# Vector cross product of $\mathbf{r}$ and $\ddot{\mathbf{r}}$ in polar coordinates

I'm struggling with the following question:

Question 6 A planet of mass $m$ moves under the gravitational attraction of a central star of mass $M$. The equation of motion of the planet is

$$\ddot{\mathbf{r}} = -\mathcal{G}\frac{(M + m)}{r^3}\mathbf{r}$$

where $\mathbf{r}$ is the position vector of the planet with respect to the star, $r = \lvert\mathbf{r}\rvert$ is the magnitude of $\mathbf{r}$ and $\mathcal{G}$ is the universal gravitational constant.

(a) Take the vector product of $\mathbf{r}$ with the above equation for $\ddot{\mathbf{r}}$ and use the standard result, $\dot{\mathbf{r}} = \dot{r}\hat{\mathbf{r}} + r\dot{\theta}\hat{\mathbf{\theta}}$ for motion in a polar coordinate system to show that $r^2\dot{\theta} = h$ where $h$ is a constant.

Part (a) asks to take vector product of $\mathbf{r}$ and $\ddot{\mathbf{r}}$. I don't know how to do this in polar coordinates.

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Is it going to be something like $-G\frac{\left(M+m\right)}{r^{3}}\left(\mathbf{r}\times\mathbf{r}\right)=0$? –  Luke May 17 '13 at 1:22
I'm guessing this is from an old exam and not something being administered as an exam presently? –  Chris White May 17 '13 at 1:25
Not that it really matter for the way you have to calculate the vector product, but should the force be equal to: $\mathbf{F} = -\mathcal{G}\frac{M m}{r^3}\mathbf{r}$ and therefore the acceleration: $\ddot{\mathbf{r}} = -\mathcal{G}\frac{M}{r^3}\mathbf{r}$. –  fibonatic May 17 '13 at 9:43

$\hat r, \hat \theta, \hat z$ form an orthogonal, right-handed triad of basis vectors in 3d. Taking the cross-product with them is no more exotic or unusual than doing so for Cartesian basis vectors. $\hat r \times \hat \theta = \hat z$, and so on.