Question

Calculate the height from the Earth's surface that a satellite must attain in order to be in geosynchronous orbit and the satellite's velocity?

Answer 1

Hint: Geostationary orbit is actually geosynchronous orbit. From the relation of centripetal force and gravitational force we will find out the height at which a satellite reaches the geostationary orbit. Then from the formula of angular velocity we will eventually find the velocity of that satellite.

Complete step-by-step solution:
Gravitational force$ = \dfrac{{G{M_e}m}}{{{{\left( r \right)}^2}}}$ where $G = $ Universal Gravitational Constant, ${M_e}$$ = $ Mass of the Earth, $m = $ mass of the body and $r = $ distance of the body from the centre of the Earth.
Centripetal force$ = mr{\omega ^2}$ where $m = $ mass of the body, $r = $ distance of the body from the centre of rotation, $\omega = $ angular velocity.
We know,
Radius of the Earth ($R$)$ = 6.38 \times {10^6}{\text{ }}m$
Gravitation Constant ($G$)$ = 6.67 \times {10^{ - 11}}{\text{ }}\dfrac{{{m^3}}}{{kg.{s^2}}}$
Mass of the Earth (${M_e}$)$ = 5.97 \times {10^{24}}\,kg$
Earth's rotational time period ($T$)$ = 24{\text{ }}hr = 24 \times 60 \times 60 = 86400{\text{ }}s$
Let the height of the orbit from Earth's surface be $h{\text{ }}m$ and a satellite of mass $m{\text{ }}kg$ rotates with uniform angular velocity $\omega {\text{ }}\dfrac{{rad}}{s}$ in the geosynchronous orbit.
So, Gravitational force$ = \dfrac{{G{M_e}m}}{{{{\left( {R + h} \right)}^2}}}$ and Centripetal force$ = m{\omega ^2}\left( {R + h} \right)$
Equating Centripetal force with the Gravitational force we can write
$\dfrac{{G{M_e}m}}{{{{\left( {R + h} \right)}^2}}} = m{\omega ^2}\left( {R + h} \right)$
Eliminating $m$ from both sides and substituting $\omega = \dfrac{{2\pi }}{T}$ we get,
$\dfrac{{G{M_e}}}{{{{\left( {R + h} \right)}^2}}} = {\left( {\dfrac{{2\pi }}{T}} \right)^2}\left( {R + h} \right)$
$ \Rightarrow \dfrac{{G{M_e}}}{{{{\left( {\dfrac{{2\pi }}{T}} \right)}^2}}} = {\left( {R + h} \right)^3}$
Simplifying the equation we get,
$\dfrac{{G{M_e}{T^2}}}{{4{\pi ^2}}} = {\left( {R + h} \right)^3}$
Cube root of the following equation we get,
${\left[ {\dfrac{{G{M_e}{T^2}}}{{4{\pi ^2}}}} \right]^{\dfrac{1}{3}}} = \left( {R + h} \right)$
Substituting the values in the above equation,
$\left( {R + h} \right) = {\left[ {\dfrac{{6.67 \times {{10}^{ - 11}} \times 5.97 \times {{10}^{24}} \times {{\left( {86400} \right)}^2}}}{{4 \times {{\left( {3.14} \right)}^2}}}} \right]^{\dfrac{1}{3}}}$
$\left( {R + h} \right) = 4224.1 \times {10^4}{\text{ }}m = 42241{\text{ }}km$
So, the height from the Earth's surface that a satellite must attain in order to be in geosynchronous orbit is,
$h = 42241 - R = 42241 - 6380 = 35861{\text{ }}km$
And the velocity of the satellite is,
$v = \left( {R + h} \right)\omega = \left( {R + h} \right)\dfrac{{2\pi }}{T}$
Substituting the values we get,
$v = 42241 \times {10^3} \times \dfrac{{2 \times 3.14}}{{86400}} = 3070.3{\text{ }}\dfrac{m}{s}$

Note: A geosynchronous orbit is a circular orbit that lies on the Earth's equatorial plane and whose period of time is equal to earth’s rotation. It follows the direction of Earth’s rotation. So, it appears to be fixed at a point. It may also be called geostationary.