A satellite used in a cellular telephone network has a mass of 2010 k g and is i

Question

A satellite used in a cellular telephone network has a mass of 2010 kg and is in a circular orbit at a height of 770 km above the surface of the earth.

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 be re = 6.38×106 m .

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

I keep getting 1.25 but apparantly that is not right, thanks!

Explanation / Answer

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 = 2010kg
radius of Earth = 6.38 * 10^6 meters
d= 6.38*10^6 meters + 7.70*10^5 meters=7.15*10^6 meters

F=16,700 Kg*m/s^2

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= (16,700Kg*m/s^2) / (2150kg * 9.8m/s^2)=0.793

Fraction=0.793

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