Question

A communication satellite of 500 kg revolves around the earth in a circular orbit of radius 4.0 $\times$ 10$^7$ m in the equatorial plane of the earth from west to east. The magnitude of angular momentum of the satellite is
A. \[\sim \;0.13{\text{ }} \times {\text{ }}{10^{14}}\;kg{\text{ }}{m^2}\;{s^{ - 1}}\]
B. \[\sim \;1.3{\text{ }} \times {\text{ }}{10^{14}}\;kg{\text{ }}{m^2}\;{s^{ - 1}}\]
C. \[\sim \;0.58{\text{ }} \times {\text{ }}{10^{14}}\;kg{\text{ }}{m^2}\;{s^{ - 1}}\]
D. \[\sim \;2.58{\text{ }} \times {\text{ }}{10^{14}}\;kg{\text{ }}{m^2}\;{s^{ - 1}}\]

Answer 1

Hint: In order to solve the above question first we will find the time period after that we will calculate the velocity of the satellite. Then we will substitute the values of mass, velocity, and radius of the earth in order to determine the angular momentum of the satellite.

Complete step by step answer:
We know that,
A geostationary satellite is an earth-orbiting satellite, placed at an altitude nearly equal to 35,800 km and directly above the equator, which revolves in the same direction as the earth rotates (from west to east).
The time period of revolution of a geostationary satellite around the earth is the same as that rotation of the earth about its own axis, i.e. 24 hours.
The satellite is revolving around earth in equatorial plane with orbital radius -
\[r = 4.0{\text{ }} \times {\text{ }}{10^7}\;m\]
Mass of satellite, $m = 500kg$
The speed of a satellite around earth is given by
$v = \sqrt {\dfrac{{Gm}}{r}} $--------------- (1)
The satellite travels around the entire circumference of the earth which is equal to $2\pi r$
If r is the radius of the orbit in the period T. This means the orbital speed must be
$v = \dfrac{{2\pi r}}{T}$--------------(2)
From equation (1) and (2), we get
$\Rightarrow \sqrt {\dfrac{{Gm}}{r}} $= $\dfrac{{2\pi r}}{T}$
On solving of time period, T we get
\[\Rightarrow T = 2\pi \sqrt {\dfrac{{{r^3}}}{{Gm}}} \]
Therefore the time period of a satellite in circular orbit is
\[\Rightarrow T = 2\pi \sqrt {\dfrac{{{r^3}}}{{Gm}}} \]
A geostationary orbit equates to an orbital velocity of 3.07 km/s or an orbital period of 1,436 minutes which equates to almost exactly one sidereal day or 23.934461223 hours, which is approximately 24 hours.
Hence the satellites are geostationary satellites.
So, the time is taken by satellite to complete it’s one revolution T = 24 h = 86400 s
Now calculate the angular momentum of the satellite by using the time period and velocity of the satellite.
Angular momentum of satellite
$L = mvr$
Using equation (2),
$\Rightarrow L = m\left( {\dfrac{{2\pi r}}{T}} \right)r = \dfrac{{2\pi m{r^2}}}{T}$
Putting the all values in above equation of angular momentum
$\Rightarrow L = \dfrac{{2 \times 3.14 \times 500}}{{86400}} \times {\left( {4 \times {{10}^7}} \right)^2}$
$\Rightarrow L = 0.58 \times {10^{14}}{m^2}{s^{ - 1}}$

The magnitude of angular momentum of the satellite is $0.58 \times {10^{14}}{m^2}{s^{ - 1}}$. So the answer is option (C).

Note:
One should take care of the direction of angular momentum in consideration because angular momentum is a vector quantity. The direction of angular momentum is perpendicular to the velocity and position vector. If the position and velocity vector are perpendicular then angular momentum comes out to be maximum, as in this case both are perpendicular.