# Dig A Hole To the Other Side

by oblivinated
Tags: hole
 P: 5 Assume you could make a tunnel through the center of the earth so you could go from one side of the planet to the other. Assume that air resistance is very minimal and that you don’t burn up due to the intense heat and radiation; describe your fall through the tunnel if you jumped into the hole on one side of the planet. Our physics Professor asked us this bizarre question and I have been pondering it ever since. Wouldn't the answer depend on the radius of the hole? My own odd logic thinks that you would be able to travel through the hole without any resistance, in fact, since there are large masses to the sides of you (and therefore gravity pulling from two sides), you would levitate from the sides which would put you right in the center of the hole. Anyone have any ideas? Thanks in advance everyone!
 Mentor P: 27,565 I'm convinced I've seen the question asked a few times on here. If you assume that the earth is a sphere with uniform mass distribution, then the dynamics of a body thrown into that hole will be identical to a simple harmonic oscillator. This is because, if you solve for the Gauss's Law equivalent of the gravitational force that the body experience, you'll get a force proportional to the distance from the center, i.e. of the form F = - kr. This is identical to the harmonic mass-spring system. Zz.
 Sci Advisor P: 2,501 I think the big thing your prof was probably trying to get you to think about was something to do with the Earth's rotation. Hint; think about the hole going from one pole to the other, then try it at the equator.
Mentor
P: 4,203

## Dig A Hole To the Other Side

 Quote by LURCH I think the big thing your prof was probably trying to get you to think about was something to do with the Earth's rotation. Hint; think about the hole going from one pole to the other, then try it at the equator.
I think he was trying to drive at the harmonic oscillator point personally
P: 15,294
 Quote by oblivinated since there are large masses to the sides of you (and therefore gravity pulling from two sides), you would levitate from the sides which would put you right in the center of the hole.
This is incorrect logic. The mass around you does not keep you centred, it cancels out. You would move side-to-side as if there was no mass around you. (Note: we are not looking at movement vertically along the tube here, only horizontally within the tube). Nonetheless, this is just getting hung up on detail. The question is clearly about your movement vertically.

I think Lurch Zz and O_S have good points. It is not clear what your prof was driving at. One aspect could be about your motion along the axis of the hole and another would be about what happens if you account for the Earths' rotation.

However, you can answer his question by stating your assumptions up front, or by including all conditions (as in Lurch's post).
 P: 154 I jotted down kepler's law and replaced the earths mass with its mass as a function of its radius (since during your fall, the only mass pulling you is that which is contained in a sphere whose radius is equal to your distance from the center). i also took earths density as a constant. my result is that your acceleration is linearly proportional to your distance from the center of the earth. a = 4/3*G*pi*ro*r a = acceleration G = gravitational constant pi = 3.14159 ro = earth density r = distance from center
 P: 5 So basically you would fall from one side of the hole to the other, having the greatest velocity at the center of the earth while slowing down as soon as you past the center. This motion would continue on forever until you grab onto something or something breaks your "fall". Hopefully I got the main idea correct! Thanks for all the wonderful responses
 P: 12 Id like to point that Newton stated clearly that the force of gravity doesnt depend on the amount of mass but on distance to the center squared so in the equator though theres more mass and radius distance to the center is bigger the force of gravity is weaker, this can be checked in reality On the other hand i have two toys able to make ellipses, a ball with an elastic string and a yoyo If i make ellipses with the string and ball the center of the ellipse or orbit would be at the center of said ellipse Instead a looping yoyo behves like a planet orbiting the hand or sun the hand is at the focus of the ellipse So if yo throw a yoyo up and let it fall when passing close to the focus will rebound in oposite sense So my answer to your teacher would be that once you reached the center of the earth you would rebound like a looping yoyo
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P: 27,565
 Quote by giordano bruno Id like to point that Newton stated clearly that the force of gravity doesnt depend on the amount of mass but on distance to the center squared so in the equator though theres more mass and radius distance to the center is bigger the force of gravity is weaker, this can be checked in reality
This is incorrect. It is only true for r>R, where R is the radius of the earth. When r<R, you have to do the Gauss's Law equivalent to solve for the potential and the gravitational force.

Zz.
 P: 14 As ZapperZ says 'This is identical to the harmonic mass-spring system.' I think it is the summary of the answer.
 P: 12 The gravity measures Im talking are taken at 0 level with respect to the sea and is proved that in the equator the gravity is lower than in the poles in this case r=R not r bigger than R
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P: 27,565
 Quote by giordano bruno The gravity measures Im talking are taken at 0 level with respect to the sea and is proved that in the equator the gravity is lower than in the poles in this case r=R not r bigger than R
Then you must be somewhere else in laa laa land, because you are not addressing the original question. Please do not hijack a thread with non-relevant post.

Zz.
Mentor
P: 40,276
 Quote by giordano bruno The gravity measures Im talking are taken at 0 level with respect to the sea and is proved that in the equator the gravity is lower than in the poles in this case r=R not r bigger than R
This has nothing to do with the topic at hand, which is about something falling through a hole drilled through the earth. You are worrying about deviations in field strength at the earth's surface--not relevant to this exercise. You should be considering the change in field strength from the surface to the center of the earth, which is linear with distance from the center as Zapper described.
 P: 689 if there is rotation, the earth might push you from the side, resulting in an "apparently" upward force (centrifugal force) in that frame of reference... the math could get complicated, but it is still doable.. other than that... harmonic motion should be the answer.

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