Question

For a satellite orbiting close to the surface of earth the period of revolution is $84$ minute. The time period of another satellite orbiting at a height three times the radius of earth from its surface will be

• A. $\left(84\right)2\sqrt{2}$ minute
• B. $\left(84\right)8$ minute
• C. $\left(84\right)3\sqrt{3}$ minute
• D. $\left(84\right)3$ minute

Answer 1

The correct option is B $\left(84\right)8$ minute

$T\propto r^{3/2}$
$\frac{T_2}{T_1}={\left(\frac{r_2}{r_1}\right)}^{3/2}={\left(\frac{3R+R}{R}\right)}^{3/2}$
$\frac{T_2}{T_1}=8$
$\therefore T_2=8T_1=8\left(84\right)minutes$