Question

The period of revolution of an Earth’s satellite close to the surface of Earth is $90\min $. The period of another Earth’s satellite in an orbit at a distance of three times Earth’s radius from its surface will be.

Answer 1

Hint: The time period of revolution of a satellite close to the surface of the Earth is considered to be the same as that of a satellite moving along the surface of the Earth. Time period of revolution of a satellite is given by Kepler’s third law planetary motion. Ratio of time period of satellite moving close to the Earth’s surface to the time period of satellite moving at a distance from the Earth’s surface can be taken to obtain the answer.

Formula used:
${{T}^{2}}=\dfrac{4{{\pi }^{2}}{{R}^{3}}}{GM}$
where
$T$ is the time period of revolution of a satellite
$R$ is the radius of the orbit in which the satellite revolves
$G$ is the gravitational constant
$M$ is the mass of the satellite

Complete step by step answer:
Kepler’s third law states that the square of time period of revolution of a satellite is directly proportional to the cube of orbital radius of the satellite. The law says that
${{T}^{2}}\propto {{R}^{3}}$.
This can be rewritten as
$T\propto {{\left( R \right)}^{\dfrac{3}{2}}}$
This is clear from the formula of time period of satellite revolution as mentioned above.
Let us apply the law to the satellite moving near the surface of the Earth. Let us call the time period of satellite ${{T}_{1}}$ and the radius of the satellite ${{R}_{1}}$. Here, ${{R}_{1}}={{R}_{earth}}$, which is the radius of the Earth. We have
${{T}_{1}}\propto {{\left( {{R}_{earth}} \right)}^{\dfrac{3}{2}}}$ (equation 1)
Now, let us apply the rule to the satellite moving at a distance from the Earth’s surface. Let us call the time period ${{T}_{2}}$and the radius ${{R}_{2}}$in this case. It is given that the satellite is at a distance three times the radius of the Earth. Hence, the orbital radius of the satellite is four times$(3+1)$ the radius of the Earth. So, we have
${{R}_{2}}=4{{R}_{earth}}$
On applying Kepler’s law,
${{T}_{2}}\propto {{\left( 4{{R}_{earth}} \right)}^{\dfrac{3}{2}}}$ (equation 2)
Now, let us take the ratio of time periods mentioned in equation 1 and equation 2.
$\dfrac{{{T}_{1}}}{{{T}_{2}}}\propto {{\left( \dfrac{{{R}_{earth}}}{4{{R}_{earth}}} \right)}^{\dfrac{3}{2}}}$
Simplifying the above equation, we have
\[\begin{align}
  & \dfrac{{{T}_{1}}}{{{T}_{2}}}={{\left( \dfrac{1}{4} \right)}^{\dfrac{3}{2}}} \\
 & \dfrac{{{T}_{1}}}{{{T}_{2}}}=\dfrac{1}{\left( {{4}^{\dfrac{3}{2}}} \right)}=\dfrac{1}{{{\left( {{2}^{2}} \right)}^{\dfrac{3}{2}}}}=\dfrac{1}{{{2}^{3}}}=\dfrac{1}{8} \\
\end{align}\]
We are provided that ${{T}_{1}}=90\min $
Substituting this value in the above equation, we have
${{T}_{2}}=8\times {{T}_{1}}=8\times 90\min =720\min $
Therefore, the time period of the satellite at a distance equal to three times the radius of the Earth away from the surface of the Earth is equal to $720\min $.

Note:
Students should take utmost care while taking the value of radius of the satellite. It is given that the distance of the satellite is three times the radius of the Earth. This means that the orbital radius of the satellite is equal to four times the radius of the Earth.
${{R}_{2}}=3{{R}_{earth}}+{{R}_{earth}}$