Question

Assuming the earth to be of uniform density, drill a hole through the Earth between any two points on the surface of the earth. Install a frictionless tube, and evacuate all the air. Show that the motion of a subway car placed in this tube is simple harmonic motion and evaluate the period of the motion. (HINT: Use gravitational equivalent of Gauss's Law.)

PROVE THE GENERAL CASE ALONG ANY CHORD BETWEEN ANY TWO POINTS. NOT JUST THE CASE WHERE IT TRAVELS ALONG A DIMATER

Answer 1

Putting the z axis pointing from the center of the Earth to the middle point of the tunnel,
and the x axis parallel to the tunnel, the tunnel equation are

z=z0
y=0

In polar coordinates,

r cos(theta) = R cos(theta0)
phi = 0

where R is Earth radius and R cos(theta0) = z0.

The tunnel goes from theta=-theta0 to theta=theta0.

Gravitational force:

Only the mass contained into a sphere of radius r contributes to the gravitational force, so

F(r) = G M(r)m/r^2 = G (rho (4/3) pi r^3)m/r^2 =
= G rho (4/3) pi m r

F(R) = g m = G rho (4/3) pi m R, then

F(r) = g m r/R

Potential energy

U(r) = g m r^2/(2R) =
= [g m z_0^2/(2R)]/cos^2(theta))


Velocity:

r cos(theta) = R cos(theta0)

then

dr/dt cos(theta) - r d theta/dt sin(theta) = 0

or

dr/dt = r d theta/dt tan(theta)

v^2 = (dr/dt)^2 + r^2 (d theta/dt)^2=
= r^2 (d theta/dt)^2 (1+tan^2(theta))=
= r^2 (d theta/dt)^2 / cos^2(theta) =
= z_0^2 (d theta/dt)^2 / cos^4(theta)

Lagrangian:

L = T-U =
= (m z_0^2/2) (d theta/dt)^2 / cos^4(theta) -
- [g m z_0^2/(2R)]/cos^2(theta))


Assuming that theta0<<1, then we can approximate the lagrangian as

L = m z_0^2/2 (d theta/dt)^2 + g m z_0^2/(2R) theta^2

and hence the equation of motion will be

d^2 theta/dt^2 + g/R theta = 0 ==> Motion is Simple Harmonic Motion

Then

theta(t) = A cos(sqrt(g/R) t + phi)

Period: T = 2pi/omega = 2 pi sqrt(R/g)