I'm doing some self study using Prusing and Conway's Orbital Mechanics. In the first chapter there is a derivation of r for relative motion that I'm having a little difficulty following. Specifically, this equation:

$$\ddot{\bf{r}} \times \bf{h} = \frac{-\mu}{\it{r^3}} \bf{r} \times \bf{h} = \frac{-\mu}{\it{r^3}} \bf{r} \times (\bf{r} \times \dot{\bf{r}}) $$

$$ = \frac{\mu}{\it{r^3}} [\dot{\bf{r}}({\bf{r}} \cdot {\bf{r}}) - {\bf{r}}({\bf{r}} \cdot {\bf{r}}) ]$$

If anybody could point me in the right direction for materials related to working through equations and flipping between cross products and dot products (as this equation seems to do?), I would be very appreciative. It's difficult for me to see how one obtains the bottom result from the rightmost side of the first equation. Is there some sort of identity relating cross products and dot products that I should know offhand?


This is in fact an identity. The statement is $$ \mathbf{a}\times(\mathbf{b}\times\mathbf{c}) = \mathbf{b}(\mathbf{a}\cdot\mathbf{c}) - \mathbf{c}(\mathbf{a}\cdot\mathbf{b}). $$

You can find a basic derivation of this identity on this wikipedia page under "Vector triple product."

Edit: If you'd like a somewhat more sophisticated derivation, here it is again with the Levi-Civita symbol, using the identity $$ \epsilon_{ijk}\epsilon_{lmk} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}, $$ where $\delta_{ij}$ is the Kronecker delta.

Here, a vector $\mathbf{a}$ may be written as $\sum_i a_i\hat{x}_i$, so that $$ \mathbf{a}\times\mathbf{b} = \sum_{ijk}\epsilon_{ijk}\hat{x}_i a_j b_k $$ and $$ \mathbf{a}\cdot\mathbf{b} = \sum_{ij}\delta_{ij} a_i b_j = \sum_i a_ib_i. $$

If you're not familiar with this notation, don't worry about it. I'm just including this derivation because I don't know what your background is and I like this derivation better than the wikipedia derivation. You can consider this 'optional reading.'

In this notation, we see that \begin{align} \mathbf{a}\times(\mathbf{b}\times\mathbf{c}) &= \sum_i \hat{x}_i(\mathbf{a}\times(\mathbf{b}\times\mathbf{c}))_i\\ &= \sum_{i}\hat{x}_i\sum_{jk}\epsilon_{ijk}a_j(\mathbf{b}\times\mathbf{c})_k\\ &= \sum_{ijk}\hat{x}_i\epsilon_{ijk}a_j\sum_{lm}\epsilon_{klm}b_lc_m\\ &= \sum_{ijk}\sum_{lm}\epsilon_{ijk}\epsilon_{klm}\hat{x}_i a_j b_l c_m\\ &= \sum_{ijk}\sum_{lm}\epsilon_{ijk}\epsilon_{lmk}\hat{x}_i a_j b_l c_m\\ &= \sum_{ij}\sum_{lm}(\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl})\hat{x}_i a_j b_l c_m\\ &= \sum_{ij}\sum_{lm}\delta_{il}\delta_{jm}\hat{x}_i a_j b_l c_m - \sum_{ij}\sum_{lm}\delta_{im}\delta_{jl}\hat{x}_i a_j b_l c_m\\ &= \sum_{ij}\hat{x}_i b_i a_j c_j - \sum_{ij}\hat{x}_i c_i a_j b_j\\ &= \sum_i b_i\hat{x}_i\sum_{j}a_jc_j - \sum_i c_i\hat{x}_i\sum_j a_j b_j\\ &= \mathbf{b}(\mathbf{a}\cdot\mathbf{c}) - \mathbf{c}(\mathbf{a}\cdot\mathbf{b}), \end{align} and we achieve the desired result.

  • $\begingroup$ Thank you so much! This is exactly what I was looking for, and learning how to do vector multiplication using the Levi-Civita symbol and Kronecker Delta has furthered my understanding of this identity. $\endgroup$
    – H. Ingram
    Apr 13 '21 at 15:47
  • $\begingroup$ @H.Ingram No problem! Glad I could help! $\endgroup$
    – Banshee
    Apr 13 '21 at 18:11

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