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(C) The kinetic energy $(K.E.)$ of a planet is given by $K.E. = \frac{1}{2} m v^2$.Let $v_P$ be the velocity at perigee and $v_A$ be the velocity at a

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$\left(\frac{1+e}{1-e}\right)^2$

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(C) The kinetic energy $(K.E.)$ of a planet is given by $K.E. = \frac{1}{2} m v^2$.Let $v_P$ be the velocity at perigee and $v_A$ be the velocity at apogee.By the law of conservation of angular momentum,the angular momentum at perigee $(L_P)$ is equal to the angular momentum at apogee $(L_A)$:$m v_P r_P = m v_A r_A$This implies:$\frac{v_P}{v_A} = \frac{r_A}{r_P}$For an elliptical orbit with semi-major axis $a$ and eccentricity $e$,the distance at perigee is $r_P = a(1-e)$ and the distance at apogee is $r_A = a(1+e)$.Substituting these values:$\frac{v_P}{v_A} = \frac{a(1+e)}{a(1-e)} = \frac{1+e}{1-e}$The ratio of kinetic energies is:$\frac{K.E._P}{K.E._A} = \frac{\frac{1}{2} m v_P^2}{\frac{1}{2} m v_A^2} = \left(\frac{v_P}{v_A}\right)^2 = \left(\frac{1+e}{1-e}\right)^2$

The ratio of kinetic energy of a planet at perigee and apogee during its motion around the sun in an elliptical orbit of eccentricity $e$ is ..........

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