What if we make a tunnel through the earth and drop a ball or any other body into it? How would the gravity behave on the ball, and , will the ball pop-up on the other side? If yes, then how much time will it take?

A question, which is usually asked in competitive exams and for general interest.

This article is a descriptive analysis of bodies inside the tunnel through earth. To do so, we assume no- frictional forces between the earth’s inner surface and the body — so there is no dissipation in the energy of the body thrown to the tunnel. It’s also assumed that the body is leaved at rest from one end of the tunnel (i.e., from the surface of earth) and the density of Earth is uniform (it’s not, actually).

When the body is on the surface of Earth, the acceleration due to gravity g is about $9.8 m/s^2$ derived from the formula $g= G \dfrac{M}{R^2}$ , where G represents the universal gravitational constant (with value $G= 6.67 \times 10^{-11} \mathrm{Newton-metre^2-kg^{-2}}$ ) M denotes the total mass of the earth and R, radius of earth. Now as body leaves the surface and enters into the Earth — the value of g decreases linearly by the formula $g= G \dfrac{M}{R^2} \left( {1-\dfrac{r}{R}} \right)$ where, r denotes the depth traveled by the body vertically. It’s clear that as the body reaches at half of the radius of Earth (i.e., r=R/2 ) and to the center of earth, (r=R), the gravitational acceleration becomes half and then ceases to zero. In other words, the weight reduces to a half and then zero at these points.

Now, the force acting on the body is F=mass * acceleration = m *(-g’). Here g’  being the reduced gravitational acceleration working on the body. Now, as g’=g'(r) (g’ is a function of r), we can easily have

$F = m \times -g'(r) = -kr$ where k being a constant which depends on m, M & R. Now, as F=-kr, the force working on the body is attractive and directly proportional to the displacement from the reference point. Such a body executes Simple Harmonic Motion. So the body we put inside, will do a simple harmonic motion through the tunnel.

The time period for this system can be given by $T= 2 \pi \sqrt{\dfrac{m}{k}} = 2 \pi \sqrt{\dfrac{mR}{mg}}$ (Remember; –mg=-kR & -mg’=-kr)

Thus, $T=2 \pi \sqrt{\dfrac{R}{g}}= 5068 \mathrm{seconds}$ or 84.47 minutes. This tells that the body will pop-up on the other side after 42.235 minutes after it was released from one side and will return back after another 42.235 minutes.

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1 comment
1. Vir Narayan Singh says:

The analysis that has been given above for the time taken by the ball for a roundtrip in the hole through the earth is true only for homogeneous earth, that is, same density (5517 kg/m^3) everywhere. However, THE EARTH IS NOT HOMOGENEOUS! Its density varies from 3350 kg/m^3 near the surface to 5560 kg/m^3 at a depth of 2891 km, jumps to 9900 kg/m^3 at this depth and increases to 13090 kg/m^3 at the centre of the earth. As a result, the value of g remains in the neighbourhood of 9.8 m/s^2 upto 2891 km depth, rises to 10.68 m/s^2 at that depth and then decreases to zero at the earth’s centre (according to Preliminary Reference Earth Models). I have performed the same analysis using a different approach, obtained a value of about 5060 seconds for the homogeneous earth and 458 seconds for the inhomogeneous earth. The value of 5060 is close to the value 5068 seconds obtained by you. This can be verified easily.

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