Question

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 earths radius? (Given : Mass of the earth, $M_E = 5.98 \times 10^{24}\, kg$, radius of the earth, $R_E = 6.37 \times 10^6\, m$, universal constant of gravitation, $G = 6.67 \times 10^{-11}\, N\, m^2\, kg^{-2}$)

• A. $2.4 \,R_E$
• B. $3.6 \,R_E$
• C. $4.8 \,R_E$
• D. $6.6 \,R_E$

Answer 1

Answer: (D)

The time period of satellite is
$T= 2\pi\sqrt{\frac{r^{3}}{GM_{E}}}$
Squaring both sides, we get
$T^{2} = \frac{4\pi^{2}r^{3}}{GM_{E}}$
$r^{3} = \frac{T^{2}GM_{E}}{GM_{E}}$
$\Rightarrow r = \left(\frac{T^{2}GM_{E}}{GM_{E}}\right)^{1/3}$
Here, $T = 24\, h = 24 \times 60 \times 60\, s$
$G = 6.67 \times 10^{-11}\,N \,m^{2}\, kg^{-2}$
$M_E = 5.98 \times 10^{24}\, kg$
$r = \left(\frac{\left(24\times 60\times 60\right)^{2}\times 6.67 \times 10^{-11}\times 5.98\times 10^{24}}{4\times \left(3.14\right)^{2}}\right)^{1/3}$
$= 42.1 \times 10^{6}\,m$
$\therefore \frac{r}{R_{E}} = \frac{42.1 \times 10^{6}}{6.37 \times 10^{6}} = 6.6$
$\Rightarrow r = 6.6R_{E}$