# Homework Help: Equations of motion for an orbit

1. Mar 8, 2007

### fasterthanjoao

Only reason I'm posting here is that i'll get more views than in the cosmology thread, I'm afraid..

(Basically, I'm working through a couple of different models and after some work I'm a bit stuck: http://trond.hjorteland.com/thesis/node21.html

I'm basically trying to integrate equation 3.33 to $$t_0$$ - without much success. Now, cosmologists will know that $$a(t)=(\frac{t}{t_0})^\frac{2}{3}$$ which I feel should be substituted, getting rid of the $$a^\frac{1}{2}da$$? Any comments on this appreciated.

2. Mar 9, 2007

### dextercioby

This is to help you find the antiderivative: Make the substitutions

$$a=\Omega_{0} , \ b=1-\Omega_{0}, \ a=y^{3/2}$$

You get the equation

$$H_{0}\int{}dt=\frac{2}{3}\int \frac{dy}{\sqrt{by^2 +a}}$$

Can you take it from here ?

3. Mar 9, 2007

### fasterthanjoao

Seems to me like it's at least definitely a log, I think roughly coming out to something like :

$$H_0t=\frac{2}{3}\log(2by+2\sqrt{a+by^2})$$

Which I'm doesn't seem right, and infact I'm almost certain the final form should be independent of y, since in a limit it should simplify to

$$H_0t_0=\frac{2}{3}$$

So it seems I need to at least have a definite integral, ill just say y from 0 to 1 since i've already normalised for y to be 1 at present.

Last edited: Mar 9, 2007
4. Mar 9, 2007

### fasterthanjoao

Integrating from 0 to 1 seems to be correct now, from Mathematica I see that it's an inverse hyperbolic sine (through I suppose theres an equivalent form in Log). Could obviously have used Mathematica at the beginning, but I'd like to be able to work through it since it's bugging me now!

If you can shed light on how this is done, that would be great. Thanks.

5. Mar 9, 2007

### fasterthanjoao

Think I have it ok now, was just a little rusty on my integration. Thanks for the help dextercioby.