Question

1)   An Earth satellite is in a circular orbit at an altitude of 500 km. Explain why the work done by the gravitational force acting on the satellite is zero. Using the work-energy theorem, what can you say about the speed of the satellite?

Answer 1

For a constant force, the differential work done is the dot product of the force and the differential displacement vector:

dw = F * dr

In the case of an satellite in a circular orbit, the force vector is always directed radially inward, toward the center of mass of the planet, while the instantaneous displacement vector is perpendicular to this. The dot product of two perpendicular vectors is identically zero, and hence the work done is also zero.

Because no work is done on the satellite, the satellite has constant energy (in the absence of other forces). The problem stipulates that the satellite is in a circular orbit, so its elevation, and hence its gravitational potential energy is constant (assuming the satellite has constant mass). The total energy is the sum of the potential and kinetic energies, so if both the total energy and the potential energy are constant, then so must be the kinetic energy. The kinetic energy is equal to 1/2 * mass * speed^2, so the orbital speed must also be constant.