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 $$M\sqrt{\frac{GMR}{x}}$$.The value of $$x$$ is ______ , where M and R are the mass and radius of earth, respectively. ( G is the gravitational constant)

A satellite of mass $$\frac{M}{2}$$ revolves around the Earth in a circular orbit at a height of $$\frac{R}{3}$$ from the Earth's surface. The angular momentum is given as $$M\sqrt{\frac{GMR}{x}}$$. We need to find $$x$$.

First we find the orbital radius, which is the distance from the center of the Earth to the satellite:

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

Next we find the orbital velocity by equating the gravitational force to the centripetal force:

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

Here $$M$$ is Earth's mass, $$m = \frac{M}{2}$$ is the satellite mass, and $$r$$ is the orbital radius. Solving for $$v$$ gives

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

Then the angular momentum of the satellite is $$L = mvr$$, where $$m = \frac{M}{2}$$, so

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

We simplify this expression step by step. First, multiply the constant factors:

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

$$ L = \frac{2MR}{3} \times \sqrt{\frac{3GM}{4R}} $$

Since

$$ \sqrt{\frac{3GM}{4R}} = \frac{\sqrt{3GM}}{2\sqrt{R}}, $$

substituting back gives

$$ L = \frac{2MR}{3} \times \frac{\sqrt{3GM}}{2\sqrt{R}} = \frac{MR}{3} \times \frac{\sqrt{3GM}}{\sqrt{R}} = \frac{M\sqrt{R} \times \sqrt{R}}{3} \times \frac{\sqrt{3GM}}{\sqrt{R}} $$

which simplifies to

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

We now rewrite this result in the required form:

$$ L = \frac{M}{3}\sqrt{3GMR} = M \times \frac{\sqrt{3GMR}}{3} = M \times \sqrt{\frac{3GMR}{9}} = M\sqrt{\frac{GMR}{3}} $$

Here we used

$$\frac{\sqrt{3GMR}}{3} = \frac{\sqrt{3GMR}}{\sqrt{9}} = \sqrt{\frac{3GMR}{9}} = \sqrt{\frac{GMR}{3}}.$$

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

$$ x = 3 $$

Thus the value of $$x$$ is $$\boxed{3}$$.