What fraction is this of the satellite\'s weight at the surface of the earth? I

Question

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

I found the F grav for the satellite which is 1.71 * 10^4 N but I don't know how to get the fraction.

A satellite used in a cellular telephone network has a mass of 2240 kg and is in a circular orbit at a height of 840km above the surface of the earth. 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 beg= 9.80m/s^2 Take the gravitational constant to be G= 6.67½10?11 N.m^2/kg^2 , the mass of the earth to be me= 5.97½1024 kg , and the radius of the Earth to be re = 6.38½106 m . I found the F grav for the satellite which is 1.71 * 10^4 N but I don't know how to get the fraction.

Explanation / Answer

g' = g(r/(r+h))^2

Therefore

g' = g*((6400)/(6400+840))^2

= 0.781 g

So Fraction is 0.781

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