<p>The formula you are using is for a point mass, but a hole through the Earth experiences less gravity than a point mass, as the comment points out.  One can then ask, what if the Earth's mass were all at the center, why wouldn't the mass fly through and go to a distance r on the other side?  It's because a mass like that is a singularity, one that doesn't matter unless you fall right through it.  If the Earth's mass was really concentrated at a point, then falling from r would still tend to just miss that point, since it's just a point anyway, and the huge forces would turn the orbit around, such that the object fell from r, went to zero, turned around, and flew back out to r.  That's a very squeezed ellipse, with a=r/2, as desired.  The motion that goes straight through and out to the opposite r, on the other hand, is not an ellipse with the mass at one focus, so it violates Kepler's laws.  Again, the violation of Kepler's laws is that the mass is singular in a way that matters if you fall through a hole in it.  Interestingly, it doesn't matter how concentrated the mass is toward the center-- if there's a hole in it that the object falls through, then the object does not obey Kepler's laws.  It will only obey Kepler's laws if the orbiting object does not enter the mass.</p>

CC BY-SA 3.0
2017-01-30T15:55:41.303Z
2017-01-30T16:01:14.187Z
2017-01-30T16:01:14.187Z
Ken G
<p>I have a problem: If I would dig a hole through the earth, and jump right in, disregarding the rotation of the earth, I should come back out on the other side.</p>

<p>However, orbital mechanics say otherwise: Specific Orbital Energy tells me:</p>

<pre><code>v^2/2 - µ/r = -µ/2a
</code></pre>

<p>Now, on the planet, my relative velocity is 0 m/s, so v^2/2 = 0. The term simplifies to:</p>

<pre><code>-µ/r = -µ/2a | / (-µ)
1/r = 1/2a | ^(-1)
r = 2a | /2
a = r/2
</code></pre>

<p>My apoapsis is 3000 km by this equation. But it should be 6000 km, since I come out of the other side.</p>

<p>Why is this, and did I do a wrong assumption?</p>

CC BY-SA 3.0
2017-01-30T15:07:48.437Z
I'm voting to close this question as off-topic because it does not appear to be about Astronomy, and it looks like a homework question.
Dean
CC BY-SA 3.0
2017-01-31T02:50:19.787Z
Small point, while @barrycarter is correct with a uniform mass, the layers of increasing density in Earth, gravity initially drops then it slowly increases before dropping again.   It's greatest just outside the outer core.   https://newtonexcelbach.wordpress.com/2010/06/29/the-hole-through-the-middle-of-the-earth-accelleration/
userLTK
CC BY-SA 3.0
2017-01-31T12:25:01.637Z
This is a very standard problem with solutions all over the internet.
ProfRob
<orbit><earth><astrophysics>
Hole through the earth
Max Willich
# Hole through the earth

(Asked by: username_0 on 2017-01-30)

I have a problem: If I would dig a hole through the earth, and jump right in, disregarding the rotation of the earth, I should come back out on the other side.
However, orbital mechanics say otherwise: Specific Orbital Energy tells me:
v^2/2 - µ/r = -µ/2a

Now, on the planet, my relative velocity is 0 m/s, so v^2/2 = 0. The term simplifies to:
-µ/r = -µ/2a | / (-µ)
1/r = 1/2a | ^(-1)
r = 2a | /2
a = r/2

My apoapsis is 3000 km by this equation. But it should be 6000 km, since I come out of the other side.
Why is this, and did I do a wrong assumption?

**Tags:** orbit, earth, astrophysics

> **Comments:**
> **username_1**: I'm voting to close this question as off-topic because it does not appear to be about Astronomy, and it looks like a homework question.  
> **username_2**: Small point, while @barrycarter is correct with a uniform mass, the layers of increasing density in Earth, gravity initially drops then it slowly increases before dropping again.   It's greatest just outside the outer core.   https://newtonexcelbach.wordpress.com/2010/06/29/the-hole-through-the-middle-of-the-earth-accelleration/  
> **username_3**: This is a very standard problem with solutions all over the internet.  

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## 1 answers

### Answer by username_4 on 2017-01-30. Score: 5

The formula you are using is for a point mass, but a hole through the Earth experiences less gravity than a point mass, as the comment points out. One can then ask, what if the Earth's mass were all at the center, why wouldn't the mass fly through and go to a distance r on the other side? It's because a mass like that is a singularity, one that doesn't matter unless you fall right through it. If the Earth's mass was really concentrated at a point, then falling from r would still tend to just miss that point, since it's just a point anyway, and the huge forces would turn the orbit around, such that the object fell from r, went to zero, turned around, and flew back out to r. That's a very squeezed ellipse, with a=r/2, as desired. The motion that goes straight through and out to the opposite r, on the other hand, is not an ellipse with the mass at one focus, so it violates Kepler's laws. Again, the violation of Kepler's laws is that the mass is singular in a way that matters if you fall through a hole in it. Interestingly, it doesn't matter how concentrated the mass is toward the center-- if there's a hole in it that the object falls through, then the object does not obey Kepler's laws. It will only obey Kepler's laws if the orbiting object does not enter the mass.


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