Question

A satellite of mass m revolving in circular orbit of radius r round the earth of mass m has binding energy e the angular momentum of sateelite is

Answer 1

Velocity Vs of the satellite is given by

GMmsr2s=nsv2srs

vs=(GMrs)−−−−−−−√

KE=12msvs=12ms(GMns)

PE=−GMmsrs

E Total energy =KE+PE-----(1)

=−GMms2rs

Angular momentum

L=msvsrs

=ms(GMrs)1/2rs

=(GMm2srs)1/2

Substituting from (1)

L=(2Emsr2s)1/2

Hence a is the correct answer.

Answer 2

Answer: The correct answer is $\sqrt{2mEr^{2} }$.

Explanation:

The expression for the kinetic energy for the satellite is as follows;

$E=\frac{1}{2} mv^{2}$

Here, m is the mass of the satellite and v is the velocity of the satellite.

Rearrange the above expression.

$mv= \sqrt{2Em}$

The expression for the angular momentum is as follows;

L=mvr

Here, r is the radius.

Put $mv= \sqrt{2Em}$ in the above expression.

$L=\sqrt{2Em}r$

Therefore, the angular momentum of satellite is $\sqrt{2Em}r$.