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The reason I want to know is because many years ago in Sunday school I learned a story that in probably 600 AD some sinner was swallowed up by a hole, and that he's still falling as of today.

I can't use the formulas I know because they all assume g as constant, and I realized that I can't assume g=9.8 the entire fall because he's getting closer and closer to the center of the earth, so I tried to use my fairly bad calculus skills and failed miserably. Can anyone do this problem? Is it even possible?

Qmechanic
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Bilal Nagi
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1 Answers1

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It would help if you provide your calculations; probably I or someone else will see and correct your errors. However, a correct approach would recognize that at any distance $r$ from the center of the Earth, Earth's radius R, the gravity the falling man experiences is only $r^3 / R^3$ times the gravity experienced at the surface of the Earth. -- that is, assuming the density of the Earth is uniform. It comes to about 21 minutes, if there is no air resistance, to reach the center.

S. McGrew
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  • Thanks! But how did you get that the gravity will be #r^3/R^3# times gravity on the surface? – Bilal Nagi Mar 30 '18 at 14:51
  • Gravity inside a hollow spherical shell is zero, so the only gravitational acceleration felt inside the Earth is due to what's left after the portion of the sphere outside r is removed. The volume of that remaining portion is proportional to r^3. – S. McGrew Mar 30 '18 at 17:33