# GRE GR 1777, Problem 018 Solution

## 18. Classical Mechanics (Angular Momentum)

##### Solution
Angular momentum in circular motion is

$L = mvr$$L = mvr$
where $L$$L$ is angular momentum, $m$$m$ is the mass of the satellite, $v$$v$ is the velocity, and $r$$r$ is the orbital radius.

In this problem, two satellites are identical. So the mass of the satellite is equal.

The orbital radius of A is the ratio of the angular momentum of A to the twice that of B, so
$r_A = 2r_B$$r_A = 2r_B$

Since satellites have circular motion, the centripetal force is equal to the gravitational force

$\frac{GMm}{r^2} = \frac{mv^2}{r}$$\frac{GMm}{r^2} = \frac{mv^2}{r}$
where $M$$M$ is the mass of the Earth.

The orbital velocity of the satellite can be obtained as

$v = \sqrt{\frac{GM}{r}}$$v = \sqrt{\frac{GM}{r}}$

Then we can wirte angular momentum as

$L = mvr = m \sqrt{\frac{GM}{r}} r = m\sqrt{GMr}$$L = mvr = m \sqrt{\frac{GM}{r}} r = m\sqrt{GMr}$
$\therefore L \propto \sqrt{r}$$\therefore L \propto \sqrt{r}$

Therefore, the ratio of the angular momentum is

$L_A : L_B = \sqrt{r_A} : \sqrt{r_B} = \sqrt{2} : 1$$L_A : L_B = \sqrt{r_A} : \sqrt{r_B} = \sqrt{2} : 1$


(C) $\sqrt{2}$$\sqrt{2}$
https://en.wikipedia.org/wiki/Angular_momentum#Scalar_—_angular_momentum_in_two_dimensions