Question:
As shown in figure, $a$ planet revolves in elliptical orbit around the sun, where $K.E.$ of the planet is maximum :
As shown in figure, $a$ planet revolves in elliptical orbit around the sun, where $K.E.$ of the planet is maximum :
Updated On: Jul 27, 2022
- at $P_{4}$
- at $P_{1}$
- at $P_{2}$
- at $P_{3}$
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The Correct Option is A
Solution and Explanation
Since angular momentum is conserved Therefore,
$I_{1} \omega_{1}=I_{2} \omega_{2}$
or $M R_{1}^{2} \omega_{1}=M R_{2}^{2} \omega_{2}$
or $M R_{1} v_{1}=M R_{2} v_{2}$
At $P_{4}$ the value of $R$ is minimum
Hence the velocity is maximum or kinetic energy will be maximum.
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Concepts Used:
Keplers Laws
Kepler’s laws of planetary motion are three laws describing the motion of planets around the sun.
Kepler First law – The Law of Orbits
All the planets revolve around the sun in elliptical orbits having the sun at one of the foci.
Kepler’s Second Law – The Law of Equal Areas
It states that the radius vector drawn from the sun to the planet sweeps out equal areas in equal intervals of time.
Kepler’s Third Law – The Law of Periods
It states that the square of the time period of revolution of a planet is directly proportional to the cube of its semi-major axis.
T2 ∝ a3