Black body radiation is proportional to the power squared:
F=k1*T^4 (1)
Differencing with respect to x:
dF/dx=4*k1*T^3*dT/DX (2)
The fraction of the power absorbed per distance is proportional to the percentage of gas:
dF/dx=k2*rho*F——- (3)
Thus:
k2*rho*F=4*k1*T^3*dT/Dx ——- (4)
The change in pressure with respect to altitude is given by:
dP/dx=rho*g*x ——- (5)
From the ideal gas law and the above equation
P=rho RT ——- (6)
Thus:
rho*(g/R)*x=R*(T*d{rho}/dx+rho*dT/dx) ——- (7)

The last piece of the puzzle is the inverse greenhouse effect as the earth must emit what it absorbs. The power absorbed over a given height is over the cross section a circle (A1) while the power emitted is over the area of a sphere (A2).

Fin/dx=-k3*rho*Fin ——- (8)

dF/dx=1/(d(A2)/dx)*Fin*(A1/A2) ——- (9)
dF/dx=1/(2*4*pi*(x+xo))*Fin*(pi*(x+xo)^2)/(4*pi*(x*xo)^2) ——- (10)
dF/dx =Fin/(32*(x+xo) ——- (11)

Thus ignoring convection and conduction which I think we can do at high altitudes, the laps rate is described by the system of differential equations (7), (4) (8) and (11). To include conduction and convection just change equation 1 and the rest follows. The fractionating of the gases can be dealt with by statistical thermodynamics.

I might try this sometime in the future to see how close an answer it gives us to the actual lapse rate and see how changing the constants k2 and k3 affect the laps rate. As of now I am not sure if the change in the lapse rate is the most significant factor or if it is the pushing of the isothermal curves to higher altitudes.

A program of the players would have been useful.

Otherwise some interesting thoughts and comments.