MHT CET · Physics · Gravitation
A satellite of mass ' \(m\) ' is revolving around the earth of mass ' \(M\) ' in an orbit of radius ' \(r\) ' with constant angular velocity ' \(\omega\) '. The angular. momentum of satellite is
( \(\mathrm{G}=\) Universal constant of gravitation)
- A \(\mathrm{m}(\mathrm{GMr})^{3 / 2}\)
- B \(\mathrm{m}(\mathrm{GMr})\)
- C \(\mathrm{m}(\mathrm{GMr})^{1 / 2}\)
- D \(\mathrm{m}(\mathrm{GMr})^{-1 / 2}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{m}(\mathrm{GMr})^{1 / 2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \frac{m^2}{r} & =\frac{G M m}{r^2} \\ \therefore \quad V & =\sqrt{\frac{G M}{r}} \\ L & =m v r=m \sqrt{\frac{G M}{r}} r \\ & =m(G M r)^{1 / 2}\end{aligned}\)
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