Deriving Total Energy of a Binary System

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SUMMARY

The total energy of a binary system consisting of two objects with masses m1 and m2 is expressed as E = (1/2)μv² - G(Mμ/r), where μ is the reduced mass. This derivation assumes both objects are in circular orbits around their common center of mass. The solution requires understanding the center of mass frame, where the net momentum is zero, and the relationship between kinetic and potential energy. The discussion emphasizes that the derivation applies generally, not just for equal masses.

PREREQUISITES
  • Understanding of kinetic energy (K = 1/2 mv²)
  • Knowledge of gravitational potential energy (U = -G(Mm/r))
  • Familiarity with the concept of reduced mass (μ)
  • Basic principles of the center of mass frame
NEXT STEPS
  • Study the derivation of the reduced mass in binary systems
  • Explore the properties of the center of mass frame in orbital mechanics
  • Learn about the implications of circular orbits on energy calculations
  • Investigate gravitational interactions in multi-body systems
USEFUL FOR

Students and educators in physics, particularly those focusing on mechanics and orbital dynamics, as well as researchers analyzing binary systems in astrophysics.

AnniB
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Homework Statement


Beginning with the kinetic and potential energies of two objects with masses m1 and m2, show that the total energy of a binary system is given by:
E=\frac{1}{2}\muv2 - G\frac{M\mu}{r}


Homework Equations


The one given
K =\frac{1}{2}mv2
U = G\frac{Mm}{r}

The Attempt at a Solution


I feel like this should be more of an explanation answer than much of an actual derivation, but I'm not sure if my logic follows correctly.

Assuming that both objects are orbiting in a circle around their common center of mass, both objects will have a radius r and a velocity v (since at any given orbit distance all objects have the same velocity - unless I'm remembering that wrong?). Because of this, we can treat the two masses as one mass revolving about the center of mass at the same r and v, and thus all you have to do is plug the reduced mass \mu into the initial expression for total energy of a system, which would give you the equation from the problem.
 
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Your logic does work, but only for the case of equal masses in a circular orbit. What they are looking for is much more general.
If you start out by writing down the total energy (i.e. adding up all of the different terms) in the center of mass frame, you can simply to reach the desired equation.
(Hint: you'll need to use some properties of the center of mass frame, e.g. the net momentum is zero).
 

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