Escaping the Gravity well of the Moon How much energy would it take to get 1 kg from the Moon's surface to Earth? Would You aim for Lagrange Point 1 or would you launch in the opposite direction that the moon is orbiting? 
 A: Assuming all bodies stationary the acceleration on the way from the earth to the moon is
$$ \rm g=\frac{G M_1}{r^2}-\frac{G M_2}{(d-r)^2} $$
where $\rm M_1$ and $\rm M_2$ is the mass of earth and moon and $\rm d$ the distance from the center of the earth to the center of the moon. At the radius
$$ \rm R_0 = \frac{d M_1-d \sqrt{M_1} \sqrt{M_2}}{M_1-M_2} $$
the acceleration in the direction of the earth and that in the direction of the moon cancel each other out, so the specific kinetic energy to get that far is
$$ \rm e=\int_{R_1}^{R_0} g \, dr $$
where $\rm R_1$ ist the radius of the earth. Assuming vacuum the initial velocity needed would be
$$ \rm v_0=\frac{c \sqrt{e} \sqrt{2 c^2+e}}{c^2+e} $$
which is a little bit lower than the escape velocity to infinity. Under ideal circumstances you would need
$$ \rm 6.13*10^7 \ J/kg $$
of energy in joule per kilogram of your cargo to make it from the surface of the earth to $\rm R_0$, and from there you gain
$$ \rm 2.58*10^6 \ J/kg $$
kinetic energy with which you will hit the surface of the moon (integrate from $\rm R_0$ to $\rm d-R_2$ where the latter is the radius of the moon). If you want to land softly, you will have to use that amount to decelerate again. If you start on the moon and want to land on the earth it is simply the other way around.
A: Approximation:
The formula for escape velocity is:
$Vesc =  \sqrt (\frac{2∗G∗M}{r})$
Where:
G = 6.67384E-11 N*m^2*kg^-2 (Gravitational Constant)
M = 7.34767E22 kg (mass of the moon)
r = 1737400 m (radius of the moon)
Plug those in and we get an escape velocity of 2375.89 m/s
To reach this speed the kinetic energy required is $Ke = \frac{1}{2} mv^2$
or   $\frac{1}{2} * 1kg * 2375.89 m/s * 2375.89 m/s $= 2,822,426joule
L1 seems reasonable, but I'm no expert. You'd probably want to consider the 4.627 m/s rotation velocity the moon has. You'll come off in a tangent.
