# 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.

• 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, 2013 at 1:22
• I'm guessing this is from an old exam and not something being administered as an exam presently?
– user10851
May 17, 2013 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: $\ddot{\mathbf{r}} = -\mathcal{G}\frac{M}{r^3}\mathbf{r}$. May 17, 2013 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.