In a solar system, the time-period of revolution of a planet tracing a circular orbit of radius R is proportional to :

According to Kepler's third law of planetary motion, the square of the time period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its elliptical orbit.

For a circular orbit, the semi-major axis is equal to the radius of the orbit $R$.

Therefore, $T^2 \propto R^3$

Taking the square root on both sides, we get:

$T \propto R^{3/2}$

Alternatively, equating the gravitational force to the required centripetal force:

$\dfrac{GMm}{R^2} = \dfrac{mv^2}{R}$

$v = \sqrt{\dfrac{GM}{R}}$

The time period $T$ is given by:

$T = \dfrac{2\pi R}{v} = \dfrac{2\pi R}{\sqrt{\dfrac{GM}{R}}} = \dfrac{2\pi}{\sqrt{GM}} R^{3/2}$

Thus, $T \propto R^{3/2}$.

Answer: $\text{R}^{3/2}$