Final answer:
The gravitational potential energy inside a uniformly dense Earth is given by U(r) = -(Gm_E m / 3R_E^3) r^2, derived by integrating the gravitational force over the distance from the center of the Earth to the object's location.
Step-by-step explanation:
To derive an expression for the gravitational potential energy (U(r)) of an object inside the Earth, we can use the concept of work done against the gravitational force. Since the gravitational force varies with distance from the center of the Earth, we use calculus to integrate the force expression over a radial path from the center outward to distance r.
Recall from physics, gravitational potential energy at a distance r from the center of a body of mass M is given by U = -GMm/r, where G is the gravitational constant, M is the mass of the body, and m is the mass of the object. However, within a body with uniform density, like our simplified Earth, the mass attracting the object is just the mass contained in a sphere of radius r. This mass (M(r)) is proportional to the volume of the sphere (proportional to r^3) and thus M(r) = (m_E / R_E^3) * r^3, where m_E is the mass of the Earth and R_E is the radius of the Earth. Therefore, we can express the force on an object inside the Earth as F = GM(r)m/r^2.
The work done in moving an object from the center of the Earth (r = 0), where we take the potential energy to be zero, to some distance r is the integral of F over that distance. This integral gives us the potential energy function:
U(r) = -\int_{0}^{r} F dr = -\int_{0}^{r} GM(r)m/r^2 dr = -\int_{0}^{r} Gm(m_E/R_E^3)r^2 dr = -Gm(m_E/R_E^3) \int_{0}^{r} r^2 dr
Solving this integral, we get:
U(r) = -Gm(m_E/R_E^3) [r^3/3] from 0 to r
Therefore, U(r) = -(Gm_E m / 3R_E^3) r^2
This is the expression for the gravitational potential energy of an object-earth system as a function of the object's distance from the center of the Earth under the assumption of uniform Earth density.