Question 252: Gravitation

Question

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)

Accepted Answer

$$ r = R + \frac{R}{3} = \frac{4R}{3} $$
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,
$$ x = 3. $$

Gravitation

Choose the correct answer: