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?

Question

Deshawn Orozco

Mathematics

20 July, 09:11

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?

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Answers (1)

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Brogan Humphrey · 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