Spiral Orbit

Consider a point in space parameterised in polar coordinates:

$x = r\cos{\theta}$ $y = r\sin{\theta}$

$\dot{x} = \dot{r}\cos{\theta} - r\dot{\theta}\sin{\theta}$ $\dot{y} = \dot{r}\sin{\theta} + r\dot{\theta}\cos{\theta}$

Assuming the particle mass to be unity, the kinetic energy of the system is:

$T = \frac{1}{2}\left(\dot{x}^2 + \dot{y}^2\right) = \frac{1}{2}\left(\dot{r}^2 + r^2 \dot{\theta}^2\right)$ $V = Ar^p + Br^q$

Applying Lagrangeé equation in terms of the $\theta$ coordinate gives:

$\frac{d}{dt}\left(\frac{\partial T}{\partial \dot{\theta}}\right) - \frac{\partial T}{\partial \theta} + \frac{\partial V}{\partial \theta}=0$ $\implies \frac{d}{dt}\left(r^2 \dot{\theta}\right) = 0$ $\implies r^2\dot{\theta} = K \ \dots \ (1)$

Where $K$ is some constant.

Now, applying the energy conservation principle gives: $T + V = K_1$

$\implies \frac{1}{2}\left(\dot{r}^2 + r^2 \dot{\theta}^2\right) + Ar^p + Br^q = K_1$

Here, $K_1$ is also constant.

Re-arranging the above gives:

$\dot{r}^2 = K_1 - Ar^p - Br^q - r^2 \dot{\theta}^2$

Consider the right-hand side of the above equation. Using (1), we get:

$\mathrm{RHS} = K_1 - Ar^p - Br^q - r^2 \dot{\theta}^2=K_1 - Ar^p - Br^q - \frac{K^2}{r^2}$

Now consider the left-hand side:

We have:

$r = c\theta^2$ $\implies \theta = K_3 \sqrt{r}$ $\dot{r}=2c\theta\dot{\theta}$

Using (1) in the equation above gives:

$\dot{r}^2 = K_4 \frac{1}{r^3}$

$\mathrm{LHS} = \dot{r}^2 = K_4 \frac{1}{r^3}$

Now, for LHS to be equal to RHS, $K_1$ must be zero and $q=-2$ and $B=-K^2$ and $p = -3$ and $K_4 =-A$

Essentially $\lvert p \rvert > \lvert q \rvert$

And the answer is $\boxed{1.5}$ .

I do not know if this is the correct approach because assuming that the total energy is zero to satisfy the condition does not seem very natural to me.

Since the force is central, $r^2\dot {\theta}=$ constant. From the equation of the path, $r$ is proportional to $\theta^2\implies \dot {r}$ is proportional to $r^{-1.5}$ . So $\ddot {r}$ is proportional to $r^{-4}$ . Since the potential is $αr^p+βr^q$ , the acceleration, which is purely radial, is proportional to $-(αpr^{p-1}+βqr^{q-1})$ . Therefore $\ddot {r}-r\dot {\theta}^2$ , which is of the form $\dfrac{a}{r^4}+\dfrac{b}{r^3}$ , is identical with $-(αpr^{p-1}+βqr^{q-1})$ . So, $p-1=-4, q-1=-3\implies p=-3, q=-2\implies \dfrac{p}{q}=\dfrac{3}{2}=\boxed {1.5}$