Question

A geo-stationary satellite orbits around the earth in a circular orbit which is at a height of $36000\,km$ from the surface of earth. Then the period of spy satellite orbiting a few hundred km above the earth’s surface (${{\text{R}}_{{\text{earth}}}} = 6400\,km$) will approximately be
A. $\dfrac{1}{2}\,hr$
B. $1\dfrac{1}{2}\,hr$
C. $2\,hr$
D. $4\,hr$

Answer 1

Hint: To solve this question, we must have a concept of Kepler’s Third law and then one can easily solve the question. Here firstly we will write all the given values and then as per Kepler’s third law we will substitute the values and solve for the period of spy satellite orbit and hence we will get our required solution.

Complete step by step answer:
According to the question we have, it is given that the height of a satellite above the earth’s surface ${R_1} = 36000\,km$. ${T_1}$ be the time period of a geo-satellite.And we know that the radius of the earth is ${R_2} = 6400\,km$ and ${T_2}$ be the time periods for the other satellite. And we know that from Kepler’s third law for any square of the time period of any orbit is directly proportional to the cube of the semi-major axis.
${T^2} \propto {R^3}$

From here we can say that
\[\dfrac{{{T_2}}}{{{T_1}}} = {\left( {\dfrac{{{R_2}}}{{{R_1}}}} \right)^{\dfrac{3}{2}}}\]
And here we can put the values and solve,
\[{T_2} = {T_1}{\left( {\dfrac{{{R_2}}}{{{R_1}}}} \right)^{\dfrac{3}{2}}} \\
\Rightarrow {T_2} = 24{\left( {\dfrac{{6400}}{{36000}}} \right)^{\dfrac{3}{2}}} \\
\Rightarrow {T_2} = 24 \times 0.08 \\
\therefore {T_2} = 2hr \\ \]
Hence, our correct option is C.

Note: The term "geo-stationary" refers to a satellite that seems to be at rest when viewed from Earth. This is because the angular velocity of the satellite's revolution is the same as the angular velocity of the earth's rotation. As a result, the time periods of the earth and satellite will be the same, i.e., 24 hours, but the linear velocities will differ due to the differing radius of revolutions.