Solution to the Two-Body Problem: Cross-Product and Dot-Product Integration

[TimeRip496]

Homework Statement

Two-body problem given as
$$\ddot{\textbf{r}}+\frac{GM}{r^2}\frac{\textbf{r}}{r}=0$$
$$\textbf{h}=\textbf{r}\times\dot{\textbf{r}}$$
where the moment of the momentum vector mh

Homework Equations

The vector solution to the above equation may be obtained by first taking the cross-product with the constant h and integrating once with respect to time. This yields
$$\dot{\textbf{r}}\times\textbf{h}=GM(\frac{\textbf{r}}{r}+\textbf{e})$$

The final solution to the equation is then obtained by taking the dot product of above equation with r, is
$$r=\frac{h^2/(GM)}{1+ecos\theta}$$

The Attempt at a Solution

I have no idea what the author is doing. How does cross product with the momentum and then dot product with r solve the equation?

I try following his step but I get a different integration result
$$\dot{\textbf{r}}\times\textbf{h}=GM(\frac{\textbf{r}}{r^3}t\times\textbf{h}+\textbf{e})$$
I have no idea how his integration w.r.t. time for the GM/r² reduces it to GM/r and how to cross product GM/r² with h since they are unknown variables?

 

[VKint]

What's the objective here? To find $r(t)$ and $\theta(t)$? If so, the solution is well-known...Landau / Lifshitz "Mechanics" has a good summary if you've never seen it.

 

[TSny]

I try following his step but I get a different integration result
$$\dot{\textbf{r}}\times\textbf{h}=GM(\frac{\textbf{r}}{r^3}t\times\textbf{h}+\textbf{e})$$

How did you get the first term on the right: $GM(\frac{\textbf{r}}{r^3}t\times\textbf{h}) \,$?

Keep in mind that $\mathbf r$ and $r$ are functions of time. You cannot treat them as constants when integrating with respect to time.