Integration/Proving help And check my work please

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SUMMARY

The discussion centers on deriving specific integral forms related to the total energy of a particle in a central conservative field, defined by the equation t-t_{0}=\int_{R_{0}}^{R(t)}\frac{du}{\sqrt{2mEL^{-2}u^4-u^2-2mL^{-2}u^4V(u)}}. For cases where E < 0, the integral simplifies to \alpha(t-t_{0})=\int_{R_{0}}^{R(\Theta)}\frac{du}{u\sqrt{a^2-u^2}}, and for E > 0, it becomes \beta(t-t_{0})=\int_{R_{0}}^{R(\Theta)}\frac{du}{u\sqrt{u^2+b}}. The participant successfully demonstrated the E = 0 case, leading to the conclusion that R(\Theta) can be expressed as R_{0}e^{(\Theta-\Theta_{0})\sqrt{s-1}} where s = m^2L^{-2}H.

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Odyssey
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Greetings,

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

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

[tex]-Hm/2u^2[/tex]

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

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

, for real numbers alpha and a.

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

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

, 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:

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

then, plugging in V(u),

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

[tex]Let s = m^2L^{-2}H[/tex]

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

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

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

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

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

[tex]\ln {R_{\Theta}/R_{0}} = \sqrt{s-1)[/tex]

then solve for [tex]R (\Theta)[/tex]
[tex]R (\Theta) = R_{0}e^{(\Theta-\Theta_{0})\sqrt{s-1}}[/tex]
 
Last edited:
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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.
 
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.
 

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