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 $g=10m{s}^{-2}$
(2) Radius of earth $R_E=6400km$. Take $\pi =3.14$

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

Answer 1

Answer: (D)

Given,
$R_e=6400km=6.4\times 10^{6}m$
$\pi =3.14,g=10m/s^2$
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, $\left(R_e+h=R_e\right)]$
So, $T=2\pi \sqrt{\frac{R_e^3}{g{R}_e^2}}$
$=2\pi \sqrt{\frac{R_e}{g}}=2\times 3.14\sqrt{\frac{6.4\times 10^6}{10}}$
$=2\times 3.14\times 0.8\times 10^3$
$=5.024\times 10^3=5024s$
and $\frac{5024}{60}=83.73min$