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

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Solution

Let us assume that semi major axis is 'a' and e= eccentricity then
distance of apogee is given by $r_{a} = a (1 + e)$
distance of perigee is given by $r_{p} = a (1 - e)$
To calculate the ratio of KE , we use the conservation of momentum

$m v_{a} r_{a}$ = $m v_{p} r_{p}$
$\frac{v_{a}}{v_{p}} = \frac{r_{p}}{r_{a}}$ ---(1)
$\frac{K E_{a}}{{K E}_{p}} = \frac{\frac{1}{2} m {v_{a}}^{2}}{\frac{1}{2} m v_{p}^{2}}$
$= \frac{{v_{a}}^{2}}{v_{p}^{2}} = \frac{r_{p}^{2}}{r_{a}^{2}}$
using value of $r_{a}$ and $r_{p}$ and eq.1
$\frac{K E_{a}}{{K E}_{p}} = \frac{({1 - e)}^{2}}{{(1 + e)}^{2}}$

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