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

 

[PJC01]

Hello, thank you for your question. It seems like you are having trouble understanding the integration step in solving Kepler's problem. Let me try to explain it to you.

First, let's look at the equation of the path that you provided: \phi=\int\frac{M dr/r^2}{\sqrt{2m[E-U(r)]-M^2/r^2}}+\mbox{constant}

This equation represents the angular position of an object in orbit around a central mass, where M is the mass of the central object, m is the mass of the orbiting object, r is the distance between them, E is the total energy of the system, and U(r) is the potential energy function.

Now, the next step involves substituting U=-\alpha/r, where \alpha is a constant related to the gravitational force between the two objects. This substitution simplifies the equation and makes it easier to integrate.

To integrate, we use a technique called u-substitution. We let u=\frac{1}{\sqrt{2m[E-U(r)]-M^2/r^2}}, which means that du=-\frac{M dr/r^2}{2\sqrt{2m[E-U(r)]-M^2/r^2}^3}. Substituting this into the original equation, we get:

\phi=\int\frac{du}{u}+\mbox{constant} = \ln|u|+\mbox{constant} = \ln|\frac{1}{\sqrt{2m[E-U(r)]-M^2/r^2}}|+\mbox{constant}

Now, using the substitution U=-\alpha/r, we get:

\phi=\ln|\frac{1}{\sqrt{2m[E+\frac{m^2\alpha^2}{M^2}]-M^2/r^2}}|+\mbox{constant}

To simplify this further, we can use the identity \ln\frac{1}{x}=-\ln x, which gives us:

\phi=-\ln|\sqrt{2m[E+\frac{m^2\alpha^2}{M^2}]-M^2/r^2}|+\mbox{constant}

Finally, using the identity \ln\sqrt{x}=\frac{1}{2}\ln x, we get:

\phi=\frac{1}{2}\ln|\frac{