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)

$7.76 \, km\,s^{-1}$

Answer 2

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, $R = 6.38 \times 10^6m$, $\, g = 9.8 \,ms^{-2}$ and
$h = 0.25 \times 10^6\, m$
$\therefore V_o (6.38 \times 10^6\,m) \sqrt \frac {(9.8 \, ms^{-2})}{(6.38\, \times \,10^6 \, m + 0.25\, \times\, 10^6 \,m)}$
$7.76 \times 10^3 \,ms^{-1} = 7.76 \, kms^{-1}$ $(\therefore 1\,km = 10^3\, m)$