Question

4. The rocket equation states that for a given Isp and propulsion system mass fraction, 2, propellant mass increased exponentially with increasing AV. If the payload mass is fixed, derive an expression for the maximum AV attainable for a given rocket with fixed Isp and 2 values. You may neglect gravity and drag losses for your analysis.

Answer 1

Newton’s third law- for every action there is an equal and opposite reaction. It means the act of ejecting molecules pushes back on the rocket: every time a water molecule flies from the nozzle, the rocket loses a little mass and in return gains a little velocity. If the rocket burns fuel at a high rate, the reaction force on the rocket will be high, and so will the acceleration. The force is approximately proportional to the rate at which propellant is consumed.

Momentum is zero in the left panel and is conserved, the total momentum in the right panel must also be zero, which means the leftward momentum of the ejected mass must equal the rightward momentum of the rocket:

dm ve = (m - dm du (1)

If the ejected mass dm is very small compared with the total mass m, we should be able to ignore the –dm term on the right-hand side. However, if we do this, it’s important to be careful in the bookkeeping and encode in the equation the fact that a decrease in the mass of the rocket results in an increase in its velocity. A consistent convention is that both +dm and +dv should indicate increases in the mass and velocity of the rocket, which means reversing the sign of dm in Equation 1 (so that dm is now a negative number) and writing

--dm ve = m du (2)

The rocket ejects small masses molecule by molecule, each one adding a small increment to the rocket’s velocity and subtracting a bit from the remaining mass. Each time we can shift to a moving frame of reference where the rocket is stationary, as in the left panel of Figure 1, then apply the conservation of momentum. As time goes on, the rocket’s mass will drop as it consumes propellant. We want to know the cumulative effect of all the small velocity changes (all the dvs) during the rocket burn, relative to all the small mass changes (all the dms), and calculate a relationship between velocity and mass. First, rewrite Equation 2 as

It looks a little strange to write it this way, but what it says is the change in the rocket’s mass, per change in velocity, is proportional to the mass divided by the exhaust velocity. This is a direct consequence of the conservation of momentum and is true at every moment in time, even though the rocket’s mass and velocity are changing during the burn (I’ll assume the exhaust velocity remains constant). Now, if you’re familiar with differential equations and calculus, you may already know what to do with Equation 3: integrate it. If not, I must ask you to accept a mathematical leap of faith: it’s a well-established result that if the change in some quantity is proportional to the quantity itself, then the quantity varies exponentially. This sort of behavior occurs frequently in nature, including the exponential decay of radioactive isotopes and the exponential growth of bacteria when food is plentiful. In the present case the change in mass, with respect to change in velocity, is proportional to the mass itself, and therefore the rocket’s mass, as a function of velocity, obeys an exponential relation called the rocket equation. Written in a convenient form it looks like this:

minitial m final (4)

where the left-hand side is the ratio of the initial mass of the rocket (at the beginning of the burn) to the final mass (after the burn is complete), delta-v is the total change in velocity caused by the burn, and exp() is the exponential function. It’s a compact and beautiful result. In English it says the ratio of initial to final mass is the exponential of the ratio of delta-v to exhaust velocity, or, even more simply, the mass ratio is the exponential of the velocity ratio. The plot below shows Equation 4, where you can see how rapidly the mass ratio rises once the velocity ratio goes beyond 3 or 4; huge increases in mass are required for each increment of velocity. Often you’ll see the rocket equation written in an equivalent form, solved for delta-v:

Av = ve In (Minitial) (5) m final

where ln() is the natural logarithm function, the inverse of the exponential function.

You may have noticed that time appears nowhere in the rocket equation. That’s because the conservation of momentum “doesn’t care” how long it takes to use up propellant—the change in momentum, and therefore the final velocity, is the same. This is certainly true in the vacuum of empty space, far away from a planet. In the presence of gravity, though, it matters how fast a rocket consumes propellant. Imagine a rocket standing on the launch pad, burning its propellant so slowly that there’s not enough thrust to lift the rocket. All the propellant will be used up, and the rocket will go absolutely nowhere. This is an extreme example of an effect called gravity drag. The absolute amount of thrust is clearly important if our rocket is to leave the pad. If we divide both sides of Equation (2) by the small time interval dt during which the mass dm was ejected, and also make use of Newton’s famous second law of motion, force equals mass times acceleration, we get

do F = ma = m dm = (6)

where the acceleration a is the change in velocity per time, dv/dt. As before, note the minus sign: since dm/dt, the rate at which the rocket’s mass changes due to consuming propellant, is a negative number, the force is positive.