QJEEM24Ph7c60: JEE Main 2024 Physics - Class 11 JEEM24Ph7c60 Solution: A light planet is revolving around a massive star in a circular …

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 $\mathrm{R}^{-3 / 2}$ .

According to Newton'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.

However, in this particular case, the force of attraction is given to be proportional to $\mathrm{R}^{-3 / 2}$ , so we can write $F \propto \frac{1}{R^{3/2}}.$

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.

Simplifying this, we get $v^2 \propto \frac{1}{R^{1/2}}.$

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: $v = \frac{2\pi R}{T}.$

Substituting this into our proportionality, we get

$\left( \frac{2\pi R}{T} \right)^2 \propto \frac{1}{R^{1/2}},$

which simplifies to

$\frac{4\pi^2 R^2}{T^2} \propto \frac{1}{R^{1/2}}.$

Solving for $T^2$ , we get

$T^2 \propto \frac{R^{2 + 1/2}}{4\pi^2},$

so $T^2 \propto R^{5/2}.$

Therefore, the correct option is Option C

$T^2 \propto R^{5/2}.$