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.