Rate at which planet "sweeps out area"

Question

I want to show that the rate at which a planet circling around a star in an orbit is "sweeps out area" is $\frac{dA}{dt}=\frac{1}{2}r^2\omega$, where $r$ is the distance from the star to the planet and $\omega$ is the planet's angular velocity ($\dot{\phi}$ in polar coordinates).

It can be seen that the infinitesimal "wedge" area is $dA=\lVert d\vec{A}\rVert=\frac{1}{2}\left\rVert\vec{r}\times(\vec{r}+d\vec{r})\right\rVert = \frac{1}{2}rdr\sin(d\phi)\approx \frac{1}{2}d\phi rdr$ (since $d\phi$ is small). Dividing by $dt$ gives: $$\frac{dA}{dt} = \frac{1}{2}\frac{d\phi}{dt}rdr=\frac{1}{2}\dot{\phi}rdr = \frac{1}{2}\omega r dr$$

Now, I believe, the next step is to integrate from $r$ to $r+dr$? I think I'm probably complicating something. Would appreciate some help.

Answer 1

I could be wrong but I believe when finding |r x (r+dr)| , it simplifies to |r x dr| where we usually take the angle between those to be close enough to a right angle ( looks like you've taken it to be phi) to approximate it to rdr. After substituting that and converting dr in terms of the angle you should get the desired result.

P.s. sorry for the lack of notation, I'm on my phone.

Answer 2

You could instead first express the element of area swept out by a radial vector in Cartesian coordinates $$dA=\|\mathbf r\times(\mathbf r+d\mathbf r)\|=\|\mathbf r\times d\mathbf r\|=\frac12|x\,dy-y\,dx|$$ and then pull back to polar coordinates: $$\begin{align}dA&=\frac12|(\sin\phi\,dr+r\cos\phi\,d\phi)\,r\cos\phi-(\cos\phi\,dr-r\sin\phi\,d\phi)\,r\sin\phi| \\ &=\frac12|(r\cos\phi\sin\phi-r\cos\phi\sin\phi)\,dr+(r^2\cos^2\phi+r^2\sin^2\phi)\,d\phi| \\ &=\frac12|r^2\,d\phi|.\end{align}$$

Answer 3

It all comes down to evaluating one cross-product, which is best done in the ${\{\hat{r},\hat{\phi},\hat{z}\}}$ basis. $$\eqalign{ r\times (r+dr) &= r\times dr \cr &= (\hat{r}\,r)\times (\hat{\phi}\,dr_\phi + \hat{r}\,dr_r) \cr &= \hat{z}\,r\,dr_\phi \cr &= \hat{z}\,r^2\,d\phi \cr }$$ Half the magnitude of the cross-product yields the swept area $$\eqalign{ dA &= \frac{1}{2}r^2\,d\phi \cr\cr \frac{dA}{dt} &= \frac{1}{2}r^2\,\frac{d\phi}{dt} \cr\cr }$$ Others have suggested that making the angle really small will make $dr$ nearly perpendicular to $r$, however that is not true.

The key is that $dr$ has components in both the $\hat{r}$ and $\hat{\phi}$ directions, but the cross product with $r$ annihilates the $\hat{r}$-component.