# Hole drilled through center of the earth

1. Oct 6, 2011

### alexmahone

Suppose that a small hole is drilled straight through the center of the earth, thus connecting two antipodal points on its surface. Let a a particle of mass m be dropped at time t = 0 into this hole with initial speed zero. Find the period of the simple harmonic motion exhibited by the particle.

I found this to be about 84 min.

Look up (or derive) the period of a satellite that just skims the surface of the earth; compare with the previous result. How do you explain the coincidence. Or is it a coincidence?

I looked this up on the internet and found the period to be 84 min as well. But I'm unable to explain the coincidence. Please help.

2. Oct 6, 2011

### HallsofIvy

Staff Emeritus
No, it is not a coincidence. The period of an orbit about the earth is determined by the maximum distance of the orbit from the center of the earth (the "apogee"). For a satellite that "just skims the earth" or one that goes through the center of the earth those are the same- the radius of the earth.

3. Oct 6, 2011

### alexmahone

I derived the formula for the orbital period:

$$T=2\pi\sqrt{\frac{a^3}{GM}}$$

where a is the the semi-major axis, which is R in this case.

But a path that goes through the center of the earth isn't an orbit, in the usual sense of the word. However, I was able to derive independently that the period in this case has the same formula. So, isn't that a coincidence?

Last edited: Oct 6, 2011
4. Oct 6, 2011

### DrStupid

No it isn't. You always get the the same values for the orbit at surface level and for a linear path through a homogeneous spherical mass distribution (it doesn't need do go through the center).

By the way: Earth is not homogeneous. With the http://geophysics.ou.edu/solid_earth/prem.html" [Broken] I get a period of 76 minutes from pole to pole and return.

Last edited by a moderator: May 5, 2017
5. Oct 6, 2011

### alexmahone

I know we get the same value, but why do you say it isn't a coincidence?

6. Oct 6, 2011

### DrStupid

According to Keplers 3rd law the period of an orbit is determined by the semi major axis and for a linear "orbit" from surface to the center of earth (assuming earth to be a point mass) it would be half the radius of earth.

7. Oct 6, 2011

### DrStupid

Because it is the inevitable result of natural laws.

8. Oct 6, 2011

### alexmahone

I guess you mean half the diameter of the earth. So you're assuming that the path, that passes through the center of the earth, is a degenerate ellipse (with semi-major axis R and semi-minor axis zero)?

What if the path does not pass through the center of the earth? Assuming that it is a degenerate ellipse to use Kepler's 3rd law would not help because the center of the earth would not be in the plane of the ellipse.

Last edited: Oct 6, 2011
9. Oct 6, 2011

### DrStupid

No, I mean half the radius. It would be a degenerate ellipse with semi-major axis R/2 and semi-minor axis zero.

That's not possible. According to Kepler's 1st law the central mass is located in a focus of the ellipse and in this special case that means at the inner end of the path.

Last edited: Oct 6, 2011
10. Oct 6, 2011

### alexmahone

But in the case of the linear "orbit" that passes through the center of the earth, the central mass is located at the center of the degenerate ellipse, and not at either of the foci.

11. Oct 11, 2011

### DrStupid

That's wrong. You can't brake natural laws.

If you want to have a linear path with a point mass in the middle you have to compose it of two elliptic orbits, each with the point mass in the inner focus. The particle starts on the first orbit and reaches the center within half the corresponding period T'. Then it changes to the second orbit (this is possible because in the center the direction is not defined due to infinite angular velocity) and reaches the opposite surface within additional T'/2. The way back takes once more T'. According to Keplers 3rd law the period of the degenerate ellipse is

$T'=\frac{T}{\sqrt{8}}$

where T is the period of the corresponding circular orbit. Therefore the period of the complete linear path is

$\frac{T}{\sqrt{2}}$

If you don't believe it you may check it with a numerical simulation.