Total energy of earth in its circular orbit around the sun

Question :
Find out the total energy of earth in its circular orbit around the sun in terms of gravitational constant
Answer:
Let R be the total distance between the earth and the sun. If \({M_e}\) and \({M_s}\) are the mass of earth and sun respectively, the gravitational force of motion of earth and sun is given by

\(F =  - \frac{{G{M_e}{M_s}}}{{{R^2}}}\)

where G is the gravitational constant. Since the centripetal force balances the gravitational force of attraction, we have

\({F_c} = |{F_G}|\),

where

\({F_c} = \frac{{{M_e}{v^2}}}{R}\)

v being the velocity with which earth is moving. Hence we have

\(\frac{{{M_e}{v^2}}}{R} = \frac{{G{M_e}{M_s}}}{{{R^2}}}\)

or

\({M_e}{v^2} = \frac{{G{M_e}{M_s}}}{R}\)

Therefore kinetic energy of earth in motion is

\(T = \frac{1}{2}{M_e}{v^2} = \frac{1}{2}\frac{{G{M_e}{M_s}}}{R}\)

As we know that , force in terms of potential energy is

\({F_G} = - \frac{{\partial V}}{{\partial R}}\)
\(V = - \int {{F_G}dR} = \int {\frac{{G{M_e}{M_s}}}{{{R^2}}}dR = - } \frac{{G{M_e}{M_s}}}{R}\)

Now total energy of the earth in the orbit around the sun is

\(E = T + V\)
\(E = \frac{1}{2}\frac{{G{M_e}{M_s}}}{R} - \frac{{G{M_e}{M_s}}}{R}\)

\(E = - \frac{1}{2}\frac{{G{M_e}{M_s}}}{R}\)

This is the required expression.