Q. A synchronous satellite goes around the earth once in every 24 h. What is the radius of orbit of the synchronous satellite in terms of the earth’s radius (Given mass of the earth, $m_{e} = 5.98 \times 10^{24}kg$. radius of earth, $r_{e} = 6.37 \times 10^{6}m$, Universal constant of gravitation, $G = 6.67 \times 10^{- 11}Nm^{2}/kg^{2})$
A) $2.4r_{e}$
B) $3.6r_{e}$
C) $4.8r_{e}$
D) $6.6r_{e}$
Explanation
(d) Time period $(T)$ of a synchronous satellite around the earth is given by
$T^{2} = \frac{4\pi^{2}r^{3}}{Gm_{e}} \Rightarrow r = \left( \frac{T^{2}Gm_{e}}{4\pi^{2}} \right)^{1/3}$
Substituting the given values, we get
$r = \left\lbrack \frac{(24 \times 60 \times 60)^{2} + 6.67 \times 10^{- 11} \times 5.98 \times 10^{24}}{4 \times \frac{22}{7} \times \frac{22}{7}} \right\rbrack^{1/3}$
$r = 42.08 \times 10^{6}m$
$\therefore\frac{r}{r_{e}} = \frac{42.08 \times 10^{6}m}{6.37 \times 10^{6}m} = 6.6 \Rightarrow r = 6.6r_{e}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.