- #1
tomwilliam2
- 117
- 2
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.
##\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.