Imagine a light planet revolving around a very massive star in a circular orbit of radius R with a period of revolution T. If the gravitational force of attraction between planet and star is proportional R⁻².⁵ , then T⁻² is proportional to

The correct option is R³.⁵.

Let us derive the relation step-by-step.

For a planet of mass $m$ revolving around a massive star in a circular orbit of radius $R$ with a period of revolution $T$, the gravitational force of attraction $F$ provides the necessary centripetal force for its circular motion.

The centripetal force is given by the formula:
$F=\frac{m{v}^2}{R}$
where $v$ is the orbital speed of the planet.

Since the planet completes one revolution of circumference $2\pi R$ in time $T$, its orbital speed is:
$v=\frac{2\pi R}{T}$

Substituting this value of $v$ into the force equation gives:
$F=\frac{m}{R}{\left(\frac{2\pi R}{T}\right)}^2=\frac{4{\pi }^{2}mR}{T^2}$

From the above expression, since $4{\pi }^{2}m$ is a constant, we have the proportionality:
$F\propto \frac{R}{T^2}$

We are given that the gravitational force of attraction is proportional to $R^{-2.5}$:
$F\propto R^{-2.5}$

By equating the two proportional relations, we get:
$\frac{R}{T^2}\propto R^{-2.5}$

To find the dependency of the time period, we rearrange the terms:
$T^2\propto \frac{R}{R^{-2.5}}$
$T^2\propto R\cdot R^{2.5}$
$T^2\propto R^{1+2.5}$
$T^2\propto R^{3.5}$

Thus, the relationship is represented by the proportionality of $T^2$ to $R^{3.5}$.