Question

A body is orbiting around the earth at a mean radius which is two times as greater as the parking orbit of a satellite, the period of the body in days is

• A. $4$
• B. $16$
• C. $2\sqrt{2}$
• D. $64$

Answer 1

The correct option is C

$2\sqrt{2}$

Step 1. Given data

A body is orbiting around the earth at a mean radius which is two times the radius of parking orbit of a satellite.

Step 2. Finding the ratio of the time period of the body, $T_2$ to time period of the satellite, $T_1$

By using the, formula of the Kepler's Third Law, the square of the time period of the orbit is directly proportional to the cube of the radius of the orbit.

$T^2$$\propto$ $R^3$ [where, $T$ is Time period , $R$ is orbital radius.]

$\Rightarrow$$\frac{{\left(T_2\right)}^2}{{\left(T_1\right)}^2}={\left(\frac{R_2}{R_1}\right)}^3$ [Given in the question, $R_1=R,R_2=2R$]

$\Rightarrow$$\frac{{\left(T_2\right)}^2}{{\left(T_1\right)}^2}={\left(\frac{2R}{R}\right)}^3$ [$R_1$ is radius of orbit of satellite, $R_2$ is radius of body]

$\Rightarrow$ ${\left(T_2\right)}^2=\frac{8R^3}{R^3}{\left(T_1\right)}^2$

$\Rightarrow$ ${\left(T_2\right)}^2=8\times 1$ [$\because T_1=1$day (geostationary)]

$\Rightarrow$ $T_2=2\sqrt{2}$

Therefore, the period of the body is $2\sqrt{2}$ $days$.

Hence, option $\left(C\right)$ is the correct answer.