I have read in my text book that for a satellite to be bound to a planet, the Binding Energy of the system must be negative. How should the binding energy of a system be calculated if the Center of Mass of the system is moving?
For a binary star system, both stars have same mass $m$, if they are seprated by a distance $d$ and have velocity $v+u$ and $v-u$ in the i-hat direction whould the binding energy be $$\frac{1}2m(v+u)^2-\frac{Gm^2}{d}$$
When we calculate a binding energy we always have to take some reference energy to compare it against, and the usual choice is to specify that the energy is zero when the two objects are separated by an infinite distance and at rest i.e. that both the potential and kinetic energies are zero. This implies that we have to do the calculation in the centre of mass frame. In that frame the total energy is just:
$$ E = -\frac{Gm_1 m_2}{d} + \tfrac{1}{2}m_1 v_1^2 + \tfrac{1}{2}m_2 v_2^2 $$
It is possible to do the calculation in a frame in which the centre of mass is moving, but then the energy at infinity won't be zero.
Suppose the centre of mass is moving at some velocity $V$, and for convenience we'll take the moment when the two velocities $v_1$ and $v_2$ are parallel to the direction of $V$, then the total energy is:
$$ E = -\frac{Gm_1 m_2}{d} + \tfrac{1}{2}m_1 (V+v_1)^2 + \tfrac{1}{2}m_2 (V-v_2)^2 $$
But now the energy at infinity is:
$$ E_\infty = \tfrac{1}{2}m_1 V^2 + \tfrac{1}{2}m_2 V^2 $$
The binding energy is $E-E_\infty$ so we get:
$$ E_\text{bind} = -\frac{Gm_1 m_2}{d} + \tfrac{1}{2}m_1 (V+v_1)^2 + \tfrac{1}{2}m_2 (V-v_2)^2 - \tfrac{1}{2}m_1 V^2 + \tfrac{1}{2}m_2 V^2$$
And if we multiply out the brackets we get:
$$ E_\text{bind} = -\frac{Gm_1 m_2}{d} + \tfrac{1}{2}m_1 v_1^2 + \tfrac{1}{2}m_2 v_2^2 + (m_1 v_1 - m_2 v_2) V$$
But we know that $m_1v_1 = m_2v_2$ because $v_1$ and $v_2$ are the velocities in the centre of momentum frame where the total momentum is zero. So our expression becomes:
$$ E_\text{bind} = -\frac{Gm_1 m_2}{d} + \tfrac{1}{2}m_1 v_1^2 + \tfrac{1}{2}m_2 v_2^2 $$
which is exactly the same as we had before.
So you can calculate the binding energy in a moving frame, but doing so just creates unnecessary extra complexity.
