What if it was possible to ride an elevator straight through earth. If we take all the heat and pressure problems away, and assume it would be possible: What would happen with gravity? Would gravity suddenly turn around as i pass through the center and hit my head into the roof? Or would i gradually get weightless as i descend?


This is not a duplicate. The other questions does not involve an elevator car. The other ones are about jumping into it.


4 Answers 4


Let's put a more precise description to the other answers, particularly Neil's.

First, note that there is a Gauss Law for static gravitational fields, owing to the inverse square nature of the static gravitational attraction. See this answer here and note that the argument it makes uses only the inverse square dependence. (Actually, the Gauss law also holds for dynamic gravitational fields in the approximation to General Relativity called gravitoelectromagnetism, but that is another story).

So now we apply the integral form of Gauss's law to the Earth, whose mass distribution is very nearly perfectly axisymmetric, i.e. depends on the distance $r$ from the Earth's centre. Thus, by symmetry and Gauss's law, we know that the gravitational field at a distance $r$ from the centre is the same as that arising from a point mass whose mass equals the total mass enclosed within a sphere of radius $r$. Thus if the density as a function of radius $r$ is $\rho(r)$ we have:

$$g(r) = 4\,\pi\,\int_0^r \,u^2 \,\rho(u)\,\mathrm{d} u\,\frac{G}{r^2}$$

where the field is of course always directed towards the Earth's centre.

Now the form of $\rho(r)$ is highly nontrivial, being determined by the different materials at different depths and the response of that material to pressure as described by e.g. the Adams-Williamson Equation.

But if we idealise the Earth so that $\rho(r) = \rho_0$ we get:

$$g(r) = \frac{4}{3}\pi\,G\,\rho_0\,r$$

so that if we, like Alice, dropped through an ideal diametrical tunnel through the Earth we would undergo simple harmonic motion with:

$$\ddot{r} = - \frac{4}{3}\pi\,G\,\rho_0\,r$$

or, in terms of Earth radius $R_\oplus$ and the value of $g_\oplus= 9.81{\rm m s^{-2}}$ at the Earth's surface:

$$\ddot{r} = - g_\oplus\frac{r}{R_\oplus}$$

so that our period is:

$$T = 2\pi\sqrt{\frac{R_\oplus}{g_\oplus}} \approx 5\,075{\rm s}$$

and it would take us about 21 minutes to fall to the centre of the Earth, whence we would keep going to the other side, and then fall back and forth sinusoidally with time.

  • $\begingroup$ Would the increase of density physics.stackexchange.com/q/99117 modify significantly the oscillation? $\endgroup$
    – jinawee
    Feb 18, 2014 at 20:28
  • $\begingroup$ @jinawee That's REALLY interesting data, thanks, and moreover it's quite a departure from the linear dependence on $r$ I used. I think I'll do a numerical integration sometime soon if I get the chance and edit the answer. I do recall the figure 20 minutes to fall to the Earth's centre, but that may well have been worked out using the same el dumbhead linear model I used above. $\endgroup$ Feb 18, 2014 at 21:31
  • $\begingroup$ It seems that someone has done the calculation: arxiv.org/abs/1308.1342. You get a couple of minutes less. $\endgroup$
    – jinawee
    Feb 28, 2015 at 9:26
  • $\begingroup$ @jinawee Interesting - that's an eerie co-incidence. Only yesterday I found out where I got twenty minutes from - my own calculation the same as the above. I was reading my daughter "Alice in Wonderland" a few years back, and, when Alice wondered how many miles she had fallen and whether she would come out the other side of the Earth, my daughter wanted to know how long it would take. She wouldn't stop interrupting until I had calculated a figure for her. Reading Alice to my little boy yesterday, I came across my calculation on a piece of paper stuck in Chapter 1 of Alice. $\endgroup$ Feb 28, 2015 at 9:38

As you already suspected, your weight would gradually decrease as you descend into earth. Technically, at the center of the earth you should feel no gravitational force (from earth, I don't know how strong additional sources of gravity affect us, but my guess would be that they are negligible). Once you start "rising" on the other side, gravity would slowly start to increase in the opposite direction.

By no means is the transition so harsh as depicted in the movie "Total Recall" (the new version with Collin Farell). There, the through-the-middle-of-earth elevator suddenly drops into a gravity-free zone once they enter the core and gravity reinstates suddenly when leaving it. This is fully made-up movie physics!

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    $\begingroup$ It's worth mentioning that in that movie, it appears as if the elevator is supposed to be in free-fall the whole time. That is, they remove the supports in England and it just keeps falling until it gets to Australia. That would work (assuming no air in the tunnel and no friction), but you would be weightless for the entire time, since you're falling at the same speed as the elevator. $\endgroup$
    – N. Virgo
    Feb 18, 2014 at 9:57
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    $\begingroup$ Additionally the Earth rotates fast enough for you - unless the hole goes from pole to pole - to move away from the center of the shaft and into the sides as your angular velocity stays unchanged but the Earths gets slower the closer you get to the core (where it is zero). As the angular velocity at the surface of the earth at equator is about 1000 miles/hour it might be a bit unpleasant :) answers.yahoo.com/question/index?qid=20100504190650AAIHTZg $\endgroup$ Feb 18, 2014 at 13:13
  • $\begingroup$ So you telling me that the movie physics is a lie? My life is ruined... $\endgroup$
    – John Odom
    Feb 18, 2014 at 16:41

Please take a look at this link Hole through the earth. It seems that the physics of passing through the earth would not be as it was implied by Total Recall in which gravity switches suddenly. While gravity is normally affected by radius squared (gravity much stronger proportionally as you approach a gravitational body), inside a gravitational body, its pull would be linear, not quadratic.

What does this mean? Well as you would approach the center of the earth, the gravitational pull would fade linearly. At half the distance to the center, the gravitational pull would be halved. At one-fourth the distance to the center, the gravitational pull would be one-fourth that of normal earth gravity, et cetera.


Here is my answer i figured out earlier when i posted this question (with no answer)

If you took an elevator through the earths core: Gravity would gradually feel weaker and weaker, and you would feel more and more weightless as you close in on the core. Because in the core: The mass of the earth is more or less equal in all directions, and it would "pull" you from all directions. As you climb back up after passing the core: You would gradually have less mass above you and more beneath you, and therefore: The gravitational pull would be stronger in the direction with more mass. And you would feel a stronger and stronger pull as you climb to the surface.

  • $\begingroup$ @dj_mummy Wow.. Thats interesting. Now, lets say that the elevator runs on a motor and wheels and moves at 200 meters per minute: How long would the journey be approximately? I am embarrassingly terrible at math. $\endgroup$ Feb 18, 2014 at 12:28

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