You want to begin with the formula for the strength of a gravitational field within a sphere, assuming a sphere of uniform density. This is given by:
g = -GMr/R^3, where G is the graivational constant, M is the mass of the sphere, r is the distance from the center of the sphere, and R is the radius of the sphere. This equation is only good when r < R. Once you're outside the sphere, the familiar equation of g = -GM/r^2 is applicable.
It helps to see where the simple harmonic motion comes from when rewriting the equation as the force on a mass traveling through the earth:
Force felt = ma = m*(-GMr/R^3)= -(GMm/R^3)*r
This looks alot like the equation for a spring:
F = ma = -kx, or accleration = -k/m*x, anytime an equation takes this form, you get solutions of sines and cosines, which gives the simple harmonic motion.
So for a mass traveling through the center of the earth, we want the accleration of the mass in term of the displacement from the center of the earth. We get this be dividing both sides of the equation by our mass m, to give:
a = -(GM/R^3)*r
The constant in front of r is equivalent to k/m term in the spring system. The sines and cosines arise from the solving the differential equation:
a = d^2r/dt^2 = -(GM/R^3)*r
Since for a spring:
a = d^2x/dt^2 = - k/m*x and x(t) = Acos(sqrt(k/m)*t + delta)
= Acos(wt + delta)
Where A, the amplitude, delta is determined by the initial conditions of the spring, and w is the angular frequency and is equal to 2*pi*frequency, 2*pi/Period, and sqrt(k/m).
With the earth, our constant term multiplying the r replaces the k/m in the spring equation. This gives:
sqrt(k/m) = sqrt(GM/R^3) = 2*pi/T, where T is the period
Multiplying both sides by T and dividing both sides by sqrt(GM/R^3) gives:
T = 2*pi/sqrt(GM/R^3) = 2*pi*sqrt(R^3/(GM)))
Taking 1/2 this number gives the time to travel through the earth, so
1/2T = pi*sqrt(R^3/(GM)) = 42 minutes
Hopefully this helps!
· 1 decade ago