Gravitational collapse

[Tach]

The earth isn't a sphere, but rather a spheroid; it has a bulge at the equator, so gravitational acceleration is greatest at the poles, where the surface is closer to the COM. So, somewhat non-intuitively, although there is more mass near the equator g is smaller there. This is because (the force of) gravity is the inverse-square of distance between centres of mass.

Actually, the net gravitational acceleration (when you take into consideration that the Earth is a geoid, not a spheroid and that it rotates) is the same at the poles and at the Equator. See here. This also explains why all clocks in the NIST tick at the same rate.

What's a good way to interpret this in terms of curvature?

Read the paper, the math is self-explanatory.

 

[arfa brane]

Actually, the net gravitational acceleration (when you take into consideration that the Earth is a geoid, not a spheroid and that it rotates) is the same at the poles and at the Equator

That appears to conflict with this:

It is worth noting that for the mass attraction effect by itself, the gravitational acceleration at the equator is about 0.18% less than that at the poles due to being located farther from the mass center. When the rotational component is included (as above), the gravity at the equator is about 0.53% less than that at the poles, with gravity at the poles being unaffected by the rotation. So the rotational component change due to latitude (0.35%) is about twice as significant as the mass attraction change due to latitude (0.18%), but both reduce strength of gravity at the equator as compared to gravity at the poles.

--http://en.wikipedia.org/wiki/Gravitational_acceleration

 

[Tach]

That appears to conflict with this:
--http://en.wikipedia.org/wiki/Gravitational_acceleration

This is precisely why wiki cannot be trusted as a scientific reference, the data in the wiki page is about gravitational acceleration on the reference ellipsoid. As I already explained, the Earth is a geoid and the geoid is gravitationally equipotential .

 

[brucep]

That appears to conflict with this:
--http://en.wikipedia.org/wiki/Gravitational_acceleration

The Kerr metric covers all that. The total mass of a rotating object is the sum of angular momentum and Mass.

g=M/r^2

g is a function of Mass and radius. Relativistic g includes gamma but gamma equals 1 for this discussion. Rotation also means the body can't be spherically symmetric thus the rotation parameter has an effect on r. It seems the folks that made the quote, you linked, figured out the details. These effects are exceedingly small as to be irrelevant when doing the physics. Tach's link concludes this and leaves it out of any derivation.

 

[arfa brane]

The wiki quote says the acceleration is different at the poles than at the equator. The explanation for this is simple: the equator is further from the earth's COM than the poles. Why is this "irrelevant"?

We have what looks like two different things: gravitational acceleration, and gravitational potential.

 

[Tach]

We have what looks like two different things: gravitational acceleration, and gravitational potential.

Mff, what do you think is the gravitational acceleration of an equipotential gravitational field? Here is some help for you.

 

[arfa brane]

Mff, what do you think is the gravitational acceleration of an equipotential gravitational field?

Any textbook will say this is g = -GM/r[sup]2[/sup]. The minus sign is because gravity is an attractive force.

Gravitational potential isn't given by the above formula. The wiki article says the acceleration (g) is smaller at the equator, it does not say this effect dissapears because the earth is rotating.

 

[OnlyMe]

This is precisely why wiki cannot be trusted as a scientific reference, the data in the wiki page is about gravitational acceleration on the reference ellipsoid. As I already explained, the Earth is a geoid and the geoid is gravitationally equipotential .

Definition of geoid, a hypothetical surface that corresponds to mean sea level and extends at the same level under the continents. So what you are saying is that the earth you refer to is hypothetical? How does that have anything to do with the real situation, involving an earth with mountains and oceans and which is not uniform in density?

As I mentioned earlier, the paper you cite treats the Earth's an ideal gravitating body. The surface of the earth is not uniform, nor is the earth of uniform composition and density. While from some distant location the earth may be treated or modeled as representing a uniform gravitational field, that does not apply at, on or near the earth's surface. The elevation of the surface of the earth relative to sea level or the hypothetical geoid, is not uniform... And even if they were the gravitational map of the earth is not uniform even at the surface of the hypothetical geoid.

Tach the paper you cited was and is a teaching tool not a definitive paper on the reality of the gravitational potential at every point on the irregular globe of the earth.., not even any real comparrison of the gravitational potential between the equator and the poles of the earth.

Based solely on Marcus' earlier statements as they relate to this issue all clocks tick at different rates at different elevations, and thus, it can also be said that all clocks tick at different rates when moved between locations of differing gravitational potentials. Your math is going to have to get a great deal more complex than what is discussed in the cited paper or anything you have yet presented, before you even get close to a real life description of the Earth's gravity at its surface, restricted to sea level or not.

 

[Tach]

Any textbook will say this is g = -GM/r[sup]2[/sup].

Nope, the textbooks tell you : $$\vec{g}=-\nabla (\Phi)$$ .

The minus sign is because gravity is an attractive force.

So, you did not understand the AmJPhys paper I linked for you?

Gravitational potential isn't given by the above formula.

I don't know how you managed to get things backwards, the link I cited for you tells you literally: "The [[gravitational field]], and thus the acceleration of a small body in the space around the massive object, is the negative [[gradient]] of the gravitational potential." So, once again, what is the gravitational acceleration for a gravitational field that is equipotential?

The wiki article says the acceleration (g) is smaller at the equator, it does not say this effect dissapears because the earth is rotating.

No one said anything about any effect "disappearing". Looks like you do the same thing every time, you ask a question but you don't really want to learn.

 

[Tach]

Definition of geoid, a hypothetical surface that corresponds to mean sea level and extends at the same level under the continents. So what you are saying is that the earth you refer to is hypothetical? How does that have anything to do with the real situation, involving an earth with mountains and oceans and which is not uniform in density?

It isn't hypothetical since it is what both NIST and the geological society use for their work.

As I mentioned earlier, the paper you cite treats the Earth's an ideal gravitating body.

Err no, quite the opposite, it is what physicists at NIST use but you seem set to cling to your fringe misconceptions.

The surface of the earth is not uniform, nor is the earth of uniform composition and density. While from some distant location the earth may be treated or modeled as representing a uniform gravitational field, that does not apply at, on or near the earth's surface.

Mainstream physicists and geologists disagree with your crackpottery. <shrug>

 

[arfa brane]

Nope, the textbooks tell you : $$\vec{g}=-\nabla (\Phi)$$

Nope, the textbooks tell you g = -GM/r[sup]2[/sup]. As usual you sem to think being evasive or obtuse is some kind of defense.

But what you posted is only one way to describe g. Nothing obtuse about that.

Then again, many textbooks say that THE MINUS SIGN is there because the acceleration is directed TOWARDS the centre, or gravity is attractive, which is the same thing.

I don't know how you managed to get things backwards,

I haven't got things "backwards", acceleration has different units than gravitational potential. Your initial argument was about this potential being equal, it was not about g being the same everywhere on the surface.

So, you did not understand the AmJPhys paper I linked for you?



I don't know how you managed to get things backwards, the link I cited for you tells you literally: "The [[gravitational field]], and thus the acceleration of a small body in the space around the massive object, is the negative [[gradient]] of the gravitational potential." So, once again, what is the gravitational acceleration for a gravitational field that is equipotential?

Why don't you tell us? Why do you THINK "read the paper" is an adequate way to address someone's question? Why should anyone believe that you understand "the paper" when you don't, or can't, explain any relevance it may have?

 

[Tach]

Your initial argument was about this potential being equal,

You mean the same, "potential being equal" is a nonsensical statement.

it was not about g being the same everywhere on the surface.

Do you think that there is a difference between the two notions?

Why don't you tell us? Why do you THINK "read the paper" is an adequate way to address someone's question?

Because the paper teaches the people who want to learn (this excludes you) that the surface of the geoid is equipotential, i.e. the same in every point, hence the resulting gravitational acceleration is the same in every point.

 

[arfa brane]

Tach, you have nothing useful to contribute, please go away.

 

[Tach]

Tach, you have nothing useful to contribute, please go away.

I was trying to correct your misconceptions. As in the past, I failed. Sorry, I'll leave you basking in them.

 

[arfa brane]

The geoid is the shape that the surface of the oceans would take under the influence of Earth's gravity alone, in the absence of other influences such as winds and tides. All points on that surface have the same scalar potential—there is no difference in potential energy between any two.

Specifically, the geoid is the equipotential surface that would coincide with the mean ocean surface of the Earth if the oceans and atmosphere were in equilibrium, at rest relative to the rotating Earth,[1] and extended through the continents (such as with very narrow canals). According to Gauss, who first described it, it is the "mathematical figure of the Earth", a smooth but highly irregular surface that corresponds not to the actual surface of the Earth's crust, but to a surface which can only be known through extensive gravitational measurements and calculations. Despite being an important concept for almost two hundred years in the history of geodesy and geophysics, it has only been defined to high precision in recent decades, for instance by works of Petr Vaníček, and others. It is often described as the true physical figure of the Earth,[1] in contrast to the idealized geometrical figure of a reference ellipsoid.

The surface of the geoid is farther away from the center of the Earth where the gravity is weaker, and nearer where it is stronger. The differences in gravity, and hence the scalar potential field, arise from the uneven distribution of the density of matter in the earth.

Wikipedia can't be trusted as a reference, but what about the references at the bottom of this page? http://en.wikipedia.org/wiki/Geoid
And note how it says "the differences in gravity"? Doesn't that mean g is not the same everywhere? (Of bloody course it bloody does) . . .

 

[Tach]

Wikipedia can't be trusted as a reference, but what about the references at the bottom of this page? http://en.wikipedia.org/wiki/Geoid
And note how it says "the differences in gravity"? Doesn't that mean g is not the same everywhere? (Of bloody course it bloody does) . . .

You are starting (albeit very slowly) to learn how to read. The "difference in gravity" (correct term: "standard deviation") on the geoid is of the order of $$10^{-5}mgal$$. Reference. You should have taken a hint when the AmJPhys paper shows the gravitational potential at the Equator being equal to the one at the pole.

 

[arfa brane]

The "difference in gravity" (correct term: "standard deviation") on the geoid is of the order of $$10^{-5}mgal$$.

Amazing. What about on the actual surface of the earth? What about the differences in the scalar potential?

Why does Wikipedia say: "So the rotational component change due to latitude (0.35%) is about twice as significant as the mass attraction change due to latitude (0.18%), but both reduce strength of gravity at the equator as compared to gravity at the poles."

?
0.18% is a lot more than 10[sup]-5[/sup]. It is definitely saying that g (the 'strength' of the field) is bigger at the poles than at the equator. Furthermore, it indicates that, despite an imaginary surface of equipotential, the strength varies considerably over the actual surface of the earth.

Are they lying?

Let's take another look at your claim:

Actually, the net gravitational acceleration (when you take into consideration that the Earth is a geoid, not a spheroid and that it rotates) is the same at the poles and at the Equator.

First point: the earth is not a geoid. Second, g is not the same at the poles and the equator (otherwise why is there an equation that corrects for latitude?).

 

[Tach]

Let's take another look at your claim: First point: the earth is not a geoid.

You are getting more entertaining by the post.

Second, g is not the same at the poles and the equator (otherwise why is there an equation that corrects for latitude?).

I suggest that you write all your "discoveries" to the guys at NIST. They will be dismayed to find out that they need to work a lot harder at synchronizing the clocks in the system because a guy called arfa brane is learning from wiki.

 

[OnlyMe]

It isn't hypothetical since it is what both NIST and the geological society use for their work.



Err no, quite the opposite, it is what physicists at NIST use but you seem set to cling to your fringe misconceptions.




Mainstream physicists and geologists disagree with your crackpottery. <shrug>

From your initial reference, A Socratic Dialog... We show how students can be led... a teaching tool!

Have you actually read the NIMA document you referenced? Yes they use the term, but they also continually reference it as, "closely associated with". Even then it is closely associated with the mean sea leavel..., not as an accurate description of the physical composition or shape, of the earth.

Again, from a distance modeling the Earth's gravitational field in this way works. But it does not and NIMA does not suggest, that the geiod even given their modified definition, represents the earth. It serves as a reference from which accurate mapping of the earth and variations in its gravitational field can be mapped.

And you are starting in with the "mainstream" card again. A poor argument for when you have one. Beyond that your last comment was utterly false. Mainstream physicists do know the earth is not of uniform composition or shape.

 

[arfa brane]

You are getting more entertaining by the post.

And you are getting less believable. But then, that's always been your problem, right?
One way you could make it more entertaining for everyone is by producing solid evidence to back up your claim: "the earth is a geoid". But you won't because that isn't your style, is it? Plus, of course, there is no such evidence. A surface of equipotential is imaginary, you see.

As to synchronising clocks, if the clocks are all at the same gravitational potential, and otherwise (mechanically) identical, then they will run at the same rate (assuming the laws of physics are identical at different locations). However, NIST has probably realised that the physical surface of the earth is not a surface of equipotential.