Question

The gravitational force, F, on a rocket at a distance, from the center of the earth is given by F = k / (r^2), where k = 10 ^3 newton*km^2. when the rocket is 10^4 km from the center of the earth, its moving away at .2 km/sec. how fast is the gravitational force changing at that moment? therefore what is the rate of change of the gravitational force in newton/ second?

Answer 1

F = k / (r^2)
dF/dt = -2k / (r^3) dr/dt

When r = 10^4 km and dr/dt = 0.2 km/sec
dF/dt = -2(10)^3 (0.2) / ((10^4)^3) = -400 / 10^7 = -0.00004 N/s

The gravitational force is changing at the rate of -0.00004 N/s