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

Answer: (C)

170 km

Answer 2

Centripetal acceleration $(a_{c}) = \frac{v^{2}}{r}$ and $T = \frac{2\pi r}{v}$

From equation (i) and (ii) $r = \frac{a_{c}T^{2}}{4\pi^{2}}$

⇒ $R + h = \frac{9.32 \times (5.26 \times 10^{3})^{2}}{4 \times \pi^{2}}$

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