Question

A satellite is revolving round the earth in circular orbit at some height above surface of earth. It takes $5.26\times {10}^3$ seconds to complete a revolution while its centripetal acceleration is $9.92m/s^2$ . Height of satellite above surface of earth is (Radius of earth $6.37\times {10}^6$ m)

• A. 70 km
• B. 120 km
• C. 170 km
• D. 220 km

Answer 1

The correct option is C

170 km

Centripetal acceleration $\left(a_c\right)=\frac{v^2}{r}andT=\frac{2\pi r}{v}$

From equation (i) and (ii) $r=\frac{a_c{T}^2}{4{\pi }^2}\Rightarrow R+h=\frac{9.32\times (5.26\times {10}^3)}{4\times {\pi }^2}$

$h=6.53\times {10}^6-R=6.53\times {10}^6-6.37\times {10}^6=160\times {10}^{3}m=160km\approx 170km$