JEE Main 2025 — Gravitation Question with Solution

$$r=R+\frac{R}{3}=\frac{4R}{3}$$

For a circular orbit, the gravitational force provides the required centripetal force:

$$\frac{GMm}{r^2}=\frac{m{v}^2}{r}$$

which simplifies to

$$v=\sqrt{\frac{GM}{r}}.$$

The angular momentum $$L$$ of the satellite is given by

$$L=mvr.$$

Substitute the values:

Satellite mass: $$m=\frac{M}{2}$$

Orbital radius: $$r=\frac{4R}{3}$$

Speed: $$v=\sqrt{\frac{GM}{\frac{4R}{3}}}=\sqrt{\frac{3GM}{4R}}$$

Thus,

$$L=\frac{M}{2}\cdot \frac{4R}{3}\cdot \sqrt{\frac{3GM}{4R}}.$$

Simplify the expression:

$$L=\frac{M\cdot 4R}{6}\sqrt{\frac{3GM}{4R}}=\frac{2MR}{3}\sqrt{\frac{3GM}{4R}}.$$

Notice that the square root can be combined as:

$$\sqrt{\frac{3GM}{4R}}=\sqrt{\frac{3}{4}}\sqrt{\frac{GM}{R}}.$$

Therefore,

$$L=\frac{2MR}{3}\cdot \sqrt{\frac{3}{4}}\sqrt{\frac{GM}{R}}=\frac{2MR\sqrt{3}}{3\cdot 2}\sqrt{\frac{GM}{R}}=\frac{M\sqrt{3}}{3}\sqrt{GMR},$$

where we used the fact that $$R\sqrt{\frac{1}{R}}=\sqrt{R}$$ to form $$\sqrt{GMR}$$.

So the final expression is:

$$L=\frac{M}{\sqrt{3}}\sqrt{GMR}.$$

It is given that the angular momentum can also be expressed as:

$$L=M\sqrt{\frac{GMR}{x}}.$$

Equate the two expressions:

$$\frac{M}{\sqrt{3}}\sqrt{GMR}=M\sqrt{\frac{GMR}{x}}.$$

Cancel $$M$$ and $$\sqrt{GMR}$$ on both sides (assuming they are nonzero):

$$\frac{1}{\sqrt{3}}=\sqrt{\frac{1}{x}}.$$

Taking squares on both sides:

$$\frac{1}{3}=\frac{1}{x}.$$

Thus,

$$x=3.$$