Question

An artificial satellite revolves round the earth at a height of $1000$km. The radius of the earth is $6.38\times {10}^3$km. Mass of the earth $=6\times {10}^{24}$kg; $G=6.67\times {10}^{-11}$ $N{m}^2$ $k{g}^{-2}$. Find the orbital speed and period of revolution of the satellite.

• A. $7364$m $s^{-1}$, $6297$s.
• B. $1064$m $s^{-1}$, $6297$s.
• C. $9364$m $s^{-1}$, $9297$s.
• D. $8364$m $s^{-1}$, $7297$s.

Answer 1

The correct option is A $7364$m $s^{-1}$, $6297$s.

Here, $h=1000$ km$=1000\times {10}^{3}m={10}^6$m
$r=R+h=6.38\times {10}^6+{10}^6=7.38\times {10}^6$m
Orbital speed,
$v_0=\sqrt{\frac{GM}{R+h}}=\sqrt{\frac{6.67\times {10}^{-11}\times 6\times {10}^{24}}{7.38\times {10}^6}}=7364$m $s^{-1}$
Time period,
$T=\frac{2\pi r}{v_0}=\frac{2\times \left(22/7\right)\times (7.38\times {10}^6)}{7.364\times {10}^3}=6297$s.