Half.
The escape velocity for an object at a distance $D$ from an object of mass $M$ is $\sqrt{2GM/D}$. The circular orbital velocity (the Moon is on an orbit that's close enough to circular that I'll just assume this) at the same distance is $\sqrt{GM/D}$. Setting the escape velocity from the Earth with it's new reduced mass $M_{\rm new}$ equal to the orbital velocity around the Earth with it's usual mass $M_{\rm old}$ gives:
$$\sqrt{\frac{2GM_{\rm new}}{D_{\rm Moon}}} = \sqrt{\frac{GM_{\rm old}}{D_{\rm Moon}}}$$
Which immediately gives:
$$\frac{M_{\rm new}}{M_{\rm old}} = \frac{1}{2}$$
Note that you'd need to remove all this mass to well outside the Moon's orbit, probably well away in the direction opposite where you're going to send the Moon. Otherwise, you'll expand the orbit of the Moon, but once it gets back outside the distribution of expelled mass, it will remain bound to the system.
This will obviously not be achieved with rockets launching satellites, chiefly because rockets work by throwing mass out of the back end... mass that would get left behind. It's pretty obvious that $0.5\,{\rm M}_{\oplus}$ in rocket fuel (even if half the Earth's mass could be made into rocket fuel!) is not enough rocket fuel to launch $0.5\,{\rm M}_{\oplus}$ in payload to escape velocity. Even if it could, you wouldn't be left with half the Earth, but instead $0.5\,{\rm M}_{\oplus}$ in rocket exhaust...