Question

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

Answer 1

- Hint: For the satellite revolving around earth in a uniform motion, the forces acting will be zero. Therefore, balance the centripetal force and gravitational force out.
Formula used: The formula for centripetal force acting radially outward:
${{F}_{c}}=\dfrac{m{{v}^{2}}}{r}$
The formula for gravitational force acting radially inward:
${{F}_{g}}=\dfrac{GMm}{{{r}^{2}}}$
The formula for speed in a uniform circular motion:

Complete step-by-step solution -

We are given the mass of Earth $M=6\times {{10}^{24}}kg$, radius of Earth $R=6400km=64\times {{10}^{5}}m$.
A satellite revolves or orbits around earth at a distance $h$ from the surface of the earth. The trajectory of the satellite is circular, therefore the radius for its motion becomes $r=R+h$. We know that satellites will experience a gravitational pull towards Earth. This is the gravitational force.
${{F}_{g}}=\dfrac{GMm}{{{r}^{2}}}$ , where $m$ is the mass of the satellite.
Another force, centripetal force is experienced when a body is moving in a circular path with radius $r$ and the velocity $v$. The expression for this force is:
${{F}_{c}}=\dfrac{m{{v}^{2}}}{r}$
Gravitational force acts radially inward, while the centripetal force acts radially outwards. For the satellite to orbit in a uniform circular motion,
\[\begin{align}
  & {{F}_{\text{gravitation}}}={{F}_{\text{centripetal}}} \\
 & \dfrac{GMm}{{{r}^{2}}}=\dfrac{m{{v}^{2}}}{r} \\
 & \text{where, v=}\dfrac{2\pi r}{T} \\
 & T=\text{Time for one revolution}=\text{1 day}=\text{24hrs}=24\times 60\times 60s \\
 & \Rightarrow \dfrac{GM}{r}={{v}^{2}} \\
 & \Rightarrow GM=r{{\left[ \dfrac{2\pi r}{T} \right]}^{2}} \\
 & \Rightarrow {{r}^{3}}=6.67\times {{10}^{-11}}\times 6\times {{10}^{24}}\times {{\left[ \dfrac{24\times 60\times 60}{2\pi } \right]}^{2}} \\
 & \Rightarrow h=r-R=4231.18\times {{10}^{4}}-640\times {{10}^{4}}m \\
 & \Rightarrow h=35912m=35.912km \\
\end{align}\]
Therefore, the answer to this question is 35.912 km.

Note: It is assumed that the satellite is geostationary. The function of such satellites is to revolve around the earth in such a way that it starts stationary with respect to the movement of Earth. Therefore, the time period of one revolution of the satellite is equal to that of the Earth, which is 24 hours.