A satellite of mass $\frac{M}{2}$ is revolving around earth in a circular orbit at a height of $\frac{R}{3}$ from earth surface. The angular momentum of the satellite is $\mathrm{M} \sqrt{\frac{\mathrm{GMR}}{x}}$. The value of $x$ is _________ , where M and R are the mass and radius of earth, respectively. ( G is the gravitational constant)

<p>$r = R + \frac{R}{3} = \frac{4R}{3}$</p> <p>For a circular orbit, the gravitational force provides the required centripetal force:</p> <p>$\frac{GMm}{r^2} = \frac{mv^2}{r}$</p> <p>which simplifies to</p> <p>$v = \sqrt{\frac{GM}{r}}.$</p> <p>The angular momentum $L$ of the satellite is given by</p> <p>$L = mvr.$</p> <p>Substitute the values:</p> <p><p>Satellite mass: $m = \frac{M}{2}$</p></p> <p><p>Orbital radius: $r = \frac{4R}{3}$</p></p> <p><p>Speed: $v = \sqrt{\frac{GM}{\frac{4R}{3}}} = \sqrt{\frac{3GM}{4R}}$</p></p> <p>Thus,</p> <p>$L = \frac{M}{2} \cdot \frac{4R}{3} \cdot \sqrt{\frac{3GM}{4R}}.$</p> <p>Simplify the expression:</p> <p>$$ L = \frac{M \cdot 4R}{6} \sqrt{\frac{3GM}{4R}} = \frac{2MR}{3} \sqrt{\frac{3GM}{4R}}. $$</p> <p>Notice that the square root can be combined as:</p> <p>$\sqrt{\frac{3GM}{4R}} = \sqrt{\frac{3}{4}} \sqrt{\frac{GM}{R}}.$</p> <p>Therefore,</p> <p>$$ L = \frac{2MR}{3} \cdot \sqrt{\frac{3}{4}} \sqrt{\frac{GM}{R}} = \frac{2MR \sqrt{3}}{3 \cdot 2} \sqrt{\frac{GM}{R}} = \frac{M \sqrt{3}}{3} \sqrt{GMR}, $$</p> <p>where we used the fact that $R \sqrt{\frac{1}{R}} = \sqrt{R}$ to form $\sqrt{GMR}$.</p> <p>So the final expression is:</p> <p>$L = \frac{M}{\sqrt{3}} \sqrt{GMR}.$</p> <p>It is given that the angular momentum can also be expressed as:</p> <p>$L = M \sqrt{\frac{GMR}{x}}.$</p> <p>Equate the two expressions:</p> <p>$\frac{M}{\sqrt{3}} \sqrt{GMR} = M \sqrt{\frac{GMR}{x}}.$</p> <p>Cancel $M$ and $\sqrt{GMR}$ on both sides (assuming they are nonzero):</p> <p>$\frac{1}{\sqrt{3}} = \sqrt{\frac{1}{x}}.$</p> <p>Taking squares on both sides:</p> <p>$\frac{1}{3} = \frac{1}{x}.$</p> <p>Thus,</p> <p>$x = 3.$</p>