# Hamiltonian of planar circular restricted 3-body problem

…in rotating coordinates

Translation of homework done for the course Computational Astrodynamics.1

The coordinates of inertial and rotating reference frame are connected

$(\xi, \eta) = R (t) (x, y) = x (\cos t, \sin t) + y (- \sin t, \cos t)$

$(x, y) = R^{- 1} (t) (\xi, \eta) = \xi (\cos t, - \sin t) + \eta (\sin t, \cos t)$

and

$(\dot{\xi}, \dot{\eta}) = (\dot{x} - y, \dot{y} + x) \cos t - (\dot{y} + x, y - \dot{x}) \sin t$

$(\dot{x}, \dot{y}) = (\dot{\xi} + \eta, \dot{\eta} - \xi) \cos t - (\xi - \dot{\eta}, \dot{\xi} + \eta) \sin t$

For the large bodies, from normalization (usage of non-dimensional coordinates)

$x_{2} - x_{1} = 1$

and from the conservation of momentum (constant velocity of the center of mass)

$(1 - \mu) x_{1} + \mu x_{2} = 0$

Solving the system

$(x_{1}, x_{2}) = (- \mu, 1 - \mu)$

In the planar circular restricted problem the orbits of them are

$(\xi _{1}, \eta _{1}) = x_{1} (\cos t, \sin t) \quad (\xi _{2}, \eta _{2}) = x_{2} (\cos t, \sin t)$

and the distances of the small body from the previous is

$r_{1}^2 = (\xi + \mu \cos t)^2 + (\eta + \mu \sin t)^2$

$r_{2}^2 = (\xi - (1 - \mu) \cos t^{})^2 + (\eta - (1 - \mu) \sin t)^2$

Utilizing the conversion relationships the distances are written

$r_{1}^2 = (x + \mu)^2 + y^2$

$r_{2}^2 = (x - (1 - \mu)_{})^2 + y^2$

The gravitational potential is written

$V = - \frac{1 - \mu}{r_{1}} - \frac{\mu}{r_{2}}$

and then the Lagrangian is written

$L = \frac{1}{2} (\dot{\xi}^2 + \dot{\eta}^2) - V$

Utilizing the conversion relationships, it is written

$L = \frac{1}{2} ( (\dot{x} - y)^2 + (\dot{y} + x)^2) - V$

The momenta are calculated differentiating this to the generalized velocities

$(p_{x}, p_{y}) = (\partial _{\dot{x}}, \partial _{\dot{y}}) L = (\dot{x} - y, \dot{y} + x)$

And therefore the Lagrangian is written

$L = \frac{1}{2} (p_{x}^2 + p_{y}^2) - V$

The Hamiltonian is found from Legendre transformation

$H = \dot{x} p_{x} + \dot{y} p_{y} - L$

Replacing and adding/subtracting the kinetical energy

$H = \frac{1}{2} (p_{x}^2 + p_{y}^2) + p_{x} (\dot{x} - p_{x}) + (\dot{y} - p_{y}^{}) p_{y} + V$

Finally

$H = \frac{1}{2} (p_{x}^2 + p_{y}^2) + p_{x} y - x p_{y} - \frac{1 - \mu}{r_{1}} - \frac{\mu}{r_{2}}$

1. Post date reflects original date; translation done 06/04/2021.