Need help with tricky gravitational force problem

AI Thread Summary
The discussion revolves around calculating the gravitational force on a point mass at different distances from the Earth's center. The initial assumption that the force at R/2 would be four times that at R is corrected, revealing that the actual ratio is two due to the effects of Earth's mass distribution. The shell theorem is highlighted as relevant for understanding gravitational forces both inside and outside a spherical mass. Participants clarify that while the theorem applies to both scenarios, different formulas are needed for calculations within the Earth's interior. The conversation concludes with an acknowledgment of the complexities involved in gravitational force calculations within a solid sphere.
PsychonautQQ
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Homework Statement


The effect of gravity on a point mass is given as F(r), where r is the radius between the point mass and the Earth's center. If you could place the point mass at R/2, where R is the radius of the earth, what would the relationship be between F(R)/F(R/2)

Homework Equations


Fg = Gm1m2 / r^2

The Attempt at a Solution


I tried just plugging R/2 and R and getting of course 4 since the radius is squared, but the correct answer is suppose to be 2. I realized the flaw in my logic was that if the point mass was half way to the center of the earth, then some of the Earth's mass would be pulling in the opposite direction as the stuff still "under" it. Help please :D
 
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The shell theorem only applies to masses outside the Earth so the formula you listed is only valid in the region external to the surface of the Earth. On the other hand, the problem wants you to compare the force of gravity at the surface of the Earth to a radius within the surface of the Earth so the formula will be different. Approximating the Earth as a uniform solid sphere, what will be the force of gravity in the interior to the surface?
 
PsychonautQQ said:
I tried just plugging R/2 and R and getting of course 4
I suggest you got 1/4, not 4.
WannabeNewton said:
The shell theorem only applies to masses outside the Earth
Not exactly. The theorem covers both inside and outside the shell, but giving different formulae: http://en.wikipedia.org/wiki/Shell_theorem.
 
haruspex said:
I suggest you got 1/4, not 4.

Not exactly. The theorem covers both inside and outside the shell, but giving different formulae: http://en.wikipedia.org/wiki/Shell_theorem.
I had originally thought the result that the gravitational force vanishes identically inside, as derived using the cone argument or what have you, only applied to a thin shell configuration as opposed to a solid sphere as in the OP's problem. Are you referring to the very last equation in the section on "inside the shell"? If so, that is quite interesting thank you muchly for the link :)

EDIT: By the way it seems the OP had also posted in advanced physics and got it answered there.
 
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WannabeNewton said:
Are you referring to the very last equation in the section on "inside the shell"?
Not sure which one you mean. I meant that inside the shell there is no field (as also for electric field from a uniform charge on a spherical shell). In consequence, given a spherically symmetric density distribution radius a, the gravitational field at radius r < a can be found by ignoring everything at radius > r and treating the rest as concentrated at the centre.
 
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Ah ok, gotcha. The ol' succession of thin spherical shells :)! Thanks and cheers.
 
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