If the body in Ex. (26) performs uniform circular motion around the earth at a hight of 3600 km from the earh's surface, what will be its gravitational potential energy?

Explanation

Here, h= 3600 Km = ` 3.6 xx 10^(8) ` m
The gravitational energy of the body = ` - ( GMm)/(R + h)`
` - ( 6.67 xx 10^(-11) N.m^(2)//kg^(2) xx 6 xx 10^(24) Kg xx 10 kg)/(( 6.4 xx 10^(6) + 3.6 xx 10^(6)) m)`
` - ( 6.67 xx 6 xx 10^(14))/(10 xx 10^(6)) J = - 4.002 xx 10^(8) J`