What are some tips/techniques that might help me solve this (brutal) differential equation?

Question

I've been working on a certain physics problem involving differential equation for two years. I've made some progress on it recently, but I've come across another roadblock, namely an integral that I have no idea how to compute. What are some tips or techniques that might help me evaluate it? I've been having trouble with it because I've nev taken a diff-eq class, although I'm also aware that this is a difficult problem in itself (non-linear second order equation). Please note: I do NOT want the solution the the problem, just the tools to solve it myself.

Here's the problem:

There is a fixed massive body at the origin and another object on the x axis with some initial velocity in the x direction. Find an equation that describes the position of the orbiting object with respect to time.

Here is my work so far. Let me know if I've made a mistake. Note that a is the derivative of v, which is the derivative of r, and G,M, and m are constants.

Start with $F=ma$

The gravitational force between two objects is $$F_g=-\frac{GMm}{r^2},$$ so $$-\frac{GMm}{r^2}=ma$$ and $$-\frac{GM}{r^2}=a.$$

Now, $$a=\frac{dv}{dt}=\frac{dv}{dr}\frac{dr}{dt}=v\frac{dv}{dr},$$ so $$-\frac{GM}{r^2}=v\frac{dv}{dr}$$ $$-\frac{GM}{r^2}dr=vdv$$ Integrating both sides I get $$\int_{r_o}^{r}-\frac{GM}{r^2}dr=\int_{v_o}^{v}vdv$$ where$r_o$ and $v_o$ are initial radius and velocity. This becomes$$\frac{GM}{r}-\frac{GM}{r_o}=\frac{1}{2}(v^2-v_o^2)$$ $$\pm\sqrt{\frac{2GMr_o-2GMr+v^2_or_or}{rr_o}}=\frac{dr}{dt}$$ $$\pm\int_{r_0}^{r}\sqrt{\frac{r_or}{2GMr_o-2GMr+v^2_or_or}}dr=\int_{0}^{t}dt$$ $$\pm\int_{r_0}^{r}\sqrt{\frac{r_or}{2GMr_o-2GMr+v^2_or_or}}dr=t$$ Which is obscene.

I just have no idea how to do this. I've tried rewriting it all sorts of ways with little to no luck. What can I do?

Answer 1

$$\ \int_{r_0}^{r}\sqrt{\frac{r_0r}{2GMr_0-2GMr+v_0^2r_0r}}dr=$$ $$\ =\sqrt{\frac{r_0}{v_0^2r_0-2GM}}\int_{r_0}^{r}\sqrt{\frac{r}{\frac{2GMr_0}{v_0^2r_0-2GM}+r}}dr$$

Where I assumed that $\ v_0^2r_0>2GM$. Now set $$\ \beta=\sqrt{\frac{r_0}{v_0^2r_0-2GM}}$$ $$\ \alpha=\frac{2GMr_0}{v_0^2r_0-2GM}$$

So you need to deal with this kind of integral: $$\ \beta\int_{r_0}^{r}\sqrt{\frac{r}{\alpha+r}}dr$$ Now put $$\ \sqrt{\frac{r}{\alpha+r}}=s \implies \frac{r}{\alpha+r}=s^2$$ Then $$\ \frac{\alpha}{(\alpha+r)^2}dr=2sds$$ Moreover: $$\ r=(\alpha+r)s^2\implies r=\frac{\alpha s^2}{1-s^2}\implies dr=\frac{2\alpha s}{(1-s^2)^2}$$ So the integral is: $$\ \beta \int_{\phi(r_0)}^{\phi(r)}s\cdot \frac{2\alpha s}{(1-s^2)^2}ds=$$ $$\ 2\alpha \beta \int_{\phi(r_0)}^{\phi(r)} \frac{s^2}{(1-s^2)^2}ds$$ Now in this form it should be easier

Answer 2

From $$2GM\left(\frac{1}{r}-\frac{1}{r_0}\right) = \dot{r}^2-v_0^2,$$ set $r = 1/u$, (this is the most important thing to remember for problems with $1/r^2$ forces) then $$ \dot{r} = -\frac{\dot{u}}{u^2},$$ so $$2GM\left(u_0-u\right) = \frac{\dot{u}^2}{u^4} + v_0^2. $$ Then, rearranging, $$ 1 = \frac{\dot{u}^2}{u^4(v_0^2+2GM(u_0-u))} $$

Now you just have to find $$ \int \frac{dx}{x^2\sqrt{1-x}} $$ and rescale to get the right integral.

(Hint: the denominator factorises as a difference of two squares, believe it or not. Or you can just do a hyperbolic substitution.)