The angular momentum of a planet of mass M moving around the sun in an elliptical orbit is L→. The magnitude of the areal velocity of the planet is :

Correct answer: (D) L/2M. For small displacement ds of the planet its area can be written as dA=1/2rdl =1/2rdsinθ A.vel =dA/dt=1/2rsinθds/dt=(Vrsinθ)/2 dA/dt=1/2(mVrsinθ)/m=L/(2 m)

Q73 / 90 · JEE Main 2021 (18 Mar Shift 2) · Physics

The angular momentum of a planet of mass $M$ moving around the sun in an elliptical orbit is $\vec{L}$ . The magnitude of the areal velocity of the planet is :

A
$\frac{4 L}{M}$
B
$\frac{L}{M}$
C
$\frac{2 L}{M}$
Correct answer
D
$\frac{L}{2 M}$

Solution

For small displacement $d s$ of the planet its area can be written as

$d A = \frac{1}{2} r d l$

$= \frac{1}{2} r d \sin θ$

A.vel $= \frac{d A}{d t} = \frac{1}{2} r \sin θ \frac{d s}{d t} = \frac{V r \sin θ}{2}$

$\frac{d A}{d t} = \frac{1}{2} \frac{m V r \sin θ}{m} = \frac{L}{2 m}$

Hence, the correct answer is option: 'D'.

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