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

Therefore kinetic energy of earth in motion is

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

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

\(E = T + V\)

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

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

This is the required expression.
\(E = \frac{1}{2}\frac{{G{M_e}{M_s}}}{R} - \frac{{G{M_e}{M_s}}}{R}\)

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