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Q. A comet revolves around the sun in an elliptical orbit. When it is closest to the sun at distance $d$ , then its corresponding kinetic energy is $k_{0}.$ If it is farthest from the sun at distance $3d$ then the corresponding kinetic energy will be

NTA AbhyasNTA Abhyas 2020Gravitation

Solution:

$v_{1}r_{1}=v_{2}r_{2}=$ constant
$\therefore v_{0}d=3dv_{1}$
$\Rightarrow v_{1}=\frac{v_{0}}{3}$
$\therefore KE=\frac{1}{2}\frac{mv_{0}^{2}}{9}$
$\Rightarrow KE=\frac{k_{0}}{9}$