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 $9.32m/s^2$. 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)

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 $v=\sqrt{9.32\left(R_e+h\right)}$
or $T=\frac{2\pi \left(R_e+h\right)}{v}=\frac{2\pi \left(R_e+h\right)}{\sqrt{9.32\left(R_e+h\right)}}$
$T=\frac{2\pi \sqrt{R_e+h}}{\sqrt{9.32}}-$
$\therefore 5.26\times 10^3=\frac{2\times 3.14\sqrt{R_e+h}}{3.05}$
$\sqrt{R_e+h}=2.55\times 10^3$
$R_e+h=6.53\times 10^6$
$h=6.53\times 10^6-6.37\times 10^6$
$=0.16\times 10^{6}m$
$=160\times 10^{3}m=160km$