JEE MAIN - Physics (2021 - 18th March Evening Shift - No. 6)
The angular momentum of a planet of mass M moving around the sun in an elliptical orbit is $${\overrightarrow L }$$. The magnitude of the areal velocity of the planet is :
$${{2L} \over M}$$
$${{L} \over 2M}$$
$${{L} \over M}$$
$${{4L} \over M}$$
Explanation
_18th_March_Evening_Shift_en_6_2.png)
Gravitational force line passes through the sun so torque about sun always zero for the planet.
$$ \therefore $$ Angular momentum about sum is constant.
$$\overrightarrow L $$ = Constant
we know, L = M$${v_ \bot }$$ r
Now,
$$dA = {1 \over 2}$$ $$\times$$ base $$\times$$ height
$$ = {1 \over 2} \times r \times {v_ \bot }dt$$
$$ \Rightarrow {{dA} \over {dt}} = {{r{v_ \bot }} \over 2} = {r \over 2} \times {L \over {Mr}} = {L \over {2M}}$$
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