Two body problem, deriving energy expression

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SUMMARY

The discussion focuses on deriving the energy expression for the two-body problem involving two rotating point masses, specifically where one mass is significantly larger than the other. The orbital energy per unit mass is defined as ##\epsilon = -\mu/2a##, with ##\mu## representing the gravitational parameter ##GM##. The participants derive the relationship between the energy and the moments of momentum at the periapsis and apoapsis, using the equations for orbital energy and moment of momentum. The discussion concludes with a suggestion to substitute for the moment of momentum in terms of ##\mu##, semi-major axis ##a##, and eccentricity ##e## for further simplification.

PREREQUISITES
  • Understanding of the two-body problem in classical mechanics
  • Familiarity with orbital mechanics concepts such as periapsis and apoapsis
  • Knowledge of gravitational parameters, specifically ##\mu = GM##
  • Proficiency in vector mathematics, particularly cross products and conservation laws
NEXT STEPS
  • Study the derivation of the gravitational parameter ##\mu = GM## in celestial mechanics
  • Learn about the conservation of angular momentum in orbital dynamics
  • Explore the implications of eccentricity ##e## on orbital shapes and energy
  • Investigate the mathematical techniques for simplifying expressions involving multiple variables
USEFUL FOR

This discussion is beneficial for physics students, astrophysicists, and engineers working on orbital mechanics and gravitational interactions, particularly those interested in the mathematical foundations of celestial dynamics.

tomwilliam2
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Homework Statement


This problem relates to the two body problem of two rotating point masses, where one is much larger than the other. Equate the orbital energy per unit mass ##\epsilon## with the moments of momenta at the apses to get:
##\epsilon = -\mu/2a##

Homework Equations


The orbital energy per unit mass is ##\epsilon = \frac{1}{2}V^2 - \frac{\mu}{r}## where I think ##\mu = GM##
Moment of momentum in massless form is, I think, ##\mathbf{h}=\mathbf{r \times}\mathbf{v}##

The Attempt at a Solution


The apses are the periapsis ##r_p = a(1-e)## and the apoapsis ##r_a = a(1+e)##. I can insert these into the energy equation above to get:
##\epsilon = \frac{1}{2}V_p^2 - \frac{\mu}{a(1-e)}=\frac{1}{2}V_a^2 - \frac{\mu}{a(1+e)}## (1)

I know that the position and velocity vectors are always perpendicular, so
##h = rv##
##h^2 = r^2v^2##
I can use the values of ##r## at the apses (and conservation of h) to get:
##\frac{h^2}{a^2(1-e)^2}=V_p^2##
##\frac{h^2}{a^2(1+e)^2}=V_a^2##
Now inserting these values into the energy equation (1) above, I get:
##\epsilon = \frac{h^2}{2a^2(1-e)^2} - \frac{\mu}{a(1-e)}=\frac{h^2}{2a^2(1+e)^2} - \frac{\mu}{a(1+e)}##

Now I'm a bit stuck...
 
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tomwilliam2 said:
Now I'm a bit stuck...

i think you should substitute for h in terms of mu , a and e and try to simplify
 
Thanks.
 

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