Gravity and bottomless pits [duplicate]

Assuming that someone is theoretically able to make a hole through the center of a large planet, and then jumps down the hole, what will happen? Given my understanding of gravity and energy, my estimate is that in the absence of any resistive forces, the person would fall right through to the other side, with at least part of their body coming through, and would continue oscillating like this. If there were resistive forces, such as air resistance, the person would oscillate similar to a spring winding down, eventually stopping in the middle of the planet. Is this what would happen, or are my calculations off somewhere?

If you use Gauss's theorem for the gravitational field of the Earth at radius $R$ and assume it has uniform density $\rho$. $$g(R) = -GM(R)/R^2 = -4\pi R G \rho /3,$$ where $M(R)$ is the mass inside radius $R$, and is given by $4\pi R^3 \rho/3$.

Thus the force the body feels towards the centre of the Earth (neglecting air resistance - see below) is proportional to $R$ and thus $$\frac{d^2 R}{dt^2} = -\frac{4\pi G \rho}{3} R,$$ which is a simple harmonic motion equation with an oscillation frequency of $(4\pi G\rho/3)^{1/2}$.

If I assume an average $\rho =5000\ kg/m^3$, this means an oscillation period of 88 minutes.

If you really did hollow out such a tunnel, it would quickly fill with air. You would reach a terminal velocity of order 200 km/h at best. Travelling at this sort of speed as you went through the centre of the Earth would mean you would only rise by a height on the other side given approximately by $$\frac{1}{2} m v^2 = m \int_0^{H} g(R)\ dR = m \frac{2\pi G \rho H^2}{3}$$ $$H = \left(\frac{3}{4\pi G \rho}\right)^{1/2} v$$ Putting in some numbers then gives $H \simeq 50\ km$. So nowhere near emerging at the antipodes.

• That's a neat way to analyze this situation. The harmonic oscillator part is only true in the approximation that the density of the Earth is constant. Jan 17, 2015 at 19:23
• Yes, it becomes anharmonic otherwise, but I would be surprised if there is a massive change from the frequency that I calculate. Jan 17, 2015 at 19:25
• Basically the same period as for low earth orbit. Not a coincidence, I think? Jan 17, 2015 at 19:41
• Jan 17, 2015 at 20:11
• @200_success It is not a coincidence, the angular frequency of an orbit skimming the surface of a body has the same formula: $\omega=\sqrt{\mu/a^3}=\sqrt{4/3\pi G\rho}$ Jan 17, 2015 at 20:17

Ignoring the engineering difficulties of constructing a tunnel through the molten core, and ignoring the effects of the rotation of the Earth which would complicate things, yes, you are correct - you would fall through and come to a stop on the other side, then go back the other way. If you add in resistive losses, you would go a little less far each time, until you came to rest at the center.

Interestingly, the trip to the other side in the absence of resistive forces would take just about 42 minutes! This mode of travel is known as a gravity train.

In fact, if you dig a tunnel from one point on the Earth to another along a particular type of curve called a hypocycloid (of which the diameter of the Earth is a degenerate example), it would take 42 minutes to get from any one point on the surface to any other.

You are correct about what you believe would happen.

Here's an important fact:

Consider yourself positioned at a random location within a solid planetary sphere. Let that point define a radius from the center of the sphere to you. Any matter that lies beyond that radius has a ZERO net gravitational pull on you. In other words, that mass does not matter at all to you. Only the matter within that radius has a net gravitational pull on you.

Also, if the object was a spherical shell and you were ANYwhere inside that shell, you would just float around. Your location within that shell wouldn't matter at all. Yes, you could be closer to one section than another, but, the closeness to the smaller amount of mass would be exactly equalized in force by the much greater amount of mass further away from you. Calculus easily proves this with either double or triple integrals.

Please don't jump in the hole. Your velocity would be tremendous as you reached the center. One brush against the side of the hole and you'd get a heck of a burn.

• Assuming there is air in the hole (a good assumption), you'd still be fighting against air resistance the same as anywhere, so the velocity nearest the center would probably be terminal velocity. Jan 17, 2015 at 20:58
• And if the hole is a decent diameter (say, 50 meters across) any competent skydiver could stay away from the walls.
– paul
Jan 18, 2015 at 1:39