# Linear motion in central field

1. May 20, 2013

### PhysStudent81

1. The problem statement, all variables and given/known data

I have two masses of finite width, m_1 and m_2. The force is newton's gravity, so U = k/r. I want to work out their relevant equations of motions r_1(t) and r_2(t) as they start off from rest and collide. I don't want to consider any rotational motion.

2. Relevant equations

$U = \frac{k}{r}$
$r = r_{1}(t) - r_{2}(t)$

0.5μ$\dot{r}^2 = E_{tot} - \frac{k}{r}$

dt = $\frac{μ}{2}$∫$\frac{dr}{\sqrt{E_{tot} - \frac{k}{r}}}$

3. The attempt at a solution

I try to integrate the above equation it gives me something very complicated (I end up integrating cosec^3 after making the substitution $\frac{1}{r} = \sin^{2}(\theta)$) which gives me t = t(r), but I can't invert this to give me r = r(t).

Am I doing something wrong? Is there another way of doing it that doesn't involve lagrangian or hamiltonian dynamics (which I haven't studied).

Another way would be to solve the the 2nd order differential equation directly:

$\frac{dr^{2}}{dt^{2}} = \frac{k}{r^{2}}$

but I can't seem to do this (I fee I'm missing something very simple here). I know that if I let $r = At^{\frac{2}{3}}$ this is a solution but it doesn't have enough constants.

Any pointers?

Thanks,

Rob

2. May 20, 2013

### bp_psy

3. May 20, 2013

### TSny

I think what you are doing is correct. Did you allow for the fact that both k and Etot are negative numbers? Anyway, as you say, you can get an expression for t(r). I agree that the result looks too complicated to invert for r(t).

4. May 20, 2013

### PhysStudent81

oh my goodness thank you.

Rob

5. May 20, 2013

### PhysStudent81

So strange that for such seemingly simple problem there is no closed form result for r(t). t(r) is actually all I wanted so that's ok!