- #1
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!
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]
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!
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]
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