Question

Orbital radius of a satellite S of earth is four times that of a communication satellite C. Period of revolution of C is
$
  A.4days \\
  B.8days \\
  C.16days \\
  D.32days \\
 $

Answer 1

- Hint: Here we will proceed by using the approach of Kepler’s third law of planet motion It will help us to find out the period of a revolution of a communication satellite C.

Complete step-by-step solution -
Here it is given that radius of the earth satellite is 4 times of a ‘communication satellite’
Larger the distance of a planet from the sun, larger will be its period of revolution around the sun.
As per Kepler’s third law, the square of the period of any planet is proportional to the cube of the semi-major axis of the orbit. The phenomena capture the relationship between the distance between the distance of planets from the sun, and the total orbital periods.
Thus,
$\therefore {T^2}\alpha {R^3}$
${\left( {\dfrac{{Ts}}{{Tc}}} \right)^2} = {\left( {\dfrac{{Rs}}{{Rc}}} \right)^3}$
$\dfrac{{Ts}}{{Tc}} = {\left( {\dfrac{{Rs}}{{Rc}}} \right)^{3 \div 2}}$
It is given that $Rs = 4Rc$ ( because radius of satellite is four times that of C )
$\dfrac{{Ts}}{{Tc}} = {\left( {\dfrac{{4Rc}}{{Rc}}} \right)^{3 \div 2}}$
Rc and Rc will be cancelled, Then
$
  \dfrac{{Ts}}{{Tc}} = {\left( {4 \times 4 \times 4} \right)^{½}} \\
  \dfrac{{Ts}}{{Tc}} = \sqrt {64} \\
  \dfrac{{Ts}}{{Tc}} = 8 \\
  Ts = 8Tc \\
 $
Time period of a communication satellite is one day
$Ts = 8days$

Note: Whenever we come up with this type of question, one must know that a satellite in a circular orbit around a celestial body moves at a velocity where the gravitational force of this body equals the centripetal force necessary to maintain the constant circular orbit.