Solving Kepler's Problem: Awkward Integration

[Piano man]

Hi,
I'm reading up on Kepler's problem at the moment, and there's a step in the book that I don't understand.

Starting off with the equation of the path $$\phi=\int\frac{M dr/r^2}{\sqrt{2m[E-U(r)]-M^2/r^2}}+\mbox{constant}$$

The step involves subbing in $$U=-\alpha/r$$, 'and effecting elementary integration' to get
$$\phi=\cos^{-1}\frac{(M/r)-(m\alpha/M)}{\sqrt{(2mE+\frac{m^2\alpha^2}{M^2}})}+\mbox{constant}$$

But it doesn't look very elementary to me

Does anyone have any idea?

Thanks.

 

[0xDEADBEEF]

First of all you could take the derivative of the thing and check the answer.

Did you check the derivative of arccos? The inverse trig functions often come up when you take integrals of square roots. I am a bit unsure about the square on the upper r. If it weren't there you could complete the square under the square root. Then do a variable substitution and integrate. The result should look very similar to the one you show.

Is the integral done in polar coordinates by any chance? That would help with the square.

 

[Gregg]

find

$$\frac{d}{dx} \cos ^{-1} (f(x))$$

Then find

$$\frac{d}{dr} \frac{\left(\frac{M}{r}-m \frac{\alpha }{M}\right)}{\sqrt{2 m E+\left(m \frac{\alpha }{M}\right)^2}}$$

 

[Gerenuk]

https://www.physicsforums.com/showthread.php?t=413574