Would someone please show me the equations that show a rocket needs something to "push against", because here is why you don't need an atmosphere for it to work.
Let us begin with Newton's second law of motion:
d (M u) / dt = F net
where M is the mass of the rocket, u is the velocity of the rocket, F net is the net external force on the rocket and the symbol d / dt denotes that this is a differential equation in time t. The only external force which we will consider is the thrust from the propulsion system.
The thrust equation is given by:
F = mdot * Veq
where mdot is the mass flow rate, and Veq is the equivalent exit velocity of the nozzle which is defined to be:
Veq = V exit + (p exit - p0) * Aexit / mdot
where V exit is the exit velocity, p exit is the exit pressure, p0 is the free stream pressure, and A exit is the exit area of the nozzle. Veq is also related to the specific impulse Isp:
Veq = Isp * g0
where g0 is the gravitational constant. m dot is mass flow rate and is equal to the change in the mass of the propellants mp on board the rocket:
mdot = d mp / dt
Substituting the expression for the thrust into the motion equation gives:
d (M u) / dt = V eq * d mp / dt
d (M u) = Veq d mp
Expanding the left side of the equation:
M du + u dM = Veq d mp
Assume we are moving with the rocket, then the value of u is zero:
M du = Veq d mp
Now, if we consider the instantaneous mass of the rocket M, the mass is composed of two main parts, the empty mass me and the propellant mass mp. The empty mass does not change with time, but the mass of propellants on board the rocket does change with time:
M(t) = me + mp (t)
Initially, the full mass of the rocket mf contains the empty mass and all of the propellant at lift off. At the end of the burn, the mass of the rocket contains only the empty mass:
M initial = mf = me + mp
M final = me
The change on the mass of the rocket is equal to the change in mass of the propellant, which is negative, since propellant mass is constantly being ejected out of the nozzle:
dM = - d mp
If we substitute this relation into the motion equation:
M du = - Veq dM
du = - Veq dM / M
We can now integrate this equation:
delta u = - Veq ln (M)
where delta represents the change in velocity, and ln is the symbol for the natural logarithmic function. The limits of integration are from the initial mass of the rocket to the final mass of the rocket. Substituting for these values we obtain:
delta u = Veq ln (mf / me)
This equation is called the ideal rocket equation. There are several additional forms of this equation which we list here: Using the definition of the propellant mass ratio MR
MR = mf / me
delta u = Veq * ln (MR)
or in terms of the specific impulse of the engine:
delta u = Isp * g0 * ln (MR)
If we have a desired delta u for a maneuver, we can invert this equation to determine the amount of propellant required:
MR = exp (delta u / (Isp * g0) )
where exp is the exponential function.
If you include the effects of gravity, the rocket equation becomes:
delta u = Veq ln (MR) - g0 * tb
where tb is the time for the burn.
And this is how rockets work.