7  Integration of vectors

The integral has the same nature as the integrand (vector or scalar) \[ \begin{aligned} & \int \mathbf{a}(u) d u=\mathbf{A}(u)+(\mathbf{b})-\text { where } \mathbf{b} \text { is a constant vector } \\ & \int_{u_{1}}^{u_{2}} a(u) d u=\mathbf{A}\left(u_{2}\right)-\mathbf{A}\left(u_{1}\right) \end{aligned} \]

Question

A small particle mass \(m\) orbits a much bigger mas \(M\) located at the origin. \[ m \frac{d^{2} \mathbf{r}}{d t^{2}}=-\frac{G Mm}{r^{2}} \hat{r} \]

Show that \(\mathbf{r} \times \frac{d \mathbf{r}}{d t}\) is a constant of motion.

Answer

First step lets take the vector product of force equation \[ \mathbf{r} \times m \frac{d^{2} \mathbf{r}}{d t^{2}}=\mathbf{r} \times\left(-\frac{G M m}{r^{2}}\right) \hat{r} \] \(m\) is a constant divide both sides \[ \mathbf{r} \times \frac{d^{2} \mathbf{r}}{d t^{2}}= \mathbf{r} \times\left(-\frac{G M}{r^{2}}\right) \hat{r} \] - \(\frac{G M}{r^{2}}\) is a constant \[ \begin{aligned} & \quad \mathbf{r} \times \frac{d^{2} \mathbf{r}}{d t^{2}}=-\frac{G m}{r^2}(\mathbf{r} \times \hat{r}) \\ & \mathbf{r} \times \hat{r}=\mathbf{0} \end{aligned} \] which means \[ \mathbf{r} \times \frac{d^{2} \mathbf{r}}{d t^{2}}=\mathbf{0}. \]

Ok but have not gotten to what we need quite yet - we have a \(\frac{d^{2} \mathbf{r}}{dt^2}\) which is one derivative higher \[ \begin{aligned} & \frac{d}{d t}\left(\mathbf{r}\times\frac{d \mathbf{r}}{d t}\right)=\frac{d \mathbf{r}}{d t} \times \frac{d \mathbf{r}}{d t}+\mathbf{r} \times \frac{d^{2} \mathbf{r}}{d t^{2}} \\ & \therefore \frac{d}{d t}\left(r \times \frac{d r}{d t}\right)=\underline{0} \\ & \text { a vector crossed with itself is $\mathbf{0}$ and we already showed that the second term is zero. } \end{aligned} \] integrate the above and we find \(\mathbf{r} \times \frac{d \mathbf{r}}{d t}=\) constant.