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  4. An earth satellite X is revolving around earth in an orbit whose radius is one-fourth of the radius of orbit of a communication satellite. Time period of revolution of X is

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Q. An earth satellite $X$ is revolving around earth in an orbit whose radius is one-fourth of the radius of orbit of a communication satellite. Time period of revolution of $X$ is

Gravitation

Solution:

Time period of a communication satellite $= 24$ hours.
Using kepler's third law,
$T^{2} \propto r^{3} $
$\Rightarrow \frac{T_{c}}{T_{x}}=\left(\frac{r_{c}}{r_{x}}\right)^{3 / 2}$
$\Rightarrow \frac{24}{T_{x}}=(4)^{3 / 2} $
$\Rightarrow T_{x}=\frac{24}{8}=3 hrs$