Question

A satellite is moving in a circular orbit at a certain height above the earth's surface. It takes $5.26\times {10}^{3}s$ to complete one revolution with a centripetal acceleration equal to 9.32 m/s${}^2$. The height of satellite orbiting above the earth is
(Earth's radius $=6.37\times {10}^{6}m$)

• A. 220 km
• B. 160 km
• C. 70 km
• D. 120 km

Answer 1

Answer: (B)

160 km

Given : $T=5.26\times {10}^{3}s$, $a=9.32m/s^2$
Centripetal acceleration $a=\frac{v^2}{r}=9.32$
or $v^2=9.32r$
or $v=\sqrt{9.32(R_e+h)}$
Time period of revolution is given by $T=\frac{2\pi (R_e+h)}{v}=\frac{2\pi (R_e+h)}{\sqrt{9.32(R_e+h)}}$
or $T=\frac{2\pi \sqrt{R_e+h}}{\sqrt{9.32}}$
$\therefore 5.26\times {10}^3=\frac{2\times 3.14\sqrt{R_e+h}}{3.05}$
$\sqrt{R_e+h}=2.55\times {10}^3$
$R_e+h=6.53\times {10}^6$
$h=6.53\times {10}^6-6.37\times {10}^6$ $(\because R_e=6.37\times {10}^{6}m)$
$h=0.16\times {10}^{6}m$
$h=160\times {10}^{3}m=160km$