Question 292: Gravitation

Question

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)

Accepted Answer

$$\begin{aligned}
\frac{\mathrm{mv}^2}{\mathrm{r}} & =\frac{\mathrm{GMm}}{\mathrm{r}^2} \
	herefore \quad \mathrm{~V} & =\sqrt{\frac{\mathrm{GM}}{\mathrm{r}}} \
\mathrm{~L} & =\mathrm{mvr}=\mathrm{m} \sqrt{\frac{\mathrm{GM}}{\mathrm{r}}} \mathrm{r} \
& =\mathrm{m}(\mathrm{GMr})^{1 / 2}
\end{aligned}$$

Answer

$\mathrm{m}(\mathrm{GMr})^{3 / 2}$

Answer

$\mathrm{m}(\mathrm{GMr})$

Answer

$\mathrm{m}(\mathrm{GMr})^{1 / 2}$

Answer

$ \mathrm{m}(\mathrm{GMr})^{-1 / 2}$

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)

Choose the correct answer: