JEE MAIN - Physics (2025 - 29th January Evening Shift - No. 23)
Two planets, $A$ and $B$ are orbiting a common star in circular orbits of radii $R_A$ and $R_B$, respectively, with $R_B=2 R_A$. The planet $B$ is $4 \sqrt{2}$ times more massive than planet $A$. The ratio $\left(\frac{\mathrm{L}_{\mathrm{B}}}{\mathrm{L}_{\mathrm{A}}}\right)$ of angular momentum $\left(L_B\right)$ of planet $B$ to that of planet $A\left(L_A\right)$ is closest to integer ________.
Answer
8
Explanation
Let a planet of mass m orbits a star of mass M in circular orbit of radius r with speed v.
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For this circular motion, centripetal force is provided by the gravitational force between both masses.
Hence, $${F_C} = {F_4}$$
$$ \Rightarrow {{m{v^2}} \over r} = {{GMm} \over {{r^2}}}$$
$$ \Rightarrow v = \sqrt {{{Gm} \over r}} $$
Angular momentum of the planet, $$L = mvr$$
$$ \Rightarrow L = m\sqrt {{{Gm} \over r}} r$$
$$ \Rightarrow L = \sqrt {GM} m\sqrt r $$
Given, $${m_B} = 4\sqrt 2 {m_A}$$ and $${R_B} = 2{R_A}$$
M is same for both.
So, $${{{L_B}} \over {{L_A}}} = \left( {{{{m_B}} \over {{m_A}}}} \right)\sqrt {{{{R_B}} \over {{R_A}}}} $$
$$ = \left( {4\sqrt 2 } \right)\sqrt 2 = 4 \times 2$$
$$ \Rightarrow {{{L_B}} \over {{L_A}}} = 8$$.
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