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A planet is revolving around the sun in an elliptical orbit. Its closest distance from the sun is \[{{r}_{\min }}\]. The farthest distance from the sun is \[{{r}_{\max }}\]. If the orbital angular velocity of the planet when it is nearest to the sun is \[\omega \], then the orbital angular velocity at the point when it is at the farthest distance from the sun is.

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Last updated date: 17th Apr 2024
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Answer
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Hint: To find the motion of the planet around the sun, the angular momentum of the planet near the sum is equal to the angular momentum of the planet farthest from the sun. Hence, the angular momentum of the planet is:
$L_x=mvR_{min}$ and $L_y = mvR_{max }$
where \[{{L}_{x}}\] is the angular momentum of the near radius planet revolution, \[{{L}_{y}}\] is the angular momentum of the far radius planet revolution, \[m\] is the mass of the planet, \[v\] is the velocity of the planet’s revolution, \[{{R}_{\min }}\] is the minimum radius, \[{{R}_{\max }}\] is the maximum radius.

Complete step by step solution:
There are no torques externally applied on the planet during minima and maxima radius of the ellipse/orbit due to force perpendicular. Due to no external torque the angular momentum remains zero. Hence, the formula for the angular momentum in the nearer radius is given as:
\[{{L}_{\min }}=mv{{R}_{\min }}\]
The angular momentum due to the farthest radius is given as:
\[{{L}_{\max }}=mv{{R}_{\max }}\]
After getting the momentum in terms of velocity, we convert the velocity in terms of angular velocity with which we get the angular momentum as:
For the planet when moving near sun:
\[{{L}_{\min }}=m{{R}_{\min }}\omega {{R}_{\min }}\, with v={{R}_{\min }}{{\omega }_{\min }}\]
For the planet when moving far from sun:
\[{{L}_{\max }}=m{{R}_{\max }}{{\omega }_{\max }}{{R}_{\max }}\, with v={{R}_{\max }}{{\omega }_{\max }}\]

Equating both the angular momentum for far and nearer radius as both are same and equal where
\[m{{R}_{\max }}{{\omega }_{\max }}{{R}_{\max }}=m{{R}_{\min }}\omega {{R}_{\min }}\]
\[{{\omega }_{\max }}=\dfrac{{{R}_{\min }}\omega {{R}_{\min }}}{{{R}_{\max }}{{R}_{\max }}}\]
\[{{\omega }_{\max }}=\dfrac{{{\left( {{R}_{\min }} \right)}^{2}}}{{{\left( {{R}_{\max }} \right)}^{2}}}\omega \]

Therefore, the orbital angular velocity at the point when it is at the farthest distance from the sun is \[{{\omega }_{\max }}=\dfrac{{{\left( {{R}_{\min }} \right)}^{2}}}{{{\left( {{R}_{\max }} \right)}^{2}}}\omega \].

Note: Another method to solve the question is by equating the velocities of the farthest and the nearer regions where the farthest velocity is equal to the nearest velocity.