# In the tunnel through the Earth thought experiment, how does total energy of a particle vary when it goes from surface of the Earth to its center?

In the typical tunnel through the center of the Earth thought experiment, a particle at the surface would oscillate between the two diametrically opposite points ignoring the rotation of the Earth. The thing I do not understand is how the conservation of energy happens if we assume the potential energy is negative at the surface of the Earth.
In the picture attached, I took four points: A at a height h from the surface of the Earth, B at the surface of Earth, C at the center of the Earth, and D at some distance r (e.g. R/2) from the center of the Earth. I can understand the conservation of energy from A to B, but as soon it enters the Earth, its total energy starts increasing. Is there some concept I am misinterpreting or missing?
I have also created a graph of the P.E. on the right according to the formulae.

• gravitational potential energy will decrease as one approaches towards centre of earth.Every objects wants to attain lowest energy thats why they fall. – Who Jun 7 at 11:23
• The $GMm/r$ formula only applies to two non-intersecting spheres. It doesn't work if one body is in a hole in the other. – Peter Shor Jun 7 at 11:25
• Yeah, I made that mistake. Thanks for clearing that – yavvee Jun 14 at 18:12

## 2 Answers

At any point (such as your D) inside the Earth, only shells of matter between D and C contribute to the field strength at D. Each of these shells behaves as if all its mass were situated at C. Thus assuming uniform density, $$\rho,$$ (not a very realistic assumption), the field strength at D is $$g=-\frac{G \rho \tfrac43 \pi r^3}{r^2}=-\tfrac43 \pi G \rho r.$$ The minus sign indicates that the field strength is towards the centre of the Earth. It is clearly also proportional to $$r,$$ so we have an shm set-up!

The work done taking a mass $$m$$ from the centre of the Earth to D is $$E_p=-\int_0^r gm dr=\int_0^r \tfrac43 \pi G \rho rm dr=\tfrac23 \pi G \rho mr^2$$ This is the potential energy of $$m$$ if we take our zero of potential energy to be at C. [It is usual with shm to take the zero of potential energy at the equilibrium position, though it's not essential to do this, as only changes in potential energy are significant.]

Now to answer the question posed in your title, the total energy is constant (ignoring resistive forces). As the potential energy goes down the kinetic energy rises, as in any case of simple harmonic motion of a single body.

• As to your last sentence: it’s worth adding that SHM isn’t special and the total energy remains constant in any motion at all in a gravitational field if one ignores friction and relativity. That’s one more thing for the OP to remember in the future and doesn’t detract in the least from your splendidly clear answer. – Martin Kochanski Jun 8 at 5:34

Your graph of PE is wrong. If the zero of PE is at infinity then in the center it has a negative value, $$-\frac{3}2 \frac{K M m}{R}$$. On the surface is $$- \frac{K M m}{R}$$. So PE keeps decreasing as you go deeper inside. And so the KE increases, reaching the maximum value in the center. Maybe you saw this graph (the one in your notes) representing the force and you confused it with the graph for PE. And from this arrised your question.

• Yeah, I actually assumed the formula for PE outside the Earth to be applicable inside the Earth too. My mistake. Thanks for clearing the confusion! – yavvee Jun 14 at 18:10