A light planet is revolving around a massive star in a circular orbit of radius $\mathrm{R}$ with a period of revolution T. If the force of attraction between planet and star is proportional to $\mathrm{R}^{-3 / 2}$ then choose the correct option :

<p>To find the correct option for the relationship between the period of revolution T and the radius of the orbit R, we will consider the force of attraction and its proportionality to $\mathrm{R}^{-3 / 2}$.</p> <p>According to Newton&#39;s law of universal gravitation, the force of attraction $F$ between two masses $m_1$ and $m_2$ separated by a distance $r$ is given by $F = \frac{G m_1 m_2}{r^2},$ where $G$ is the gravitational constant.</p> <p>However, in this particular case, the force of attraction is given to be proportional to $\mathrm{R}^{-3 / 2}$, so we can write $F \propto \frac{1}{R^{3/2}}.$</p> <p>Since the planet is in a circular orbit around the star, the centripetal force required to keep the planet in orbit must be provided by this gravitational force. Hence, we can write that $\frac{m v^2}{R} \propto \frac{1}{R^{3/2}},$ where $m$ is the mass of the planet and $v$ is its orbital speed.</p> <p>Simplifying this, we get $v^2 \propto \frac{1}{R^{1/2}}.$</p> <p>Now, the speed $v$ can be related to the period T through the circumference of the orbit, which is given by $2\pi R$. The orbital speed is the circumference divided by the period: $v = \frac{2\pi R}{T}.$</p> <p>Substituting this into our proportionality, we get <br/><br/>$\left( \frac{2\pi R}{T} \right)^2 \propto \frac{1}{R^{1/2}},$ <br/><br/>which simplifies to <br/><br/>$\frac{4\pi^2 R^2}{T^2} \propto \frac{1}{R^{1/2}}.$</p> <p>Solving for $T^2$, we get <br/><br/>$T^2 \propto \frac{R^{2 + 1/2}}{4\pi^2},$ <br/><br/>so $T^2 \propto R^{5/2}.$</p> <p>Therefore, the correct option is Option C <br/><br/>$T^2 \propto R^{5/2}.$</p>