Question

What is a period of revolution of earth satellite ? Ignore the height of satellite above the surface of earth.
Given : (1) The value of gravitational acceleration $g= 10\,ms^{-2}$
(2) Radius of earth $R_E = 6400\, km$. Take $\pi = 3.14$

• A. 156 minutes
• B. 90 minutes
• C. 85 minutes
• D. 83.73 minutes

Answer 1

Answer: (D)

Given,
$R_{e}=6400 \,km =6.4 \times 10^{6} \,m$
$\pi=3.14, \,g=10 \,m / s ^{2}$
We know that the period of revolution of the earth satellite
$T=2 \pi \sqrt{\frac{\left(R_{e}+h\right)^{3}}{g R_{\theta}^{2}}}$
[if $h << R_{e}$, then, $\left.\left(R_{e}+h=R_{e}\right)\right]$
So, $T=2 \pi \sqrt{\frac{R_{e}^{3}}{g R_{e}^{2}}}$
$=2 \pi \sqrt{\frac{R_{e}}{g}}=2 \times 3.14 \sqrt{\frac{6.4 \times 10^{6}}{10}}$
$=2 \times 3.14 \times 0.8 \times 10^{3}$
$=5.024 \times 10^{3}=5024 s$
and $\frac{5024}{60} =83.73\, min$