Question

A satellite is launched into a circular orbit of radius $R$ around the earth. A second satellite is launched into an orbit of radius $4R$. The ratio of their respective periods is

• A. $4:1$
• B. $1:8$
• C. $8:1$
• D. $1:4$
• E. $1:2$

Answer 1

The correct option is B

$1:8$

Step 1. Given data:

Radius of first satellite = $R$

Radius of second satellite = $4R$

Let the Time Period of the first satellite is $T_1$ and Time Period of second setellite is $T_2$,

Step 2. The ratio of Time Period of satellite $\frac{T_1}{T_2}$,

We know that, Kepler's third law

$T^2\propto R^3$

$\Rightarrow T^2=\alpha R^3$ (where, $\alpha$is proportionality constant.)

Therefore,

Time Period of the first satellite is $T_1$, $T_1^2=\alpha R^3\dots \left(1\right)$

Time Period of the first satellite is $T_2$, $T_2^2=\alpha {\left(4R\right)}^3\dots \left(2\right)$

The ratio of Time Period of satellite $\frac{T_1}{T_2}$,

Dividing equation $\left(1\right)$ by equation $\left(2\right)$ we have,

$$\begin{gathered} \frac{{T^2}_1}{{T^2}_2}=\frac{\alpha R^3}{\alpha {\left(4R\right)}^3} \\ \Rightarrow \frac{{T^2}_1}{{T^2}_2}=\frac{R^3}{{\left(4R\right)}^3} \\ \Rightarrow \frac{T_1}{T_2}=\frac{1}{8} \end{gathered}$$

Hence the ratio of Time Period of satellite $T_1:T_2=1:8$,

Hence, Option B is correct.