# Newton's Laws with one body inside another

Hello All

What would be the motion of a weight dropped into a hole drilled all the way through an Earth-sized planet?

Would the weight accelerate all the way to the centre and then decelerate until it got to the far side, and repeat this motion forever?
OR
Would it accelerate initially and then start to decelerate as more of the planet was "behind" it and therefore slowing it down, and come to rest gently at the centre?
OR
Something else?

Assume the planet isn't spinning and the hole is in a vacuum.

Best regards. Stef

2022 Award
If you truly want answers at A level this is something you should have no trouble doing for yourself - it's a textbook application of Gauss' Theorem. I suspect you want to pick I level.

In idealised frictionless circumstances the weight accelerates towards the core and decelerates once it passes. The acceleration profile depends on your model of the Earth.

The application of Gauss' theorem that I mentioned is to use it to realize that (for a spherically symmetric mass distribution) the mass outside the radius you are at is irrelevant. The gravitational acceleration is therefore ##GM(r)/r^2## where ##M(r)## is the mass of the Earth inside radius ##r##. If you assume constant density this is just the density times the volume of a sphere of radius ##r##. A realistic density profile requires some integration. However, note that ##M(0)=0## for any mass distribution, so the acceleration always falls to zero at the core.

Many thanks for your answer. Yes I was assuming that the planet had a uniform mass distribution.

Does the weight therefore stop accelerating at the centre and starts to decelerate until it stops at the opposite opening and repeats the process ad infinitum?

• sophiecentaur
2022 Award
In idealised frictionless circumstances, yes. The acceleration may increase for some of the fall (depending on the mass distribution, and not in the case of a uniform mass distribution) but will always drop to zero at the core. Spherical symmetry means that the deceleration profile is the reverse of the acceleration profile and the weight must come to rest at the surface again.

Gold Member
Yes I was assuming that the planet had a uniform mass distribution.
And that results in a simple harmonic motion because the 'restoring force' towards the centre of the Earth is proportional to the displacement from the centre. Practicalities like enormous temperatures and pressures as you go down (and the Earth's rotation), mean that the experiment is a non-starter BUT: the time period for a simple harmonic oscillator is independent of amplitude so you could repeat the experiment for real if you got a spherical asteroid with a hole through it and a small rock. (the outside bits of the Earth don't count).

Interestingly, the period is (nearly) the same as the period of a satellite in low Earth orbit.

Edit PS there is a vast amount of information about SHM but it nearly all gets Mathematical very quickly. I suggest you do a search and find yourself a hit that suits your level.

Just imagine a future Richard Branson setting up such an extravagant experiment just for fun and to show how rich he is.

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Richard R Richard
The period is a constant to cross the planet from side to side, it has been calculated in 42 minutes, both to cross any string and to cross the diameter.
By the way, another idealization that has to be abstracted is that the planet should not rotate, and if it does, it will only reach the other ends without colliding with the walls if the hole coincides with the axis of rotation.