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)

Given, $T=5.26 \times 10^{3} s , a=9.32\, m / s ^{2}$
Centripetal acceleration $a=\frac{v^{2}}{r}=9.32$
or $v^{2}=9.32\, r$
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 =160 \,km $