Question

The period of revolution of an earth satellite close to the surface of the earth is $90 \,min$. The time period of another earth satellite in an orbit at a distance of three earth radii from its surface will be :

• A. 90 min
• B. $ 90\times \sqrt{8}\,min $
• C. 270 min
• D. 720 min

Answer 1

Answer: (D)

The period of revolution of a satellite depends only upon its height above earths surface, for a satellite very close to the surface of the earth, period of revolution is
$T_{1}=2 \pi \sqrt{\frac{R}{g}}$ .... (i)
For a height h above the earths surface,
$T_{2}=\frac{2 \pi}{R} \sqrt{\frac{(R+h)^{3}}{g}}$ ... (ii)
Given, $h=3 R $
$\therefore T_{2}=\frac{2 \pi}{R}=\sqrt{\frac{(R+3 R)^{3}}{g}}$
$=2 \pi \times 8 \sqrt{\frac{R}{g}}$ ... (iii)
From Eqs. (i) and (iii), we have
$\frac{T_{1}}{T_{2}}=\frac{1}{8}$
$ \Rightarrow T_{2}=90 \times 8=720 \,min$