# Integration/Proving help! And check my work please!

1. Oct 4, 2004

### Odyssey

Greetings,

I am given the following equation.
$$t-t_{0}=\int_{R_{0}}^{R(t)}\frac{du}{\sqrt{2mEL^{-2}u^4-u^2-2mL^{-2}u^4V(u)}}$$

This is the total energy of a prticle moving in a central conservative field. m = mass, E = energy, L = angular momentum. The force the particle experiences is F = -Hmu^-3, where H is some constant, m is the mass of the particle, and u the distance. V(u), the potential, is just the negative integral of the force, and it is

$$-Hm/2u^2$$

How can I show that the energy, E, if E < 0, then

$$\alpha(t-t_{0})=\int_{R_{0}}^{R(\Theta)}\frac{du}{u\sqrt{a^2-u^2}}$$

, for real numbers alpha and a.

And similarly, for E > 0, how can I show it's

$$\beta(t-t_{0})=\int_{R_{0}}^{R(\Theta)}\frac{du}{u\sqrt{u^2+b}}$$

, for some real numbers beta and b?

I really need help on this!

I did show, for the E = 0 case, how it should be done. Please take the time check my work for this part.

Since E = 0, the total energy equation simplifies to:

$$t-t_{0}=\int_{R_{0}}^{R(\Theta)}\frac{du}{\sqrt{-u^2-2mL^{-2}u^4V(u)}}$$

then, plugging in V(u),

$$t-t_{0}=\int_{R_{0}}^{R(\Theta)}\frac{du}{\sqrt{-u^2+m^2L^{-2}Hu^2}}$$

$$Let s = m^2L^{-2}H$$

$$t-t_{0}=\int_{R_{0}}^{R(\Theta)}\frac{du}{\sqrt{-u^2+su^2}}$$

$$t-t_{0}=\int_{R_{0}}^{R(\Theta)}\frac{du}{u\sqrt{-1+s}}$$

$$t-t_{0}=\int_{R_{0}}^{R(\Theta)}\frac{du}{u\sqrt{-1+s}}$$

Since s is only a bunch of constants, we can factor it out.

$$t-t_{0}=\frac{1}{\sqrt{-1+s}}\int_{R_{0}}^{R(t)}\frac{du}{u}$$

$$\ln {R_{\Theta}/R_{0}} = \sqrt{s-1)$$

then solve for $$R (\Theta)$$
$$R (\Theta) = R_{0}e^{(\Theta-\Theta_{0})\sqrt{s-1}}$$

Last edited: Oct 4, 2004
2. Oct 4, 2004

### Tide

I don't know whether you did your integrations correctly but the problems as you stated them only involve factoring numbers out from the radicals and rearranging terms.

3. Oct 4, 2004

### Odyssey

Yes...I was asked to get the equation...carry out the integral and arrange the terms so that it resembles the ones above with the alpha and beta.