JEE Mains · Physics · STD 11 - Gravitation
- JEE Mains 2025
- Medium
- Numerical
Two planets, \(A\) and \(B\) are orbiting a common star in circular orbits of radius \(R_A\) 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 \(\left(\frac{L_B}{L_A}\right)\) of angular momentum \(\left(L_B\right)\) of planet \(B\) to that of planet \(A\left(L_A\right)\) is closest to integer ________.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) 8
Step-by-step Solution
Detailed explanation
\(\mathrm{L}=\mathrm{mv}_0 \mathrm{R}=\mathrm{m} \sqrt{\frac{\mathrm{GM}}{\mathrm{R}}} \mathrm{R}=\mathrm{m} \sqrt{\mathrm{GMR}}\)
here M is mass of star
\(\begin{aligned}
& \frac{\mathrm{L}_{\mathrm{B}}}{\mathrm{~L}_{\mathrm{A}}}=\frac{\mathrm{m}_{\mathrm{B}}}{\mathrm{~m}_{\mathrm{A}}} \sqrt{\frac{\mathrm{R}_{\mathrm{B}}}{\mathrm{R}_{\mathrm{A}}}} \\
& =4 \sqrt{2} \sqrt{\frac{2}{1}} \\
& \frac{\mathrm{~L}_{\mathrm{B}}}{\mathrm{~L}_{\mathrm{A}}}=8
\end{aligned}\)
here M is mass of star
\(\begin{aligned}
& \frac{\mathrm{L}_{\mathrm{B}}}{\mathrm{~L}_{\mathrm{A}}}=\frac{\mathrm{m}_{\mathrm{B}}}{\mathrm{~m}_{\mathrm{A}}} \sqrt{\frac{\mathrm{R}_{\mathrm{B}}}{\mathrm{R}_{\mathrm{A}}}} \\
& =4 \sqrt{2} \sqrt{\frac{2}{1}} \\
& \frac{\mathrm{~L}_{\mathrm{B}}}{\mathrm{~L}_{\mathrm{A}}}=8
\end{aligned}\)
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