JEE Main 2024 — Gravitation Question with Solution

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

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