Question

Imagine a light planet revolving around a very massive star in a circular orbit of the radius $R$ with a period of revolution $T$. If the gravitational force of attraction between the planet and the star is proportional to $R^{\raisebox{1ex}{$-5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$, then

• A. $T^2$ is proportional to $R^3$
• B. $T^2$ is proportional to $R^{7/2}$
• C. $T^2$ is proportional to $R^{3/2}$
• D. $T^2$ is proportional to $R^{3/73}$

Answer 1

The correct option is B

$T^2$ is proportional to $R^{7/2}$

Step 1: Given data and assumptions:

Gravitational force is proportional to $R^{-5/2}$

Let gravitational force = $\frac{K}{R^{5/2}}$

Where $K$ is the proportionality constant,

Step 2: Formula used

Gravitational force acts as a centripetal force for the circular motion of one body around the massive body ⇒ $\frac{m{v}^2}{R}=\frac{K}{R^{5/2}}$

Where $m$ is mass, $v$ is velocity, $R$ is the radius of circular orbit and $K$ is the proportionality constant.

$$\begin{gathered} V^2=\frac{K*R}{m*R^{5/2}} \\ {V}^2=\frac{K}{m*R^{5/2}*R^{-1}} \end{gathered}$$

$v^2=\frac{K}{m{R}^{3/2}}$

Step 3: Calculating time period

Time period $T=\frac{2\mathrm{\pi R}}{v}$

⇒ $T^2=\frac{4{\mathrm{\pi }}^2{\mathrm{R}}^2}{v^2}$

Where $v$ is velocity and $R$ is the radius of circular orbit

On substituting the value of $v^2$:

$$\begin{gathered} T^2=\frac{4{\mathrm{\pi }}^2{\mathrm{R}}^2}{v^2} \\ {T}^2=\frac{4{\mathrm{\pi }}^2{\mathrm{R}}^2*{\mathrm{mR}}^{{\displaystyle \frac{3}{2}}}}{K}\Rightarrow \{R^2*R^{\frac{3}{2}}\}\Rightarrow \left\{R^{\frac{4+3}{2}}\right\}\Rightarrow \left\{R^{7/2}\right\} \end{gathered}$$

$T^2=\frac{4{\mathrm{\pi }}^2{\mathrm{mR}}^{7/2}}{K}$

$T^2$ is proportional to $R^{7/2}$

Hence, the correct answer is option B.