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, $\frac{K^3}{V}=\frac{6\times 10^{10}g}{R}$

So, $T=\sqrt{\frac{2\times 6\times 10^{10}}{3}}$

$=2\times 10^{5}sec$