A particle of mass m is moving in a stable circular orbit of radius r₀ with angular momentum L. For a potential energy V(r) = βr^k  (β>0 and k>0), which of the following options is correct?

Correct option is B

Solution:

​$$V(r) = \beta r^k \quad \text{force} \quad V_{\text{effective}} = \frac{L^2}{2 m r^2} + \beta r^k$$

For Circular orbit $$\frac{\partial V_{\text{effective}}}{\partial r} = 0 \Rightarrow -\frac{L^2}{m r^3} + \beta k r^{k-1} = 0 \Rightarrow r = r_0 = \left( \frac{L^2}{m \beta k} \right)^{\frac{1}{k+2}}$$

For Stable Orbit $$\frac{\partial^2 V_{\text{effective}}}{\partial r^2} \bigg|_{r = r_0} > 0$$

Stability condition after taking second derivative:

$$\frac{3 L^2}{m r^4} + \beta k (k - 1) r^{k-2} > 0 \Rightarrow 3 \beta k r^{k-2} + \beta k (k - 1) r^{k-2} > 0$$

​$$3 + k - 1 > 0 \Rightarrow k + 2 > 0 \Rightarrow k > -2$$

k=2

​​​So, Answer is Option b.