A 2150kg satellite used in a cellular telephone network is ina circular orbit at a height of 780km above the surface ofthe earth. What is the gravitational force of the sattilite? What fraction is this force of the satellite's weight at thesurface of the earth? A 2150kg satellite used in a cellular telephone network is ina circular orbit at a height of 780km above the surface ofthe earth. What is the gravitational force of the sattilite? What fraction is this force of the satellite's weight at thesurface of the earth?
Explanation / Answer
mass of sateellite m = 2150 kg mass of earth M = 5.98 * 10 ^ 24 kg radius of earth R = 6.38 * 10 ^ 6 m height h = 780 * 10 ^ 3 m = 0.78 * 10 ^ 6 m Gravitational force F = GMm / ( R + h ) ^ 2 where G = Gravitational constant = 6.67 * 10 ^ -11 N m ^2 / kg ^ 2 plug the values weget F = 1672.78 * 10 ^ 1 = 16727.8 N (b). Gravitational force on surface of earth F ' =GMm / R ^ 2 = 21068 N So, required fraction f = F / F ' = 0.7939