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We have an object of mass $m$ which can go through matter like a ghost. It's only affected by gravity, not by any other force, so it's in free fall.

At the instant $t=0$ it is at the surface of the Earth which is assumed to be a perfect sphere of radius $R=6.371 \times 10^6$ m, of mass $M=5.972 \times 10^{24}$ kg and of uniform density $\rho=5.515 \times 10^3$ kg/m$^3$. $G=6.674 \times 10^{-11}$ m$^3$ kg$^{-1}$ s$^{-2}$.

The object has an initial velocity $v_0$ directed precisely at the center of the Earth. Unlike previous questions the velocity need not be zero.

My question: depending on the variables $m$ and $v_0$, how much time will it take to reach the opposite side of the Earth?

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  • $\begingroup$ As soon as it goes past the centre, the gravitational force would pull it back towards the centre, so you would get some sort of simple harmonic motion occurring. DOnt think it would reach the other side. $\endgroup$ – surelyourejoking Feb 17 '15 at 12:26
  • $\begingroup$ I don't think it will be in SHM, since in SHM the restoring force increases with the distance from the mean position, whereas the gravitational force decreases? $\endgroup$ – Hritik Narayan Feb 17 '15 at 12:33
  • $\begingroup$ @HritikNarayan Gravitational force increases linearly with distance from the center so long as you remain inside the gravitating body (assuming constant density). It only starts decreasing with the inverse square law outside the body. So inside Earth, this absolutely would be SHM $\endgroup$ – Jim Feb 17 '15 at 12:35
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    $\begingroup$ But even if it had a zero initial velocity, its velocity would increase during the first half of the route. Then, once it gets pass the center of the Earth, gravity would slow it down, but since its velocity would have been previously increased it should be able to reach the opposite side of the Earth. $\endgroup$ – Quantum Force Feb 17 '15 at 13:01
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    $\begingroup$ No you are wrong. SHM is defined by the fact that the restoring force is proportional to the displacement. The initial conditions are irrelevant. What is relevant is that you probably should not assume that the Earth has a constant density (depending how accurate you want the answer). It is not such a terribly wrong assumption if you are interested in getting it to say the nearest 5 minutes. $\endgroup$ – Rob Jeffries Feb 17 '15 at 14:20