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Wednesday, November 14, 2012

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
\({F_c} = |{F_G}|\),
\({F_c} = \frac{{{M_e}{v^2}}}{R}\)
v being the velocity with which earth is moving. Hence we have
\(\frac{{{M_e}{v^2}}}{R} = \frac{{G{M_e}{M_s}}}{{{R^2}}}\)
\({M_e}{v^2} = \frac{{G{M_e}{M_s}}}{R}\)

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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