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I would like to know the expected behavior of the Gravity under the following mentioned imaginary experiment:

What if we dig a well or a boar or a straight hole (say, its diameter is 100 meter) throughout the earth (i.e. Start digging from the one end of the Earth until you reach the exact opposite end of the Earth) and then one heavy and rigid spherical stone (say, its diameter is 1 meter) is to be fallen down into that well or boar or hole? What would happen to that sphere?

Assume all the best conditions for this experiment: a perfect round-shaped well with no inner disturbances, sphere is made from the very hard material, a perfect vertical free-fall of the sphere, etc.

How long would it travel through this well?

Will sphere stop its journey by reaching the center of the earth?

If yes, then could you justify this situation by the various forces applying on that sphere at that time?

Please share your views and justifications on this experiment.

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marked as duplicate by Qmechanic Jun 10 '13 at 12:56

This question was marked as an exact duplicate of an existing question.

Hint: think about simple harmonic motion. – Will Jun 10 '13 at 4:19

Well If the ball will not take any type of resistance then it will just oscillate from one end of the well to other end forever.

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Let the center of the earth lie at the origin, and let $r$ denote radial position. The gravitational force in the sphere will be $$ \mathbf F = -\frac{G\mu(r)m}{r^3}\mathbf r $$ where $\mu(r)$ represents the mass contained in the subsphere of the Earth with radius less than $r$; $$ \mu(r) = \frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi R^3}M = \frac{r^3}{R^3}M $$ Using Newton's second law, this gives the following equation of motion: $$ \ddot{\mathbf r} = -\frac{GM}{R^3} \mathbf r $$ Now contemplate Will's comment.

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The mass of Earth, $M$, is missing in the last two formulas, right? – jkej Jun 10 '13 at 10:19
@jkej Wow woops, yes thanks! – joshphysics Jun 11 '13 at 5:12

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