Question

A satellite in a circular orbit of radius $R$ has a period of $4h$. Another satellite with an orbital radius of $3R$ around the same planet will have a period

(in hours)

• A. $16$
• B. $4$
• C. $4\sqrt{27}$
• D. $4\sqrt{8}$

Answer 1

The correct option is C

$4\sqrt{27}$

Step 1. Data given:

The radius of the satellite orbit $R_1=R$

The time period for the satellite $T_1=4h$

The radius of orbit for the second satellite $R_2=3R$

Let the time period of second satellite is $T_2$.

Step 2. Finding the Time period of the second satellite:

If $T$ is the time period of a satellite and $R$ is the orbital radius then,

According to Kepler's third law, $T^2\propto R^3$

Assume the time period and radius of orbit for the first and second satellites as $T_{1,}{R}_1$and $T_{2,}{R}_2$respectively, then

$\Rightarrow$ ${\left(\frac{T_2}{T_1}\right)}^2={\left(\frac{R_2}{R_1}\right)}^3$

$\Rightarrow$ $\frac{T_2}{T_1}={\left(\frac{R_2}{R_1}\right)}^{\frac{3}{2}}$

$\Rightarrow$ $\frac{T_2}{4}={\left(\frac{3R}{R}\right)}^{\frac{3}{2}}$

$\Rightarrow$ $T_2=4\sqrt{27}h$

Hence the time period of the second satellite is $T_2=4\sqrt{27}h$.

Therefore, option (C) is correct.