Falling through the rotating Earth

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.

• The usual problem is with a hole through the center and then the motion is oscillatory physicscentral.com/explore/poster-earth.cfm . yours is assuming a material not interacting electromagnetically . The velocity due to the rotation is too small,imo, to avoid ending going in a more complex trajectory oscillating about the center , from surface to surface across. Feb 2, 2013 at 7:16
• You will love the recent movie "Total Recall", don't miss it. They travel that way between UK and Australia, although they use some mechanism to add extra acceleration for part of the way, but they are in free fall across the Earth core. Feb 2, 2013 at 8:54
• Are you a president? Feb 3, 2013 at 6:01
• May 14, 2018 at 22:24

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.

• For a quadratic potential, the orbits are ellipses, with the center of the ellipse at the point of attraction and not the focus; see Wikipedia. This explains why you always emerge at the opposite point. In fact, quadratic potentials just give a classical harmonic oscillator, so the 42 minutes time to reach the other side is independent of the latitude. (To me, this sounds like it might have been a homework problem.) Feb 3, 2013 at 13:53
• @PeterShor Thanks! I did get to the Wikipedia page on Bertrand's theorem (en.wikipedia.org/wiki/Bertrand's_theorem) about half an hour after finishing my fight above with polar coordinates... Feb 3, 2013 at 15:12
• @PeterShor It's actually not a homework problem and neither was my toilet roll unrolling question that was closed on that basis. If a problem I'm considering would make a good homework problem, does that mean it shouldn't be answered? Feb 3, 2013 at 18:56
• @Barack Obama: no, it should be answered (and it was). Feb 3, 2013 at 19:19
• Not that it matters much, but if you plug in the parameters for the closest approach to the center of the earth, you don't get something close to R/2, but, instead something less than R/10. See linked question on imaginary neutrino balls. May 15, 2018 at 1:04

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.

• A rather superficial treatment. Feb 2, 2013 at 20:17
• Pretty sure the link you provided doesn't take into account the rotation of the Earth. Feb 3, 2013 at 6:02

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.

• Ps. I'm guessing you're a dark matter being and I think the Earth could capture you under the right circumstances;-) Feb 3, 2013 at 11:36