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# The planet mercury is revolving in an elliptical orbit around the sun as shown in the figure. The kinetic energy of mercury will be greatest at (A) A(B) B(C) C(D) D

Last updated date: 19th Jul 2024
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Hint: We will be solving this question using the conservation laws in planetary motion. We will also be using the relation between the velocity and the distance of mercury from the sun.
(1) Conservation of angular momentum $mvr$ = constant
Where $m$ is the mass, $v$ is the velocity and $r$ is the radius.
(2) The formula for the kinetic energy $K=\dfrac{1}{2}m{{v}^{2}}$
Where $m$ is the mass, $v$ is the velocity.

$mvr$ = constant
Here, $m$ is the mass of the mercury, $v$ is the velocity of mercury and $r$ is the radius, i.e., the distance between the sun and the planet mercury.
We will now discuss the kinetic energy of the planet mercury. As we can see from the formula for kinetic energy, $K=\dfrac{1}{2}m{{v}^{2}}$ the kinetic energy is directly proportional to the velocity of the planet. And as we have concluded above, the velocity is inversely proportional to the distance. Therefore, for the kinetic energy to be maximum, the velocity should be maximum, and the only way it is possible is when the distance of mercury from the sun is minimum.