Question

A satellite used in a cellular telephone network has a mass of 2050 kg and is in a circular orbit at a height of 880 km above the surface of the earth.

Part A

What is the gravitational force Fgrav on the satellite?

Take the gravitational constant to be G = 6.67×10⁻¹¹ N⋅m2/kg2 , the mass of the earth to be me = 5.97×10²⁴ kg , and the radius of the Earth to be re = 6.38×10⁶ m .

part A equals 1.55*10^4 N

Part 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 .

Answer 1

Part A

F = G (m1*m2)/(d^2)

G= gravitational constant (6.67*10^-11 Nm^2/kg^2)
m1 and m2 = the masses object 1 and object 2 respectively
d=the distance between objects

mass of Earth = 5.97 × 10^24 kilograms
mass of satellite = 2050kg
radius of Earth = 6.38 * 10^6 meters
d= 6.38.1*10^6 meters + 8.80*10^5 meters=7.26*10^6 meters

F=15487.48 Kg*m/s^2

Part B

At surface of earth, you could use radius of Earth = 6.38 * 10^6 meters for distance(d). But you know that F=ma and at the surface of the earth a = 9.8m/s^2, so...

Fraction= (15487.48 Kg*m/s^2) / (2050kg * 9.8m/s^2)=0.7709