Periodic time of a satellite revolving above Earth’s surface at a height equal to R, radius of Earth, is
[g is acceleration due to gravity at Earth’s surface]
Answer & explanation
Correct answer: option 2
$\frac{GMm}{(R+h)^2}=m(R+h) \omega^2 ; \omega=\sqrt{\frac{GM}{(R+h)^3}}=\sqrt{\frac{gR^2}{(R+h)^3}}$
or $\frac{2 \pi}{R}=\sqrt{\frac{gR}{(R+h)^3}} ; T=2 \pi \sqrt{\frac{(R+h)^3}{gR^2}}$
here h = R
$T=2 \pi \sqrt{\frac{(R+R)^3}{gR}}=2 \pi \sqrt{\frac{8 R}{g}}$
$=(2 \sqrt{2}) 2 \pi \sqrt{\frac{R}{g}}=4 \sqrt{2} \pi \sqrt{\frac{R}{g}}$