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Let’s say, as a thought experiment, that we drilled a hole straight through the Earth and out the other side. Let’s ignore things like heat and what a massive undertaking this would be.

We would have a hole with no bottom. If you could look through the hole and somehow see out the other side clearly, you would see the sky.

We then drop an object into the hole. Does the object fall all the way to the center and out the other side at an extreme speed due to the acceleration of gravity? Or does it fall to the center, rise “up” the other way, only to fall back “down” again? (and if so, how long would that yo-yo action continue for?)

Or would something else happen? How does gravity work in this situation?

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  • $\begingroup$ To clarify, do you know what differential equations are? $\endgroup$
    – Philip
    Aug 15 '20 at 18:13
  • $\begingroup$ I do, but I neither need nor prefer the answers to be too mathematical. I'm just looking for an informal explanation! :) $\endgroup$
    – Michael
    Aug 15 '20 at 18:22
  • $\begingroup$ I would suggest you to watch this youtu.be/urQCmMiHKQk $\endgroup$ Aug 15 '20 at 18:41
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/7346/2451 and links therein. $\endgroup$
    – Qmechanic
    Aug 15 '20 at 20:39
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As the object falls into the tunnel there is no longer the entire mass of the Earth attracting it, instead the part behind it (closer to the surface) is pulling it backwards. So the acceleration rate would drop as it approached the center. It would whiz past the center then gradually slow down as past until it came to a stop near the other side, though not quite reaching the surface.

Note - this assumes you have a vacuum, otherwise the wind resistance will slow it down.

One other thing I'd like to point out - if you drilled a hole through the Earth, forgetting about magma, you would have a huge amount of pressure on that tunnel form all the rock along it's walls. The tunnel would collapse.

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  • $\begingroup$ Are you saying that the object would pass through the center and then gradually slow down until it came to a full stop in midair, though not quite reaching the surface? Wouldn't it then fall back "down"? I am ignoring magma, wind, etc. Really spherical-cowing here :) $\endgroup$
    – Michael
    Aug 15 '20 at 18:20
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    $\begingroup$ It would oscillate in simple harmonic motion, like a mass on a spring :) $\endgroup$
    – Philip
    Aug 15 '20 at 18:22
  • $\begingroup$ Very cool, thank you! $\endgroup$
    – Michael
    Aug 15 '20 at 18:23
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The other answer is correct, but it turns out that it becomes a little interesting if you work out the mathematics. I'm not sure how much you know of differential equations, but this is a fun problem to solve (with certain assumptions).

So let's consider an object falling through the Earth along the $y-$axis. At some arbitrary point (at a distance of, say, $y$) from the center, what will the force on it be? Well, it has mass all around it. It will be pulled towards the centre by the mass "below" it, and pulled back up by the mass "above" it. This seems like a complicated problem to solve, but we have an interesting theorem that can help us: Gauss's Law (for Gravity).

We can use this law by first drawing a closed sphere of radius $y$ from the centre of the Earth (the object will therefore lie on the surface of this sphere). The law states that the force experienced by the particle will just be:

$$\mathbf{F} = - G\, \frac{m_\text{obj} M_\text{enc}}{y^2}\mathbf{\hat{y}},$$

where $M_\text{enc}$ is the mass enclosed by the sphere. Obviously, as the object falls deeper and deeper into the Earth, this "enclosed mass" becomes smaller and smaller (since $y$ is reducing) and so, so is the force! Keep in mind that this law takes into account the effects of all the mass (even the stuff "above" the object you've dropped. It just turns out magically (mathematically?) given the symmetry of the problem to only depend on the enclosed mass.)

Using this, we can actually determine how the object will "move". The first thing to do is to find a general formula for the enclosed mass. I'm going to make the following assumption: the density of the earth is a constant $\rho$. It's not too bad an assumption, frankly, and no less realistic than actually drilling a hole through the Earth!

It should be easy to show that the mass of the Earth is related to its radius $R_E$ by:

$$M_\text{E} = \frac{4}{3}\pi R_E^3 \rho,$$

just as the enclosed mass is related to $y$ by

$$M_\text{enc} = \frac{4}{3}\pi y^3 \rho.$$

Given that we're assuming the density to be constant, it's very easy to show that

$$M_\text{enc} = M_E \left(\frac{y}{R}\right)^3,$$ which makes sense in all the limits.

We can now just plug this into Newton's Second Law to find the acceleration of the object: $$a = \frac{F}{m_\text{obj}},$$

and it's easy to show that

$$a = \frac{\text{d}^2 y}{\text{d} t^2} = -\frac{GM_E}{R^3} y.$$

This should be an equation that any first year physics undergraduate should recognise in their sleep, it's the equation for Simple Harmonic Motion!

So in other words, the object will oscillate in SHM, with a frequency given by $$\omega^2 = \frac{G M_E}{R^3},$$

and an amplitude $R_E$ (given that you dropped it from the surface on one end at rest). You could have found the time period using dimensional analysis as well, but I think it's cool to see this as an example of how ubiquitous the equation for SHM is in physics!

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Once an object goes below the surface of Earth the net gravity acting on it is the radius of the mass from the object to the Earth's center. As the object goes deeper the radius gets smaller, and the object has less net gravity pulling it downward. At the Earth's center there would be no net gravity from the Earth. Assuming you had a maintainable clear path through Earth, with no air resistance, then the object accelerate to the center. Once it's inertia carried it through the center of Earth the net gravity acting on it would slow it until it started falling back through the center. It would ten oscillate falling back and forth.

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