ab]
Earth is not exactly a body of constant density, but not to the extent as to disrupt the overall scenario described.
======
Constant density and near liquid consistency seem like reasonable approximations. If this process occurs in nature it will occur very early in the life of a planet, while it is still hot enough to shrink significantly as it cools (scenarios involving the over spin of present-day Earth are fun but not very realistic). A crust may have formed/be forming as even spin is approached. No biosphere - probably no liquid water - and lots of gas.
No one has actually seen a planet do this, but we do have some pretty good computer simulations. Perhaps it is time for someone with some experience with these simulations to say a few words.
Regards,
LB
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- AgoraBasta
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22 years 2 months ago #2832
by AgoraBasta
Replied by AgoraBasta on topic Reply from
Here I could agree with you, Larry. Still I'll try some more analysis.
Any real config should be somewhere between a sphere and a flat disk. Consider gravity within a flat disk: it does NOT depend on distance from its centre, it depends only on the thickness and density. So gravity can't balance out the centrifugal force above certain radius as function of thickness. Hence we'd have freely orbiting rings above that radius. Any thickening towards the centre makes the equivalent max radius smaller, since a spheric form is totally non-stable to overspin. (If we allow accretion to the central area, the max equatorial diameter would shrink less than one order of magnitude, more like by a factor of about 2, but the thickness at polar axis shall stay by yet another factor, of the same order, smaller than the equatorial diameter.) Thus we come to what I intended to prove - overspin is shed immediately into a dissipating ring. The process goes the other way in the nature - as excessive spin is shed, accretion may go on. Tom's model resembling separation of one or two liquid droplets is generally invalid. Quod erat demonstrandum.
(BTW, your "rounded triangles" become "sharpened triangles" well before overspin condition reached, that's due to internal pressure; if without pressure - exactly at "evenspin")
Any real config should be somewhere between a sphere and a flat disk. Consider gravity within a flat disk: it does NOT depend on distance from its centre, it depends only on the thickness and density. So gravity can't balance out the centrifugal force above certain radius as function of thickness. Hence we'd have freely orbiting rings above that radius. Any thickening towards the centre makes the equivalent max radius smaller, since a spheric form is totally non-stable to overspin. (If we allow accretion to the central area, the max equatorial diameter would shrink less than one order of magnitude, more like by a factor of about 2, but the thickness at polar axis shall stay by yet another factor, of the same order, smaller than the equatorial diameter.) Thus we come to what I intended to prove - overspin is shed immediately into a dissipating ring. The process goes the other way in the nature - as excessive spin is shed, accretion may go on. Tom's model resembling separation of one or two liquid droplets is generally invalid. Quod erat demonstrandum.
(BTW, your "rounded triangles" become "sharpened triangles" well before overspin condition reached, that's due to internal pressure; if without pressure - exactly at "evenspin")
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22 years 2 months ago #3217
by Entropic
Replied by Entropic on topic Reply from
By capture theory I am assuming what is meant is the moon was a free object that was captured by the earth's gravity. One could infer this is the case with all moons. A coincidence all were captured and there are none anywhere but in orbit around large planets (Pluto could be a free one)? Couldn't the formation of the moon from earth matter involve a collision?
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22 years 2 months ago #3090
by n/a1
Replied by n/a1 on topic Reply from john duff
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
I suspect that the phase change of "normal" Iron to high density Iron would be a self-accelorating reaction, with the rate of reaction increasing expotentially.
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Nope. This reaction contains a negative feedback built-in. Pressure triggers shrinkage - shrinkage drops the pressure in surroundings down. The reaction being exothermic doesn't help since the reverse process is endothermic by the same margin.
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John Duff
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
I suspect that the phase change of "normal" Iron to high density Iron would be a self-accelorating reaction, with the rate of reaction increasing expotentially.
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Nope. This reaction contains a negative feedback built-in. Pressure triggers shrinkage - shrinkage drops the pressure in surroundings down. The reaction being exothermic doesn't help since the reverse process is endothermic by the same margin.
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
John Duff
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21 years 7 months ago #5276
by kingdavid
Replied by kingdavid on topic Reply from David King
Hi tom
I am sure youre aware of these lunar anomalies:
" www.keelynet.com/unclass/luna.htm "
I have just finished an old book about the possibility of our moon being artificial and was surprised to know that much of this information of anomalies of the moon are not answered.
What are your views on this.
Also why if gravity can push through earth from space can it not go straight through the opposite end as this 'Levity' that people talk about.
Using the inverse square law it would mean that levity would be a third of the incoming gravitational force and would also explain why we can pick up thing as easly as we can regardless of gravity - something i picked up while watching a documentary on M Theory.
Thanks,
David
I am sure youre aware of these lunar anomalies:
" www.keelynet.com/unclass/luna.htm "
I have just finished an old book about the possibility of our moon being artificial and was surprised to know that much of this information of anomalies of the moon are not answered.
What are your views on this.
Also why if gravity can push through earth from space can it not go straight through the opposite end as this 'Levity' that people talk about.
Using the inverse square law it would mean that levity would be a third of the incoming gravitational force and would also explain why we can pick up thing as easly as we can regardless of gravity - something i picked up while watching a documentary on M Theory.
Thanks,
David
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