1. Condition for Circular Orbit of Light

For a light ray to travel in a circular path concentric with the planet, the bending of the ray due to the refractive index gradient must exactly match the curvature of the circle. In a spherically symmetric medium, the generalized Snell’s Law (Bouguer’s formula) is: $$ \mu(r) \cdot r \cdot \sin \theta = \text{constant} $$ For a circular path at a constant radius $r = R+h$, the angle of incidence with the radial direction is always $\theta = 90^\circ$. Thus, $\sin \theta = 1$, and the condition becomes: $$ \mu r = \text{constant} $$

2. Deriving the Height

For the trajectory to be stable at radius $r$, the derivative of the quantity $\mu r$ with respect to $r$ must be zero (or strictly, the gradient must provide the necessary curvature). Differentiating $\mu r$ with respect to $r$: $$ \frac{d}{dr} (\mu r) = \mu + r \frac{d\mu}{dr} = 0 $$ $$ \frac{d\mu}{dr} = -\frac{\mu}{r} \quad \text{…(1)} $$

We are given the refractive index profile as a function of altitude $h$: $$ \mu = \mu_0 – kh $$ Since $r = R + h$, we have $dr = dh$. Thus: $$ \frac{d\mu}{dr} = \frac{d\mu}{dh} = -k $$

Substituting this into equation (1): $$ -k = -\frac{\mu_0 – kh}{R + h} $$ $$ k(R + h) = \mu_0 – kh $$ $$ kR + kh = \mu_0 – kh $$ $$ 2kh = \mu_0 – kR $$