Circularization Energy

Let us consider a particle of mass $m$ orbiting a point mass $M$ at some eccentric orbit. Suppose that said particle can lose energy, but cannot rid itself of its angular momentum. Under this restriction there is a minimal energy that a particle can attain. The angular momentum is

$L = m v r$

In the terminal circular orbit, the relation between the radius and the angular velocity is

$v = \sqrt{\frac{GM}{r}}$

Solving for $v$ and $r$ yields

$v = \frac{G M m}{L}$

$r = \frac{L^2}{G M m^2}$

and the terminal energy is

$E = -\frac{1}{2} m v^2 = - \frac{1}{2} \frac{G M^2 m^3}{L^2}$