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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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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 $1/r^2$