Question

The ratio of cube of circumference of the orbit of a satellite to the volume of the earth is $6 \times 10^{10} (g/R)$, where $g$ is the acceleration of gravity. Find the time period of satellite (in seconds). $R$ is the radius of earth

• A. $2 \times 10^5$
• B. $10^5$
• C. $4 \times 10^4$
• D. $10^4$

Answer 1

Answer: (A)

$T =\frac{2\pi r}{R}\sqrt{\frac{r}{g}} = \frac{K}{R}\sqrt{\frac{K}{2\pi g}}$ If $K = 2\pi r$
$= \sqrt{\frac{\frac{2K^{3}}{3}R}{\left(\frac{4}{3}\pi R^{3}\right)g}} = \sqrt{\frac{2}{3} \frac{K^{3}R}{Vg}} $
Given, $\frac{K^{3}}{V} = \frac{6\times10^{10}g}{R}$
So, $T =\sqrt{\frac{2\times 6\times10^{10}}{3}} $
$= 2 \times10^{5} sec$