# Conservation Of Angular Momentum

1. Jan 27, 2014

### BOAS

Hello :)

1. The problem statement, all variables and given/known data

The Earth of mass $m_{e}$, moves with an approximately circular orbit of radius $r = 1.5 * 10^{8}km$ around the sun of mass $M_{s} = 2 * 10^{30}kg$.

(a) Determine the numerical value of the speed of the earth.

(done, got an answer of $29821 ms^{-1}$ by equating the gravitational equation with the centripetal force equation)

(b) Is the angular momentum $L$ of the earth conserved? Why? Show that its module is given by $L = m_{e} \sqrt{GM_{s}r}$

2. Relevant equations

3. The attempt at a solution

I do not understand how to show that angular momentum is conserved, one of the older students that help in our workshops gave an 'explanation' but I don't actually follow his argument.

I'll try to explain what I think he said.

$L = I \omega$

$\frac{dL}{dt} = \tau = f * r$

$\tau = o$

$\frac{dL}{dt} = 0$

∴ Angular momentum is conserved

The step that confuses me somewhat is why $\tau = 0$

Thanks!

2. Jan 27, 2014

### consciousness

Make a small diagram of the earth and the sun. Make the force vector with proper direction.
What is the moment of this force about the axis of rotation?

3. Jan 27, 2014

### lightgrav

from an experimental perspective, the speed of Earth ought to be its travel distance (2 pi r) divided by travel time (1 year = 31.557 Megasec).
Angular momentum L = r x p = r x mv = r m v sin(theta) = r m |v| ... do any of these values change?

4. Jan 28, 2014

### BOAS

There isn't one, the force acts towards the axis of rotation.

That makes sense.

5. Jan 28, 2014

### BOAS

Assuming I understand your symbols correctly, no.

So, the derivative of a constant is 0, hence angular momentum is conserved.

Thank you.