The ratio of kinetic energy of a planet at pe | vedclass

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Question

(c)

$K.E$ of a planet $=\frac{1}{2} m v^2$

$K.E$ at perigee $=\frac{1}{2} m v_P^2$

$K.E$ at apogee $=\frac{1}{2} m v_A^2$

Using conservati

Vedclass · Accepted Answer

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

Vedclass · Answer

(c)

$K.E$ of a planet $=\frac{1}{2} m v^2$

$K.E$ at perigee $=\frac{1}{2} m v_P^2$

$K.E$ at apogee $=\frac{1}{2} m v_A^2$

Using conservation of angular momentum at $P$ and $A$

$\Rightarrow m v_P r_P=m v_A r_A$

$\Rightarrow \frac{v_P}{v_A}=\frac{r_A}{r_P}=\frac{a(1+e)}{a(1-e)}$

$\Rightarrow \frac{ K \cdot E _P}{ K \cdot E _A}=\frac{v_P^2}{v_A^2}=\left(\frac{1+e}{1-e}\right)^2$

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

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