Let's simplify the scenario slightly - imagine an object at the distance of the Moon, that has had its angular momentum slowed sufficiently that it will approach the Earth to a distance less than the radius of the Earth plus the radius of the Moon.
Distance Moon-Earth ~ 400,000 km
Radius Moon ~ 1,700 km
Radius Earth ~ 6,300 km
Mass Moon ~ 7.3E22 kg
Mass Earth ~ 6.0E24 kg
The orbit we are interested in has a semimajor axis $a$ less than (400,000+1,700+6,300)/2 = 204,000 km.
The specific orbital energy is given by
$$\epsilon = -\frac{GM}{2a}$$
This means that the energy of the "grazing" orbit of the moon is $6.7\cdot 10^{-11}\cdot 6.0\cdot 10^{24}\cdot 7.3\cdot 10^{22}/(4\cdot 10^8) = 7.2\cdot 10^{28}~\rm{J}$
If you assume that you would need to increase the semimajor axis by 20 km to make sure that the moon just clears the Earth (recognizing that it would give you a horrific tidal pull), then the difference in energy is given by
$$\begin{align}
\Delta E &= GMm\left(\frac{1}{2a_1}-\frac{1}{2a_2}\right)\\
&=\frac{GMm}{2a_1}\left(1-\frac{a_1}{a_2}\right)
\end{align}$$
So for a change in orbit of 20 km / 400,000 km, you need to add about $3.6\cdot 10^{24}~\rm{J}$. To put this in perspective, the entire world population's energy consumption in 2012 was estimated to be about $5.2\cdot 10^{20}~\rm{J}$; so even nudging the moon a little bit would the world's total energy use for about 7000 years.
As for the time taken - the orbit I described would have a period of
$$T=2\pi\sqrt{\frac{a^3}{GM}}\approx 10.6~ \rm{days}$$
It would take half that time to drop from apogee to perigee; you have about 5 days to get that kind of energy together and shoot it to the moon. Not gonna happen.