Question

A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the ${\text{n}}^{th}$power of R. If the period of rotation of the particle is T, then

• A. $\text{T}\propto {\text{R}}^{n/2}$
• B. $\text{T}\propto {\text{R}}^{(\text{n}+1)/2}$
• C. $\text{T}\propto {\text{R}}^{3/2}\text{for any}\text{n}$
• D. $\text{T}\propto {\text{R}}^{\frac{\text{n}}{2}+1}$

Answer 1

The correct option is B $\text{T}\propto {\text{R}}^{(\text{n}+1)/2}$

Find the value of angular speed $\left(\omega \right)$
Given:
Force $\propto \frac{1}{{\text{R}}^{\text{n}}}$
As we know,
Force = $\text{m}{\omega }^2\text{R}$
Hence,
$\text{m}{\omega }^2\text{R}\propto \frac{1}{R^{\text{n}}}\Rightarrow {\omega }^2\propto \frac{1}{{\text{R}}^{\text{n}+1}}$
$\Rightarrow \omega \propto \frac{1}{{\text{R}}^{\frac{\text{n}+1}{2}}}$
Find the value of Time period (T).
Therefore,
Time period,
$\text{T}=\frac{2\pi }{\omega }$
Hence, from given equation we get,
$\text{T}\propto \frac{1}{\omega }$
By putting the value of $\omega$ we get,
$\text{T}\propto {\text{R}}^{\frac{\text{n}+1}{2}}$
Final Answer: Option (d)