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Gravitation question

2021 · 18 Mar · Shift 2 · Q49
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  5. /2021 · 18 Mar · Shift 2 · Q49

Gravitation question

2021 · 18 Mar · Shift 2 · Q49

JEE MainPhysicsGravitationMCQ+4 / −1
The angular momentum of a planet of mass M moving around the sun in an elliptical orbit is L→{\overrightarrow L }L. The magnitude of the areal velocity of the planet is :
  1. A
    2LM{{2L} \over M}M2L​
  2. B
    L2M{{L} \over 2M}2ML​
  3. C
    LM{{L} \over M}ML​
  4. D
    4LM{{4L} \over M}M4L​
View written solutionFree

Correct answer: B

  1. Areal velocity definition

For a particle moving in a plane, the areal velocity is the rate at which the radius vector sweeps area:

rac{dA}{dt} = \frac{1}{2} r^2 \dot{\theta}

  1. Angular momentum of the planet

The angular momentum magnitude for a planet of mass MMM is

L=Mr2θ˙L = M r^2 \dot{\theta}L=Mr2θ˙

So,

r2θ˙=LMr^2 \dot{\theta} = \frac{L}{M}r2θ˙=ML​

  1. Relate areal velocity and angular momentum

Substitute into the formula for areal velocity:

dAdt=12(LM)\frac{dA}{dt} = \frac{1}{2} \left(\frac{L}{M}\right)dtdA​=21​(ML​)

Hence,

dAdt=L2M\frac{dA}{dt} = \frac{L}{2M}dtdA​=2ML​

  1. Match with the options

The correct option is:

L2M\boxed{\frac{L}{2M}}2ML​​

So, Option B is correct.

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