Question

A 2150 kg satellite used in a cellular telephone network is in a circular orbit at a height of 780 km above the surface of the earth. (a) What is the gravitational force on the satellite? (b) What fraction is this force of the satellite’s weight at the surface of the earth?

Answer 1

Answer:

a)$F=16741.9N$

b)$\frac{F}{W}=0.795$

Explanation:

The gravitational force on the satellite is calculated with Newton's Gravitation Law:

$F=\frac{GMm}{r^2}$

Where $M=5.97\times10^{24}kg$ is Earth's mass, $m=2150kg$ is the satellite mass, $r=R+h$ is the distance between their centers, where $h=780000m$ is the height of the satellite (from Earth's surface) and $R=6371000m$ is Earth's radius, and $G=6.67\times10^{-11}Nm^2/kg^2$ is the gravitational constant.

a) With these values we then have:

$F=\frac{GMm}{r^2}=\frac{(6.67\times10^{-11}Nm^2/kg^2)(5.97\times10^{24}kg)(2150kg)}{(6371000m+780000m)^2}=16741.9N$

b) And the fraction this force is of the satellite’s weight W=mg is:

$\frac{F}{W}=\frac{GMm}{mgr^2}=\frac{GM}{gr^2}=\frac{(6.67\times10^{-11}Nm^2/kg^2)(5.97\times10^{24}kg)}{(9.8m/s^2)(6371000m+780000m)^2}=0.795$