Falling through the rotating Earth

Question

Suppose you were standing on the rotating Earth (not necessarily Equator or the poles) and suddenly your body lost the ability to avoid effortlessly passing through solid rock.

Because the earth's rotation at the surface is considerably below escape velocity, you would slip below the earth's surface. If the earth's gravity were a consequence of a central point mass, you'd have an elliptical orbit (mostly) within the earth.

With a planet of constant density, the gravity you feel underground is equivalent to standing on the surface of an identically dense planet with a radius equal to your current distance from the centre. So effectively, as you fall the gravity you experience lessens.

1. What would be the shape of the trajectory?

2. How close would you get to the centre?

3. How long would it take before your orbit brought you back to the surface (assuming no losses and a stationary planet)?

4. What complications would arise from the planet being in orbit around a star?

5. Bonus points for finding two well-known places on the Earth that you could travel between in one "orbit" using this method.

Answer 1

The force you experience is of the form $\vec{F} = - Gmr\vec{u_r}$, and we also know that in the surface, $r=R$, it is $\vec{F}=- gm\vec{u_r}$, so

$$\vec{F} = -gm\frac{r}{R}\vec{u_r}$$

This is a conservative force that can be derived from a potential

$$U = \frac{1}{2}gm\frac{r^2}{R}$$

Because this is a central force, angular momentum is conserved, so $r^2 \dot{\theta} = L$, and if $\Omega$ is the rotational velocity of earth,

$$r^2 \dot{\theta} = R^2 \Omega$$

And of course we have conservation of energy,

$$\frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2)+\frac{1}{2}gm\frac{r^2}{R} = E$$

but we also know the initial conditions, $r=R$, $\dot{\theta}=\Omega$, $\dot{r}=0$, so

$$E = \frac{1}{2}m(R^2\Omega^2)+\frac{1}{2}gmR$$

and conservation of energy can be rewritten as

$$\dot{r}^2+r^2\dot{\theta}^2+g\frac{r^2}{R} = R^2\Omega^2+gR$$

and including conservation of angular momentum as

$$\dot{r}^2+ \frac{R^4 \Omega^2}{r^2}+g\frac{r^2}{R} = R^2\Omega^2+gR$$

If you set $\dot{r} = 0$ and solve for $r$, there are two solutions, marking the annular region in which motion will happen. One is the obvious $r=R$, the other comes out to

$$r = \Omega R \sqrt{\frac{R}{g}}$$

which with the Earth parameters at the equator, comes out to $r=3740\ \mathrm{km}$.

You can rearrange the equation of energy as

$$\frac{dr}{\sqrt{R^2\Omega^2+gR - \frac{R^4 \Omega^2}{r^2}-g\frac{r^2}{R}}} = dt$$

which you could integrate to get a probably implicit relation between $r$ and $t$, which you could use in the conservation of angular momentum to get $\theta$ as a function of $r$ and/or $t$.

I have done that numerically, and again, for the point on the Equator, it would take about 21 minutes to reach the point closest to the Earth's center, and 21 more to get back up at the surface.

One neat result I don't fully understand where it comes from is that, at the minimum point, the angle $\theta$ has changed by $\pi / 2$, independently of what the rotation speed is, so that you always emerge at a point opposite where you went down. Since the Earth is rotating, you wouldn't actually come out at the antipodal point, but some $1175\ \mathrm{km}$ from it.

Away from the equator you would have a reduced $\Omega$, and the movement will happen in a plane perpendicular to the meridian going through that point.

Answer 2

This is a pretty fun topic of classical mechanics. You should check this article out: here It has a really detailed analysis of how to calculate the time it takes to fall through the earth.

Answer 3

I don't think you would come out directly opposite. This is due to the Earth having the Moon orbiting around it. There are tides in rocks to think about too.