Question

Time period of revolution of polar satellite is around
(A). 6 minutes
(B). 100 minutes
(C). 8 hours
(D). 24 hours

Answer 1

Hint: Polar satellites orbit the earth in the lower earth orbit which is around 2000 km above the earth’s surface. Find the mathematical expression for the time period of revolution of satellite. Put the values regarding the polar satellite to find the answer to the above question.
Formula used:
$T=2\pi \sqrt{\dfrac{{{\left( R+h \right)}^{3}}}{GM}}$
Where T is the time period of revolution, R is the radius of earth, h is the distance to satellite from the surface of earth, G is the gravitational constant and M is the mass of earth.

Complete step by step answer:
Polar satellites can be described as the type of satellite that will orbit the earth from pole to pole. Polar satellite orbits have an inclination of about 90 degrees to the equator.
Polar satellites orbit the earth at the lower earth orbit. Polar satellites are often used to monitor earth, earth mapping etc.
A polar satellite travels over the poles from north to south in approximately an hour and a half.
These satellites orbit in the lower earth orbit. The lower earth orbit has a smaller altitude of about 2000 km.
We can find the time period of revolution of a satellite from the following equation
$T=2\pi \sqrt{\dfrac{{{\left( R+h \right)}^{3}}}{GM}}$
Where T is the time period of revolution, R is the radius of earth, h is the distance to satellite from the surface of earth, G is the gravitational constant and M is the mass of earth.
$\left( R+h \right)$ can also be called the radius of the satellite orbit.
It can also be expressed as
$T=2\pi \sqrt{\dfrac{{{\left( R+h \right)}^{3}}}{g{{R}^{2}}}}$
 Where, g is acceleration due to gravity $=\text{ }\dfrac{GM}{{{R}^{2}}}$
Now,
$\begin{align}
  & \text{R = 6300 km = 6}\text{.3 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{6}}}\text{ m} \\
 & \text{h = 2000 km = 2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{6}}}\text{ m} \\
 & \text{g = 9}\text{.8 m}{{\text{s}}^{-2}} \\
\end{align}$
Hence,
$\begin{align}
  & T=2\pi \sqrt{\dfrac{{{\left( R+h \right)}^{3}}}{g{{R}^{2}}}} \\
 & T=2\pi \sqrt{\dfrac{{{\left( \left( 6.3+2 \right)\times {{10}^{6}} \right)}^{3}}}{9.8\times {{\left( 6.3\times {{10}^{6}} \right)}^{2}}}} \\
 & T=2\pi \sqrt{\dfrac{571.787\times {{10}^{18}}}{388.962\times {{10}^{12}}}} \\
 & T=2\pi \times 1.212\times {{10}^{3}} \\
 & T=7618.04\text{ seconds} \\
 & \text{T= 126 minutes} \\
 & \text{T}\approx \text{ 100 minutes} \\
\end{align}$
So, the polar satellites approximately take 100 min to orbit around the earth.
The correct option is (B)

Note: Different kinds of satellites have different time periods of revolution depending on in which orbit they are orbiting earth. If a satellite is in medium earth orbit their time period of revolution will be higher. If a satellite is moving in a high earth orbit their time period of revolution will be the largest.