I am going to change the notation in order to make the equations more compact. The counter weight is $M$ and the payload is $m$. The length of the bar is $L$ and the distance of the center of gravity to the counterweight is $a=\frac{m}{M+m}\,L$ and from the payload $b=\frac{M}{M+m}\,L$ such that $L=a+b$. Note I have said nothing about the pivot yet.
The distance between the pivot and the center of gravity is $c$ and it is an independent variable we wish to optimize. The pivot is between the center of gravity and the payload (for positive $c$). The angle of the bar is $\theta$ with $\theta=0$ when horizontal.
The height of the pivot from the ground is $h$ such that when the counterweight hits the ground the payload is launched at $\theta_{f} = -45^\circ$. So $h=(a+c) \sin(- \theta_f$). As a consequence the initial angle is $\sin \theta_i = \frac{a+c}{b-c} \sin(- \theta_f )$ in order for the payload to rest in the ground initially. This is valid for $c<\frac{b-a}{2}$, otherwise the things sits vertically with $\theta_i=\frac{\pi}{2}$.
Doing the dynamics using Newtons's Laws, or Langrange's equations will yield the following acceleration formula
$$ \ddot{\theta} = -\,\frac{c g (M+m) \cos(\theta)}{\frac{m M}{M+m} L^2 + (M+m) c^2} $$
The denominator being the moment of inertia about the pivot. Here is the fun stuff.
The above can be integrated since the right hand side is a function of $\theta$ only with a constant $\alpha$:
$$ \ddot{\theta}=\frac{{\rm d}\dot\theta}{{\rm d}t} =-\alpha \cos(\theta) $$
$$ \frac{{\rm d}\dot\theta}{{\rm d}\theta} \frac{{\rm d}\theta}{{\rm d}t} = -\alpha \cos(\theta) $$
$$ \frac{{\rm d}\dot\theta}{{\rm d}\theta} \dot\theta = -\alpha \cos(\theta) $$
$$ \int \dot\theta {\rm d}\dot\theta =-\int \alpha \cos(\theta) {\rm d}\theta + K$$
$$ \frac{{\dot \theta}^2}{2} = -\alpha \sin \theta + K $$
with $K$ based on the initial conditions ($\theta=\theta_i$, $\dot\theta=0$)
$$ \dot\theta = \sqrt{2 \alpha (\sin(\theta_i)-\sin(\theta))}$$
and final rotational velocity
$$ \dot\theta_f = \sqrt{2 \alpha (\sin(\theta_i)-\sin(\theta_f))}$$
tangentially the payload launch velocity is
$$ v_{B_f} = (b-c) \dot\theta_f = (b-c) \sqrt{2 \alpha (\sin(\theta_i)-\sin(\theta_f))} $$
with both $\alpha$ and $\theta_i$ depending on the variable $c$.
To optimize we set $ \frac{{\rm d}v_{B_f}}{{\rm d}c}=0 $ which is solved for:
$$ \frac{c}{L} = \frac{\sqrt{m} \left( \sqrt{M+m}-\sqrt{m} \right)}{M+m } $$
For example, a $m=20 {\rm lbs}$ payload, with a $M=400 {\rm lbs}$ counter weight on a $L=20 {\rm ft}$ bar, requires the pivot to be $c=20\;\frac{\sqrt{20} \left( \sqrt{420}-\sqrt{20} \right)}{420 } = 3.412\;{\rm ft} $ from the center of gravity. The c.g. is $a=\frac{20}{420}\,20 = 0.952 {\rm ft}$ from the counterweight.
Edit 1
Based on comments made by the OP the launch velocity is
$$
v_{B_{f}}=\left(\frac{M}{M+m}L-c\right)\sqrt{\frac{2cg(M+m)}{\frac{M}{M+m}L^{2}+(M+m)c^{2}}\left(\sin\theta_{i}-\sin\theta_{f}\right)}
$$
where $g=9.80665\,{\rm m/s}$ is gravity.
With infinite counterweight the maximum launch velocity is $\max(v_{B_{f}})=\sqrt{\frac{2g(L-c)^{2}}{c}}$ so to reach $v_{B_{f}}=6000\,{\rm m/s}$ from earth if $c=1\,{\rm m}$ then $L>1355.8\,{\rm m}$.
With infinite bar length the maximum launch velocity is $\max(v_{B_{f}})=\sqrt{\frac{2Mcg}{m}}$ so to reach $v_{B_{f}}=6000\,{\rm m/s}$ from earth if $c=1\,{\rm m}$ then $M>18.3\,10^{6}\,{\rm kg}$.
So lets consider $L=2000\,{\rm m}$ and $M=40.0\,10^{6}\,{\rm kg}$ then we choose pivot location at $c=1.500\,{\rm m}$ to get
$$
v_{B_{f}}=(1999.999-1.5)\sqrt{2\;4.526\left(\sin\theta_{i}-\sin\theta_{f}\right)}
$$
which is solved for $v_{B_{f}}=6000\,{\rm m/s}$ when $\sin\theta_{i}-\sin\theta_{f}=0.9957$
with $\theta_{i}>0$ and $\theta_{f}<0$.
The extra force in a lever comes from the distribution of energy over a smaller distance of action. We can look at it very generally from the perspective of conservation of energy, or the lossless transmission of work through a machine.
When we move something against a resistance, with a fixed force, we are putting in energy and doing work. This work is the product of force and distance: $w = F \times d$.
Devices which use pulleys, levers or hydraulics to generate a larger force all work by "trading" to a smaller $d$ and larger $F$. Since the machine is efficient (little energy is lost in the machine), the amount of work $w$ done on the machine by you is almost fully transmitted to the load. Since the distance $d$ by which the load is moved is smaller, the force is greater.
For instance a lifting machine whereby we push against a force, and move a meter, such that the load lifts by 1 cm, gives us a 100 fold advantage. The tradeoff is that although we moved something by a meter, the load only moved one centimeter. We paid for more force by sacrificing range of action.
We do not have to appeal to energy to see that large forces are generated; another way is to draw free-body diagrams of the parts of the machine, and the force applied to it, and to the load. Balancing the forces requires that a large force must be present on the load.
Best Answer
In deep space, the mass of $2m$ does not exert any force on the lever because it has no weight, so nothing happens.