Question

A shaft is drilled from the surface to the center of the earth as in the example 12.10 (section 12.6), make the unrealistic assumption that the density of the earth is uniform. With this approximation, the gravitational force on an object with mass m, that is inside the earth at a distance r from the center, has magnitude Fg=GmEmr/R3E (as shown in the example 12.10) and points toward the center of the earth.

Part A

Derive an expression for the gravitational potential energy U(r) of the object-earth system as a function of the object's distance from the center of the earth. Take the potential energy to be zero when the object is at the center of the earth.

Answer 1

The expression is U(r) = -G * m * E^2 / (2 * R_E) * (r^2 / R_E^2 - 3).

The gravitational potential energy (U) of the object-earth system in the given scenario is derived by considering the gravitational force acting on an object with mass m at a distance r from the center of the Earth. The gravitational force (Fg​) is given by
`F_g=G_mE_mr/R^3_E`, where G is the gravitational constant, mE is the mass of the Earth, r is the distance from the center of the Earth, and RE is the radius of the Earth.

To obtain the gravitational potential energy, we integrate this force with respect to distance (r) from the center of the Earth, taking into account the negative sign since the force points toward the center. The integration yields U(r) = -G * m * E^2 / (2 * R_E) * (r^2 / R_E^2 - 3). The potential energy is taken to be zero when the object is at the center of the Earth, resulting in a gravitational potential energy expression that depends on the distance (r) from the center of the Earth.