Question

A synchronous satellite goes around the earth once in every $24h$. 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, $G=6.67\times 10^{-11}N{m}^{2}k{g}^{-2}$)

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

Answer 1

Answer: (D)

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