Question

A 2070 kg satellite used in a cellular telephone network is in a circular orbit at a height of 800 km above the surface of the earth.

Part A

What is the gravitational force on the satellite?

Part B

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

Answer 1

Part A

Fg = G * M * m ÷ d^2

d is the distance from the center of the earth to the satellite. This is the sum of the radius of the earth and the distance above the surface of the earth.

r = 6.38 * 10^6

d = 6.38 * 10^6 + 800,000 = 7.18 * 10^6 meters

Fg = 6.67 * 10^-11 * 5.98 * 10^24 * 2070 ÷ (7.17 * 10^6)^2

This is approximately 1.60 * 10^4 N.

Part B

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

Let’s use the following equation to determine the value of g at this altitude.

m * g = G * M * m ÷ d^2

g = G * M ÷ d^2

g = 6.67 * 10^-11 * 5.98 * 10^24 ÷ (7.18 * 10^6)^2

This is approximately 7.737 m/s^2.

To determine the fraction, divide by 9.8 m/s^2

Fraction = 7.737÷ 9.8

This is approximately 0.789