Question

Calculate the height of the communication satellite.
[Given: $G=6.67\times {{10}^{-11}}\text{ N}{{\text{m}}^{2}}/k{{g}^{2}},M=6\times {{10}^{24}}kg,R=6400km$]

Answer 1

Hint: A communication satellite will always be at a height ‘h’ from the surface of the earth. The centripetal acceleration of the satellite in orbit is balanced by the gravitational force acting between the earth and the satellite. If we apply this concept, we can find the height at which the satellite is rotating about the earth. Also, we can assume that the communication satellite is in a geostationary orbit.

Complete step by step answer:
So, as I mentioned in the hint, the gravitational force between the satellite and the earth provides the necessary centripetal acceleration for the satellite to revolve around the earth.
If ‘r’ is the distance between the earth’s centre and the satellite, the distance can be split into the distance of the earth plus the height at which the satellite is orbiting.
So, we can write $r=R+h$, where, r is the total distance between the earth and satellite. R is the radius of the earth. h is the height from the surface of the earth at which the satellite is rotating.

The centripetal acceleration associated with a moving body of mass m around a circular path of radius r is given by,

${{a}_{c}}=\dfrac{m{{v}^{2}}}{r}$

So, the centripetal acceleration for the satellite revolving around the earth is given by,

${{a}_{c}}=\dfrac{m{{v}^{2}}}{r}=\dfrac{m{{v}^{2}}}{(R+h)}$ …. Equation (1)

Where m is the mass of the satellite. R is the radius of the earth. h is the height at which the satellite is rotating above the earth.

The gravitational force between the satellite and the earth is,

${{F}_{g}}=\dfrac{GMm}{{{(R+h)}^{2}}}$ … equation (2)

Where G is the gravitational constant. M is the mass of the earth.
So these two forces must be equal, so equating equation (1) and (2), we get,

$\begin{aligned} & {{F}_{c}}={{F}_{g}} \\\ & \Rightarrow \dfrac{m{{v}^{2}}}{(R+h)}=\dfrac{GMm}{{{(R+h)}^{2}}} \\\ \end{aligned}$

$v=\sqrt{\dfrac{GM}{(R+h)}}$

Where, v is the orbital velocity of the satellite around the earth.

Since, we are dealing with a communication satellite, the time period of the communication satellite will always be 24 hours because it occupies a geostationary orbit in the sky.
So the time period of the communication satellite is given by,

$T=\dfrac{2\pi r}{v}=\dfrac{2\pi \left( R+h \right)}{v}$

Where, v is the orbital velocity we have derived above. Substituting the value of v in the above equation we get,
$T=2\pi \left( R+h \right)\sqrt{\dfrac{(R+h)}{GM}}$
We can substitute for T as 24 hours, so we get

$24hrs=2\pi \sqrt{\dfrac{{{(R+h)}^{3}}}{GM}}$

From the above equation, we can find h as,

$h={{\left( GM{{\left( \dfrac{24hr}{2\pi } \right)}^{2}} \right)}^{\dfrac{1}{3}}}-R$

Substituting the values of G, M and R in the above equation we get,

$\therefore h\approx 35911km$

So, the height at which the communication satellite orbits the earth is approximately 35911km.

Note: A communication satellite is mostly a geostationary satellite, which means the time period of the satellite is equal to the time period of the earth. It is focused at a fixed point on the earth and rotates along with that fixed point. Communication satellites can be deployed in orbits other than the geostationary orbits.
The acceleration due to the gravity of the earth is given by the formula, $g=\dfrac{GM}{{{R}^{2}}}$ and it has a value of $9.8\text{ m/}{{\text{s}}^{2}}$ near the surface of the earth. Its value decreases as we go to higher altitudes above the surface of the earth. Also, if we go from the surface of the earth to a depth ‘d’ into the earth’s surface, the acceleration due to gravity decreases, reaching the value zero at the centre of the earth.