# 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
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.

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