MHT CET · Physics · Gravitation
A satellite of mass \(m\) is moving in a circular orbit of radius \(r\) around the earth. The angular momentum of the satellite about the center of orbitis ( \(M=\) mass of earth, \(G=\) gravitational constant)
- A \(\left(G M m^2 r^2\right)^{1 / 2}\)
- B \((G M m r)\)
- C \(\left(G M m^2 r\right)^{1 / 2}\)
- D \(\left(G M^2 m r\right)^{1 / 2}\)
Answer & Solution
Correct Answer
(C) \(\left(G M m^2 r\right)^{1 / 2}\)
Step-by-step Solution
Detailed explanation
Let the satellite revolves in a circular orbit of radius \(r\) with a linear velocity \(v\).
The centripetal force of acting on the satellite is balanced by the gravitational force due to the earth.
\(\begin{aligned} & \therefore F_{\text {centripetal }}=F_{\text {gravitational }} \\ & \Rightarrow \frac{m v^2}{r}=\frac{G M m}{r^2} \\ & \Rightarrow v=\sqrt{\frac{G M}{r}}\end{aligned}\)
Angular momentum of satellite about the center of orbit, is
\(L=m v r=m\left(\frac{G M}{r}\right)^{1 / 2} r=\left(G M m^2 r\right)^{1 / 2}\)
The centripetal force of acting on the satellite is balanced by the gravitational force due to the earth.
\(\begin{aligned} & \therefore F_{\text {centripetal }}=F_{\text {gravitational }} \\ & \Rightarrow \frac{m v^2}{r}=\frac{G M m}{r^2} \\ & \Rightarrow v=\sqrt{\frac{G M}{r}}\end{aligned}\)
Angular momentum of satellite about the center of orbit, is
\(L=m v r=m\left(\frac{G M}{r}\right)^{1 / 2} r=\left(G M m^2 r\right)^{1 / 2}\)
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