Question

The time of revolution of a satellite close to earth in 90 min. The time of another satellite in an orbit at a distance of three times the radius of earth from its surface will be:
A) \[90\sqrt 8 \min \]
B) $360 min$
C) \[270\sqrt 3 \min \]
D) $720 min$

Answer 1

Hint: An orbit is defined as a regular and repeating path such that one object in space takes around another one and the object in an orbit is called a satellite. When an object is moving in a circle then orbital motion occurs.

Complete step by step solution:
Given data:
The time of revolution of the satellite \[{T_1} = 90\min \]
The time of revolution of another satellite \[{T_2} = ?\]
We know that from Kepler’s law, ${T^2} \propto {R^3}$
$ \Rightarrow T \propto {R^{\left( {\dfrac{3}{2}} \right)}}$
$ \Rightarrow \dfrac{{{T_1}}}{{{T_2}}} = {\left( {\dfrac{{{R_1}}}{{{R_2}}}} \right)^{3/2}}\_\_\_\_\_\_\_\_\_\left( 1 \right)$
Where T is the time-period and R is the radius
It is given that when ${R_1} = R$ then ${T_1} = 90\min $
So when ${R_2} = 4R,$ we have to find out the value of ${T_2}$
Substituting the values of ${T_1},{R_1},{R_2}$ in equation 1, we get
$\Rightarrow \dfrac{{90}}{{{T_2}}} = {\left( {\dfrac{R}{{4R}}} \right)^{3/2}}$
$ \Rightarrow {T_2} = 720\min $
Thus the time required by another satellite= 720 min

Hence the correct option is D.

Note: 1. There are three Kepler’s laws of planetary motion. Kepler’s third law states that the square of the period of any planet is directly proportional to the cube of the semimajor axis of the orbit.
2. The orbital period for the earth is 365 days while the orbital period of Mars is 687 days. Revolution can be defined as the act of turning round on an axis. The earth’s revolution effects include the different seasons, variations in the length of days and nights.
3. Planets move in rotation and revolution. Revolution and rotation take place due to gravity, centrifugal momentum, and angular momentum. Mostly the rotation comes from the conservation of angular momentum. Thus if the radius of the orbit decreases then the angular velocity increases and the mass will remain constant.