# I Trying to solve the restricted two-body problem

1. Aug 29, 2016

### tomwilliam2

I'm just trying to solve the restricted two-body problem mentioned in passing in my textbook:

$\mathbf{\ddot{r}}+\frac{GM}{r^2}\left(\frac{\mathbf{r}}{r}\right)=\mathbf{0}$

I understand the way to do it is to take the cross-product with the constant angular momentum $\mathbf{h}$ then integrate wrt time:

$\mathbf{\dot{r} \times h}= -\frac{GM}{r^2}\int \left(\frac{\mathbf{r}}{r}\right)\mathbf{\times h}\ dt$

But I'm not sure what to do next. Can I use the fact that $\mathbf{h}=\mathbf{r \times \dot{r}}$ to make it:

$\mathbf{\dot{r} \times h}= -\frac{GM}{r^2}\int \left(\frac{\mathbf{r}}{r}\right)\mathbf{\times r \times \dot{r}}\ dt$
Can anyone help?
The answer I'm trying to get to is:
$\mu\left(\frac{\mathbf{r}}{r} + \mathbf{e}\right)$, where $\mathbf{e}$ is a constant vector of integration and presumably $\mu$ is some constant related to G.

2. Aug 29, 2016

Hi,

After you manage to use the angular momentum conservation, you need to do variable change from r to u = 1/r.
Also there are many online texts which solve this problem.

BR,

3. Aug 30, 2016

### tomwilliam2

Thanks. The problem I'm having is that the cross product of a vector with itself is zero, so I seem to get 0 in the integrand. Actually, I think the $r^{-2}$ term I factored out should also be in there, but that would still give me an integrand of:
$\int \frac{\mathbf{r \times r\times \dot{r}}}{r^3} dt$
Isn't the inside of the integrand then zero?

4. Aug 30, 2016

### vanhees71

It's simpler to work with differential equations. Starting from your EoM
$$\ddot{\vec{r}}=-\frac{G M}{r^3} \vec{r},$$
you can very easily derive the conservation of angular momentum (here everything is divided by $m$, the mass of the "test particle"):
$$\vec{r} \times \ddot{\vec{r}}=0.$$
But on the other hand
$$\frac{\mathrm{d}}{\mathrm{d} t} (\vec{r} \times \dot{\vec{r}})=\dot{\vec{r}} \times \dot{\vec{r}} + \vec{r} \times \ddot{\vec{r}}=\vec{r} \times \ddot{\vec{r}}.$$
Thus you have with $\vec{h}=\vec{r} \times \dot{\vec{r}}$
$$\dot{\vec{h}}=0 \; \Rightarrow \; \vec{h}=\text{const}.$$
For $\vec{h} \neq 0$ it's clear that the motion is in a plane perpendicular to $\vec{h}$. The standard way to proceed further is to use the conservation of energy as a 2nd first integral and solve for the orbit $r=r(\phi)$, where $r$ and $\phi$ are polar coordinates in the plane.

Obviously your textbook follows another very elegant way, using the huge symmetry of the Kepler problem. The point is that for the $1/r$ potential there is an extra symmetry, which leads to another conserved vector, the Runge-Lenz vector in the plane of motion. To derive it we use
$$\frac{\mathrm{d}}{\mathrm{d} t} (\dot{\vec{r}} \times \vec{h})=\ddot{\vec{r}} \times \vec{h}=-\frac{GM}{r^3} \vec{r} \times \vec{h} \qquad (*).$$
On the other hand we have
$$\frac{\mathrm{d}}{\mathrm{d} t}=\frac{\vec{r}}{r} = \frac{r \dot{\vec{r}}-\dot{r} \vec{r}}{r^2}.$$
Now we have (check it!)
$$\dot{\vec{r}} = \frac{\mathrm{d}}{\mathrm{d} t} \sqrt{\vec{r} \cdot \vec{r}}=\frac{\vec{r} \cdot \dot{\vec{r}}}{r}=-\frac{\vec{r} \times \vec{h}}{r^3}.$$
Multiplying with $GM$ and subtracting from (*) you get
$$\frac{\mathrm{d}}{\mathrm{d} t} \left [\dot{\vec{r}} \times \vec{h} - G M \frac{\vec{r}}{r} \right]=0,$$
i.e., the Runge-Lenz vector
$$\vec{e}= \dot{\vec{r}} \times \vec{h} - G M \frac{\vec{r}}{r}=\text{const} \qquad(**).$$
From this it's very easy to find the shape of the orbit by introducing polar coordinates in the orbital plane with $\vec{e}$ as the polar axis. Multiplying (**) with $\vec{r}$ you get
$$\vec{r} \cdot \vec{e} = r e \cos \varphi.$$
On the other hand from the definition of the Runge-Lenz vector) you have
$$\vec{r} \cdot \vec{e} = \vec{r} \cdot (\dot{\vec{r}} \times \vec{h})-GM r = (\vec{r} \times \dot{\vec{r}}) \cdot \vec{h} - G M r =h^2-G M r.$$
Together with the previous equation you find
$$r=\frac{h^2}{GM+e \cos \varphi}=\frac{h^2/(GM)}{1+e/(GM) \cos \varphi}.$$
This is a conic section with the excentricity $\epsilon=e/(GM)$, i.e., an ellipse for $0<\epsilon<1$, a circle for $\epsilon=0$, a parabola for $\epsilon=1$, and a hyperbola for $\epsilon>1$.

5. Aug 30, 2016

### tomwilliam2

Thanks, that's an awesome answer. I'm going to come back to it to work through properly, and make sure I understand it fully.
Thanks for taking the time out.