Question

A 2190 kg satellite used in a cellular telephone network is in a circular orbit at a height of 730 kmabove the surface of the earth.

What is the gravitational force on the satellite?

What fraction is this force of the satellite's weight at the surface of the earth?

Answer 1

Here,

gravitational force on the satellite = G * m * M/(R + h)^2

gravitational force on the satellite = 6.673 *10^-11 * 5.98 *10^24 * 2190/(6371 *10^3 + 730 *10^3)^2

gravitational force on the satellite = 17331 N

the gravitational force on the satellite is 17331 N

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ratio of force to weight at the surface = force / weight at the surface

ratio of force to weight at the surface = 17331/(2190 * 9.8)

ratio of force to weight at the surface = 0.81

the ratio of force to weight at the surface is 0.81