QJEEM25Ph96b2: JEE Main 2025 Physics - Class 11 JEEM25Ph96b2 Solution: A satellite of mass is revolving around earth in a circular…

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

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

$\frac{GMm}{r^2} = \frac{mv^2}{r}$

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,

$x = 3.$