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.





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