Question

Write the equation for orbital velocity of a satellite orbiting the earth at a height of h from the surface of the earth.

Answer 1

Hint: Define orbital velocity. For a satellite orbiting earth, the required centripetal force is given by the gravitational attraction force of earth. We can find the orbiting velocity of the satellite by equating these two forces.

Complete step-by-step answer:
Consider the mass of the satellite is m and the mass of the earth is M. the satellite orbits around the earth due to the net centripetal force as a result of the gravitational attraction of earth on the satellite.
The net centripetal force on the satellite orbiting the earth at a height h from the earth’s surface is given by the mathematical expression,
${{F}_{c}}=\dfrac{m{{v}^{2}}}{(R+h)}$
Where, m is the mass of the satellite, v is the orbital velocity of the satellite around earth and $R+h$ is the orbital radius of the satellite around earth, R is the radius of the earth.
Now the gravitational force on the satellite due to earth is given as,
${{F}_{g}}=\dfrac{GMm}{{{\left( R+h \right)}^{2}}}$
Where G is the gravitational constant with value $G=6.67\times {{10}^{-11}}N{{m}^{2}}k{{g}^{-2}}$ and M is the mass of the earth.
Now, the centripetal force on the satellite is a result of the gravitational attraction of earth on the satellite. So, these two forces are equal. We can write,
$\begin{align}
  & {{F}_{g}}={{F}_{e}} \\
 & \dfrac{GMm}{{{\left( R+h \right)}^{2}}}=\dfrac{m{{v}^{2}}}{\left( R+h \right)} \\
 & \dfrac{GM}{\left( R+h \right)}={{v}^{2}} \\
 & v=\sqrt{\dfrac{GM}{\left( R+h \right)}} \\
\end{align}$
So, orbital velocity of the satellite around earth at a height h above the surface of earth is given by,
$v=\sqrt{\dfrac{GM}{\left( R+h \right)}}$

Note:
The orbiting velocity of a satellite around earth depends on the distance of the satellite from the earth. With the increase in the distance of the satellite from earth the orbiting velocity will decrease and with the decrease in the distance of the satellite from the earth, the orbiting velocity will increase.