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A geostationary satellite is orbiting the earth at a height of 5R above the surface of the earth, R being the radius of the earth. The time period of another satellite in hours at a height of 2R from the surface of the earth is
A
$\frac{6}{\sqrt2}$
B
5
C
10
D
$6\sqrt2$
Detailed Solution
$T^2 \propto r^3$
Geostationary satellite: T = 24 h, $r_1 = R + 5R = 6R$. Other satellite: $r_2 = R + 2R = 3R$.
$\frac{24}{T} = \left(\frac{6R}{3R}\right)^{3/2} = 2^{3/2} = 2\sqrt2$
$T = \frac{24}{2\sqrt2} = 6\sqrt2$ h
Geostationary satellite: T = 24 h, $r_1 = R + 5R = 6R$. Other satellite: $r_2 = R + 2R = 3R$.
$\frac{24}{T} = \left(\frac{6R}{3R}\right)^{3/2} = 2^{3/2} = 2\sqrt2$
$T = \frac{24}{2\sqrt2} = 6\sqrt2$ h