Understanding Kepler's $2^{nd}$ law in terms of angular momentum conservation

A) Explain how Kepler's $2^{nd}$ law - "The radius vector from the Sun to a planet sweeps out equal areas in equal time intervals" - can be understood in terms of angular momentum conservation.

I know that:

Angular momentum is conserved and therefore $\vec{L}=\vec{r} \times \vec{p}=\vec{r} \times m\vec{v}=constant$ and $L=mrv\sin\theta$.

Kepler's $2^{nd}$ law means $\frac{dA}{dt}=constant$

Somehow this comes out to be $dA=(\frac{1}{2})(\frac{L}{m})dt$ but I'm having a hard time getting there.

B) Explain how circular motion can be described as simple harmonic motion.

I know that:

For circular motion $m\vec{a}=\vec{F}_{c}=-m\frac{v^2}{R}\vec{r}=-m\omega^2R\vec{r}$

However, I'm fairly lost on this equation. Where does the negative sign come from, and where does the $\vec{r}$ come from?

• Hi Ground Clouds. Welcome to Physics.SE. Please have a look at the definition of homework tag. It still applies to your question ;-) Apr 28, 2013 at 19:08
• Hi ground.clouds1. Echoing @CrazyBuddy's comment: If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. Apr 28, 2013 at 19:10
• Gotcha. Odd definition of homework, but works for me.. As long as I can get a little guidance here. :) Apr 28, 2013 at 19:14

Answer to question B is quite simple,

You are denoting the centripetal force as a vector, which acts inward(towards the center), in terms of the radius vector r, which is pointing outward(away from the center) hence the '-' sign.

consider that in time $dt$ the planet covers an angle of $d\theta$ around the sun. The area it covers in this time is given by

$dA = \frac12 r^2d\theta$

so

$\frac{dA}{dt} = \frac12r^2\frac{d\theta}{dt}$

where $r$ is the distance of the planet from the sun. We can substitute $\frac{d\theta}{dt}$ as $\omega$ or $\frac{vsin\theta}{r}$ where $vsin\theta$ is the perpendicular component of velocity to the radius vector. Thus we get

$\frac{dA}{dt} = \frac12 rvsin\theta$

But $rvsin\theta = \frac{L}{m}$,

Therefore,

$\frac{dA}{dt} = \frac12\frac Lm$

• I'd add a little hint about describing circular motion as SHM - if you try writing circular motion as the (vector) sum of motion along two orthogonal axes... well, you'll see. Jun 27, 2013 at 12:45
• @Kyle , I don't think that is required for explaining why the negative sign occurs... Jun 27, 2013 at 15:42
• the asker said they were fairly lost... wasn't sure, but thought it might be more than just the -ve sign that was confusing them. Jun 27, 2013 at 15:48