$A$ 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 orbit is ($M=$ mass of earth,$G=$ gravitational constant).

  • A
    $(G M m^2 r^2)^{1/2}$
  • B
    $(G M m r)$
  • C
    $(G M m^2 r)^{1/2}$
  • D
    $(G M^2 m r)^{1/2}$

Explore More

Similar Questions

Two satellites of mass $m$ and $9m$ are orbiting a planet in orbits of radius $R$. Their periods of revolution will be in the ratio of

$A$ planet moving along an elliptical orbit is closest to the sun at a distance $r_1$ and farthest away at a distance of $r_2$. If $v_1$ and $v_2$ are the linear velocities at these points respectively,then the ratio $\frac{v_1}{v_2}$ is

Imagine a light planet revolving around a very massive star in a circular orbit of radius $R$ with a period of revolution $T$. If the gravitational force of attraction between the planet and the star is proportional to $R^{-5/2}$,then,

Difficult
View Solution

If the Earth is considered a point mass of $6 \times 10^{24} \ kg$ revolving around the Sun at a distance of $1.5 \times 10^8 \ km$ in a period of $T = 3.14 \times 10^7 \ s$,then the angular momentum of the Earth around the Sun will be:

The time period $T$ of a satellite is related to the density $(\rho)$ of the planet around which it is orbiting close to the surface as:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo