Hole drilled through center of the earth

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.

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?

__My attempt:__

I derived the formula for the orbital period:

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

where $\displaystyle a$ is the the semi-major axis, which is $\displaystyle R$ 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, is that a coincidence?

Re: Hole drilled through center of the earth

for a mass m at a point r inside the Earth, $\displaystyle 0 < r < R_e$

$\displaystyle F = -\frac{GM_i m}{r^2}$ where $\displaystyle M_i$ is the mass of that part of the earth inside a radius $\displaystyle r$

assuming uniform density ...

$\displaystyle \frac{M_i}{\frac{4}{3}\pi r^3} = \frac{M_e}{\frac{4}{3}\pi R_e^3} \implies M_i = \frac{Mr^3}{R_e^3}$

substituting for $\displaystyle M_i$ in the gravitational force equation ...

$\displaystyle F = -\frac{GM_e m}{R_e^3} \cdot r = -kr$

where $\displaystyle k = \frac{GM_e m}{R_e^3}$

for SHM, $\displaystyle T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{k}}$

back sub for $\displaystyle k$ ...

Re: Hole drilled through center of the earth

Quote:

Originally Posted by

**skeeter** for a mass m at a point r inside the Earth, $\displaystyle 0 < r < R_e$

$\displaystyle F = -\frac{GM_i m}{r^2}$ where $\displaystyle M_i$ is the mass of that part of the earth inside a radius $\displaystyle r$

assuming uniform density ...

$\displaystyle \frac{M_i}{\frac{4}{3}\pi r^3} = \frac{M_e}{\frac{4}{3}\pi R_e^3} \implies M_i = \frac{Mr^3}{R_e^3}$

substituting for $\displaystyle M_i$ in the gravitational force equation ...

$\displaystyle F = -\frac{GM_e m}{R_e^3} \cdot r = -kr$

where $\displaystyle k = \frac{GM_e m}{R_e^3}$

for SHM, $\displaystyle T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{k}}$

back sub for $\displaystyle k$ ...

Actually, I did that and found the time period. But my question is, whether it is a coincidence that the time period equals that of a satellite that just skims the surface of the earth.

For an ellipse, $\displaystyle T=2\pi\sqrt\frac{a}{g}$, where $\displaystyle a$ is the semi-major axis. I wonder if the path, that passes through the center of the earth, can be considered to be a degenerate ellipse (with semi-major axis $\displaystyle R$ and semi-minor axis zero). But then, the central mass would be located at the center of the degenerate ellipse, and not at either of the foci (where it really should be).

Re: Hole drilled through center of the earth

I don't think it's a coincidence ... the velocity of the mass thru the tube is the same as one component of the velocity as as the mass orbits the earth.

Re: Hole drilled through center of the earth

Quote:

Originally Posted by

**skeeter** I don't think it's a coincidence ... the velocity of the mass thru the tube is the same as one component of the velocity as as the mass orbits the earth.

The speed of the particle as it passes through the center of the earth is equal to that of a satellite that just skims the surface of the earth.

As I had asked earlier, is there any way the path, that passes through the center of the earth, can be considered to be a degenerate ellipse?

Re: Hole drilled through center of the earth

Quote:

Originally Posted by

**alexmahone** The speed of the particle as it passes through the center of the earth is equal to that of a satellite that just skims the surface of the earth.

As I had asked earlier, is there any way the path, that passes through the center of the earth, can be considered to be a degenerate ellipse?

according to this definition ... The Most Marvelous Theorem in Mathematics , I guess you can.

Re: Hole drilled through center of the earth

Quote:

Originally Posted by

**skeeter**

But then, the central mass would be located at the center of the degenerate ellipse, and not at either of the foci (where it really should be).

Re: Hole drilled through center of the earth

Quote:

Originally Posted by

**alexmahone** But then, the central mass would be located at the center of the degenerate ellipse, and not at either of the foci (where it really should be).

but if the ellipse is a circle, then the foci are at the center ...

Re: Hole drilled through center of the earth

Quote:

Originally Posted by

**skeeter** but if the ellipse is a circle, then the foci are at the center ...

But in this case, the degenerate ellipse is the line segment (that passes through the center of the earth), and the foci are at the end points.

Re: Hole drilled through center of the earth

then I guess the degenerate ellipse idea is bogus ... I will have to plead ignorance on how this argument answers the original question, anyway.

My problem is that I'm looking at the problem from a physics viewpoint. As I said earlier, circular orbits in two dimensions can be viewed as linear SHM when looking at the movement in only a single dimension.

Re: Hole drilled through center of the earth

Quote:

Originally Posted by

**skeeter** As I said earlier, circular orbits in two dimensions can be viewed as linear SHM when looking at the movement in only a single dimension.

But isn't it a coincidence that the motion of the particle through the hole mirrors the projection of the motion of the satellite along that dimension?