Solution :
Let satellite of mass $$m$$ be revolving closely around the earth of mass $$M$$ and radius $$R.$$
Total energy of satellite $$ = PE + KE = - \frac{{GMm}}{R} + \frac{1}{2}m{v^2}$$
$$\eqalign{
& = - \frac{{GMm}}{R} + \frac{m}{2}\frac{{GM}}{R}\,\,\left[ {{\text{as}}\,v = \sqrt {\frac{{GM}}{R}} } \right] \cr
& = - \frac{{GMm}}{{2R}} \cr} $$
∴ Total energy of satellite $$ = - \frac{1}{2}m{v^2}$$
Releted MCQ Question on Basic Physics >> Gravitation
Releted Question 1
If the radius of the earth were to shrink by one percent, its mass remaining the same, the acceleration due to gravity on the earth’s surface would-
If $$g$$ is the acceleration due to gravity on the earth’s surface, the gain in the potential energy of an object of mass $$m$$ raised from the surface of the earth to a height equal to the radius $$R$$ of the earth, is-
A geo-stationary satellite orbits around the earth in a circular orbit of radius $$36,000 \,km.$$ Then, the time period of a spy satellite orbiting a few hundred km above the earth's surface $$\left( {{R_{earth}} = 6400\,km} \right)$$ will approximately be-