Question

A remote-sensing satellite of earth revolves in a circular orbit at a height of $0.25\times 10^{6}m$ above the surface of earth. If earth's radius is $6.38\times 10^{6}m$ and $g=9.8m{s}^{-2}$, then the orbital speed of the satellite is

• A. $9.13km{s}^{-1}$
• B. $6.67km{s}^{-1}$
• C. $7.76km{s}^{-1}$
• D. $8.56km{s}^{-1}$

Answer 1

Answer: (C)

The orbital speed of the satellite is
$v_o=R\sqrt{\frac{g}{(R+h)}}$
where $R$ is the earth's radius, $g$ is the acceleration due to gravity on earth's surface and $h$ is the height above the surface of earth.
Here, $R=6.38\times 10^{6}m,g=9.8m{s}^{-2}$
and $h=0.25\times 10^{6}m$
$\therefore v_o=\left(6.38\times 10^{6}m\right)\sqrt{\frac{\left(9.8m{s}^{-2}\right)}{\left(6.38\times 10^{6}m+0.25\times 10^{6}m\right)}}$
$=7.76\times 10^{3}m{s}^{-1}=7.76km{s}^{-1}$
$\left(\because 1km=10^{3}m\right)$