Question

EARTH SATELLITE
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, $M_E=5.97\times {10}^{24}$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)

$6.6R_E$

The time period of satellite is $T=2\pi \sqrt{\frac{r^3}{G{M}_E}}$

Squaring both sides, we get $T^2=\frac{4{\pi }^2{r}^3}{G{M}_E}$

$r^3=\frac{T^{2}G{M}_E}{4{\pi }^2}\Rightarrow r={\left(\frac{T^{2}G{M}_E}{4{\pi }^2}\right)}^{1/3}$

Here, $T=24h=24\times 60\times 60s$

$G=6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}$

$M_E=5.98\times {10}^{24}kg$

$r={\left(\frac{{(24\times 60\times 60)}^2\times 6.67\times {10}^{-11}\times 5.98\times {10}^{24}}{4\times {\left(3.14\right)}^2}\right)}^{1/3}$

$=42.1\times {10}^{6}m$

$\therefore \frac{r}{R_E}=\frac{42.1\times {10}^6}{6.37\times {10}^6}=6.6\Rightarrow r=6.6R_E$