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why we would weigh less at earth's center(as opposed to sea level). Looking around the net, I have seen different approaches to this problem, each with different solutions.

According to Newton's universal law of gravitation:

$$F= \frac{Gm_1m_2}{r^2}$$

If this is the case, then as r approaches nearly zero as the object gets closer to the center of the earth, the denominator gets smaller and smaller, making the quotient approach infinity.

On the other hand, I've heard the explanation that all the mass around you cancels each other out at the center of the earth.

This explanation also seems to make sense, but both explanations contradict each other..

johnson316
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  • Possible duplicates: http://physics.stackexchange.com/q/2481/2451 , http://physics.stackexchange.com/q/18446/2451 and links therein. – Qmechanic Dec 06 '14 at 06:49

1 Answers1

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It is true that the shell outside the current radius does not contribute, so you are only left with the force from the mass inside:

$$F=G\frac{mM_{inside}}{r^2}$$

but the M inside becomes smaller as the radius becomes smaller.Assuming the dendity, $\rho$, is a constant, then $$M_{inside}=\rho\frac{4}{3}\pi r^3 $$

which leaves you with:

$$F=G\rho\frac{4}{3}\pi m r $$

Thus the force decreases linearly with $r$ and is zero at the center. The reason is that the mass inside the radius diminishes with r faster than the increase caused by $1/r^2$

  • Except once you're inside the earth, then there's mass above you as well as below that's creating a gravitational pull upwards, which will then decrease the net pull downwards. Basically, when outside of a body, you can treat it as a point source of gravity, but when inside, the math is not so simple. – MattS Dec 06 '14 at 02:51
  • yes it is so simple, there is a theorem that shows that the gravity inside a shell gets cancelled, regardeless of where inside the shell you are (you do not need to be at the center). Let me see if I fidn a link. –  Dec 06 '14 at 02:53
  • I've heard of that before, but technically speaking doesn't that only apply if we assume it's a body of uniform density? And also doesn't it only apply to perfect spheres? – MattS Dec 06 '14 at 02:57
  • It applies to perfect spheres, that is true, but the density doesnt need to be constant, it can be a function of the radius but not of the angle. That is, you can divide the shell in thin spheres of siffernt density, and the contribution of every sphere is zero. In a real case like Earth, of course ther might be some residual gravity at the center, but the idea is that it is very far from infinite. –  Dec 06 '14 at 03:01