Question

A satellite is revolving around a planet in an orbit of radius$R$. Suddenly radius of orbit becomes $1.02R$, then what will be the percentage change in its time period of revolution?

• A. $2\%$
• B. $3\%$
• C. $5\%$
• D. $7\%$

Answer 1

The correct option is B

$3\%$

Step 1. Given data:

Radius$=R$

Step 2. Find the percentage change in the satellite time period of revolution:

According to Kepler's law,

$T^2\propto R^3$

$\Rightarrow$$T\propto R^{(3/2)}$ ......(1)

Where,

$T=$ time period

$R=$ radius

Now, differentiating both sides we get

$dT=\left(\frac{3}{2}\right)R^{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}dR$

Now, Dividing both sides by equation (1), we get

$\Rightarrow$$\frac{dT}{T}=\frac{\left({\displaystyle \frac{3}{2}}\right)R^{({\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.})}}{R^{\left({\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}}dR$

$\Rightarrow$$\frac{dT}{T}=\frac{3}{2}\frac{dR}{R}$​ .......(2)

Now,

Step 3. Percentage difference in radius is

$$\begin{gathered} =\frac{dR}{R}\times 100\% \\ =\frac{1.02R-R}{R}\times 100\% \\ =2\% \end{gathered}$$

So, Percentage difference in time period,

$$\begin{gathered} =\frac{dT}{T}\times 100 \\ =23\times 2 \\ =3\% \end{gathered}$$

Thus, the percentage change in its time period of revolution is $\mathbf{3}\mathbf{\%}$

Hence, (B) is correct option.