Question

A satellite is moving very close to a planet of density $8\times {10}^{3}kg{m}^{-3}$. If $G=6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}$, then the time period of the satellite is nearly.

• A. $420s$
• B. $4200s$
• C. $1$ hour
• D. $1$ day

Answer 1

The correct option is B $4200s$

Given,

$\rho =8\times {10}^{3}kg/m^3$

$G=6.67\times {10}^{-11}N{m}^2/k{g}^2$

The time period of the satellite,

$T=2\pi \sqrt{\frac{r^3}{GM}}$. . . . . . .(1)

where, $r=$ radius of the orbit

$M=$ mass of the planet

As the satellite is revolving very close to the planet

thus, $r=R$

Mass of the planet, $M=\rho \times \frac{4}{3}\pi R^3$

From equation (1),

$T=2\pi \sqrt{\frac{R^3}{G\rho \times \frac{4}{3}\pi R^3}}=\sqrt{\frac{3\pi }{\rho G}}$

$T=\sqrt{\frac{3\times 3.14}{8\times {10}^3\times 6.67\times {10}^{-11}}}$

$T=4200sec$

The correct option is B.