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

Question

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?

Answer 1

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.

Answer to A:

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,

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