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I am solving central force problem to deduce equations of orbit of planets. During the calculation, I am stuck over an integral which I am unable to solve. Can anyone help or guide me in this?

The integral is:

$$\int \dfrac{dr}{r \left(2\mu Er^2 + 2 \mu Cr - l^2 \right)^{1/2}} .$$

Here, $\mu$, $E$, $C$ and $l$ are constants.

Thanks in advance!

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1 Answer 1

up vote 6 down vote accepted

Make the substitution $\frac1r = x$. Then, $$ \begin{align*} \int \frac{dr}{r \left(2\mu Er^2 + 2 \mu Cr - l^2 \right)^{1/2}} &= \int \frac{dr}{r^2 \left( 2\mu E + \frac{2 \mu C}{r} - \frac{l^2}{r^2} \right)^{1/2}} \\ &= - \int \frac{dx}{\left( 2\mu E + 2 \mu C x - l^2x^2 \right)^{1/2}} \\ &= - \int \frac{dx}{\left( 2\mu E+ \frac{\mu^2 C^2}{l^2} - \left( lx - \frac{\mu C}{l} \right)^2 \right)^{1/2}} \end{align*} $$

Substitute $lx - \frac{\mu C}{l} = \sqrt{2 \mu E + \frac{\mu^2 C^2}{l^2}} \sin \theta$. Then $l dx = \sqrt{2 \mu E + \frac{\mu^2 C^2}{l^2}} \cos \theta d \theta$. Therefore, the integral simplifies to $$ \begin{align*} - \frac{1}{l} \int \frac{\sqrt{2 \mu E + \frac{\mu^2 C^2}{l^2}} \cos \theta d \theta}{\sqrt{2 \mu E + \frac{\mu^2 C^2}{l^2}} \cos \theta} &= - \frac{1}{l} \int d \theta = - \frac{\theta}{l} \\ &= - \frac{1}{l} \arcsin \left( \frac{lx - \frac{\mu C}{l}}{\sqrt{2 \mu E + \frac{\mu^2 C^2}{l^2}}} \right) \\ &= - \frac{1}{l} \arcsin \left( \frac{\frac{l}{r} - \frac{\mu C}{l}}{\sqrt{2 \mu E + \frac{\mu^2 C^2}{l^2}}} \right) \end{align*} $$


Intuition for some of the steps. The above answer might appear a bit too slick, but it really is natural. In the step where we got $$ - \int \frac{dx}{\left( 2\mu E+ \frac{\mu^2 C^2}{l^2} - \left( lx - \frac{\mu C}{l} \right)^2 \right)^{1/2}} , $$ the natural thing to do is to substitute $lx - \frac{\mu C}{l} = y$. Moreover, we will denote the constant $2\mu E+ \frac{\mu^2 C^2}{l^2}$ by $A^2$. Then the integral becomes $$ - \frac{1}{l} \int \frac{dx}{\sqrt{A^2 - y^2}} . $$ The reason for calling that quantity $A^2$ instead of $A$ is to make the expression inside the square root look homogeneous (in physical terms, now $A$ has the same dimension as $y$). The last integral is a standard integral. It is fairly common to try substitutions involving trigonometric or hyperbolic functions. After some trial and error, we find that $y = A \sin \theta$ and $y = A \cos \theta$ will both work. [Of course, in this case, substituting $y$ as $A \tan \theta$, $A \sec \theta$, $A \cosh \theta$, $A \sinh \theta$ etc. do not seem to give anything.] My solution uses $y = A \sin \theta$.

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@orion It was a typo and it's fixed now. Thanks. –  Srivatsan Dec 16 '11 at 5:16
    
@ Srivatsan. Thank you very much. –  orion Dec 16 '11 at 5:21
    
@ Srivatsan: Just one more doubt. What made you assume, $lx - \dfrac{\mu C}{l} = \sqrt{2 \mu E + \dfrac{\mu^2 C^2}{l^2}} \sin{\theta}$? It didn't occured to my mind? :( –  orion Dec 16 '11 at 5:43
    
@orion, If the answer appears too slick, I will make the steps more explicit now. –  Srivatsan Dec 16 '11 at 6:05
    
@orion I have expanded my answer. Hope this helps. –  Srivatsan Dec 16 '11 at 6:19

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