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galben [10]
4 years ago
5

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 e

arth. (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?
Physics
1 answer:
Tom [10]4 years ago
5 0

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 <em>W=mg</em> 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

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