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Find the flux of the earth's gravitational field ,g, through the whole of the earth's surface?? Assume the earth to be a perfect shpere with a uniform gravitational filed all aver it's surface ???

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I think it's a "trick" question, in the sense that you might not have to actually do a lot of math to figure it out.

Since this is clearly homework, you probably shouldn't expect much explicit help.

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Quote:
Original post by Omid Ghavami
Using Gauss' Law this problem is solved very easily.

Isn't it solved even more easily using the Divergence Theorem?

Admiral

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Original post by TheAdmiral
Quote:
Original post by Omid Ghavami
Using Gauss' Law this problem is solved very easily.

Isn't it solved even more easily using the Divergence Theorem?

Admiral


Assuming he's going with the E&M meaning of flux, then that would be the appropriate method of attack, but since it's a special case of a closed surface the problem is reduced. It would be like breaking out the quadratic formula for x2+x.

And, assuming means otherwise, the answer comes out the same. :)

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Quote:
Original post by TheAdmiral
Quote:
Original post by Omid Ghavami
Using Gauss' Law this problem is solved very easily.

Isn't it solved even more easily using the Divergence Theorem?

Admiral


They are the same (Well, Gauss' Law is what you get when you apply Gauss Theorem (A.K.A Divergence Theorem) in for example Electrostatics, but also other branches of classical physics, amongst others this very example of a gravitational field). What it basically says is that the flux of such a field out of a closed surface is related to sources and sinks within the surface. In other words that the size or shape of the closed surface is irrelevant, if I have a point source electric charge the flux out of a spherical surface surrounding that point will be the same regardless of the radius of this sphere.

As for the problem, if you still haven't solved it F@ke, here's another hint: You can solve it by setting up a spherical Gaussian surface with an arbitrary radius. We choose a spherical one since we have spherical symmetry for the gravitational force, since it is only dependent on distance, which means that it can be factored out of the surface integral since R is constant for this integral. The surface integral will now be nothing more than the surface area of the Gaussian sphere with radius R.

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Quote:
Original post by Omid Ghavami
Quote:
Original post by TheAdmiral
Quote:
Original post by Omid Ghavami
Using Gauss' Law this problem is solved very easily.

Isn't it solved even more easily using the Divergence Theorem?

Admiral


They are the same (Well, Gauss' Law is what you get when you apply Gauss Theorem (A.K.A Divergence Theorem)

I see.
IANAP - the Law/Theorem bit threw me [rolleyes].

Admiral

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