Integration/Proving help! And check my work please!

[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! :confused:

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}}

[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.

[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.