Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
Hi Ground Clouds. Welcome to Physics.SE. Please have a look at the definition of homework tag. It still applies to your question ;-) –  Waffle's Crazy Peanut Apr 28 '13 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. –  Qmechanic Apr 28 '13 at 19:10
Gotcha. Odd definition of homework, but works for me.. As long as I can get a little guidance here. :) –  ground.clouds1 Apr 28 '13 at 19:14

1 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

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


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

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


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

share|improve this answer
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. –  Kyle Oman Jun 27 '13 at 12:45
@Kyle , I don't think that is required for explaining why the negative sign occurs... –  udiboy1209 Jun 27 '13 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. –  Kyle Oman Jun 27 '13 at 15:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.