TS EAMCET · Physics · Gravitation
A planet of mass \(m\) moves in a elliptical orbit around an unknown star of mass \(M\) such that its maximum and minimum distances from the star are equal to \(r_1\) and \(r_2\) respectively. The angular momentum of the planet relative to the centre of the star is
- A \(m \sqrt{\frac{2 G M r_1 r_2}{r_1+r_2}}\)
- B 0
- C \(m \sqrt{\frac{2 G M\left(r_1+r_2\right)}{r_1 r_2}}\)
- D \(\sqrt{\frac{2 G M m r_1}{\left(r_1+r_2\right) r_2}}\)
Answer & Solution
Correct Answer
(A) \(m \sqrt{\frac{2 G M r_1 r_2}{r_1+r_2}}\)
Step-by-step Solution
Detailed explanation
According to the law of conservation of angular momentum, \( \begin{aligned} m v_1 r_1 & =m v_2 r_2 \\ \Rightarrow \quad v_2 & =\frac{v_1 r_1}{r_2} \end{aligned} \) From the law of conservation of total mechanical energy.…
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