Motion through the center of the Earth.

Question

Imagine there is a hole drilled through the center of the Earth. We drop an object of mass m through hole so at time t=0 the distance of the object from the Earth's center is R (radius of Earth). Derive the equation y(t) for the distance for the object from the Earth's center as a function of time.

Answer 1

When the object is a distance y from the center the Earth the gravitational force on the object is -mg(y) where g(y) = GV(y)ρ/y², V(y) = 4πy³/3, ρ = M/V(R), M is the mass of the Earth, and G is the gravitational constant. V(y) is the volume of a sphere with radius y so ρV(y) is the mass of the Earth within a radius y.

Thus -mg(y) = -4πGMmy³/(3y²V(R)) = -GMmy/R³.

From Newton's 2nd Law we have md²y/dt² = -GMmy/R³. Define GM/R³ ≡ ω² ⇒ d²y/dt² = -ω²y.

This is the equation for simple harmonic motion which has the solution y(t) = Asin(ωt) + Bcos(ωt).

Applying our initial condition y(0) = R gives y(t) = Rcos(ωt).