1. Given data

• Mass of satellite: $m = \dfrac{M}{2}$
• Radius of Earth: $R$
• Height above Earth’s surface: $h = \dfrac{R}{3}$

So, orbital radius from Earth’s center is $r = R + \frac{R}{3} = \frac{4R}{3}$

2. Speed of satellite in circular orbit

For a circular orbit, $v = \sqrt{\frac{GM}{r}}$

Substitute $r = \dfrac{4R}{3}$: $v = \sqrt{\frac{GM}{4R/3}} = \sqrt{\frac{3GM}{4R}}$

3. Angular momentum of the satellite

For circular motion, $L = mvr$

Substitute $m = \dfrac{M}{2}$, $v = \sqrt{\dfrac{GM}{r}}$, and $r = \dfrac{4R}{3}$: $L = \frac{M}{2} \cdot \sqrt{\frac{GM}{r}} \cdot r$

This becomes $L = \frac{M}{2}\sqrt{GMr}$

Now substitute $r = \frac{4R}{3}$: $L = \frac{M}{2}\sqrt{GM\cdot \frac{4R}{3}}$

$L = \frac{M}{2}\cdot 2\sqrt{\frac{GMR}{3}}$

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

4. Compare with given form

Given, $L = M\sqrt{\frac{GMR}{x}}$

Comparing with $L = M\sqrt{\frac{GMR}{3}}$ we get $x = 3$

5. Comparison with stored answer

Stored correct answer = $3$

Our derived answer also gives $x = 3$.

Hence, the answer agrees with the stored correct answer.