A satellite used in a cellular telephone network has a mass of 2290 kg and is in

Question

A satellite used in a cellular telephone network has a mass of 2290 kg and is in a circular orbit at a height of 820 km above the surface of the earth.

A) What is the gravitational force Fgrav on the satellite?

Take the gravitational constant to be G = 6.67×1011 Nm2/kg2 , the mass of the earth to be me = 5.97×1024 kg , and the radius of the Earth to bere = 6.38×106 m .

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

Take the free-fall acceleration at the surface of the earth to be g = 9.80 m/s2 .

Explanation / Answer

earth's mass=5.97*10^24 kg
earth's radius=6.38*10^6 m

distance of satellite from earth's core=6.38*10^6+820*10^3=7.2*10^6 m

then gravitational force=G*mass of satellite*mass of earth/(distance from earth)^2

=6.67*10^(-11)*2290*5.97*10^(24)/(7.2*10^6)^2=1.759*10^4 N

part B)

weight of the satellite on earth=2290*9.8=22442 N

then force as fraction of weight on earth=1.759*10^4/22442=0.7838

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