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(B) The total energy of a satellite in a circular orbit of radius $r$ is given by $E = -\frac{GMm}{2r}$.The initial energy in orbit $R_1$ is $E_1 = -\

Vedclass · Accepted Answer

$\frac{GMm}{2} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$

Vedclass · Answer

(B) The total energy of a satellite in a circular orbit of radius $r$ is given by $E = -\frac{GMm}{2r}$.The initial energy in orbit $R_1$ is $E_1 = -\frac{GMm}{2R_1}$.The final energy in orbit $R_2$ is $E_2 = -\frac{GMm}{2R_2}$.The change in total energy required is $\Delta E = E_2 - E_1 = -\frac{GMm}{2R_2} - (-\frac{GMm}{2R_1}) = \frac{GMm}{2} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.Since the potential energy change is $\Delta U = U_2 - U_1 = -\frac{GMm}{R_2} - (-\frac{GMm}{R_1}) = GMm \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$,and $\Delta E = \Delta K + \Delta U$,the required additional kinetic energy is $\Delta K = \Delta E - \Delta U = \frac{GMm}{2} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) - GMm \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = -\frac{GMm}{2} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.However,standard problems of this type usually ask for the energy required to change the orbit,which is $\Delta E$. Given the options,the correct expression for the energy difference is $\frac{GMm}{2} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.

The additional kinetic energy to be provided to a satellite of mass $m$ revolving around a planet of mass $M$ to transfer it from a circular orbit of radius $R_1$ to another of radius $R_2$ $(R_2 > R_1)$ is:

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