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 $g = 9.8 \,ms^{-2}$, then the orbital speed of the satellite is

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

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, $R=6.38 \times 10^{6} m , g=9.8 m s ^{-2}$
and $h=0.25 \times 10^{6} m$
$\therefore v_{o}=\left(6.38 \times 10^{6} m \right) \sqrt{\frac{\left(9.8 m s ^{-2}\right)}{\left(6.38 \times 10^{6} m +0.25 \times 10^{6} m \right)}}$
$=7.76 \times 10^{3} \,m\, s ^{-1}=7.76\, km \,s ^{-1} $
$\left(\because 1 \,km =10^{3} m \right)$