Created: September 06, 2022
Modified: September 06, 2022

rocket equation

This page is from my personal notes, and has not been specifically reviewed for public consumption. It might be incomplete, wrong, outdated, or stupid. Caveat lector.

Deriving here just for my own edification.

At each timestep a rocket ejects mass $\Delta m$ at velocity $-v_e$ relative to its current reference frame. At step $t$ the rocket itself has mass $M_t$ and gains velocity $\Delta v_t$ . By conservation of momentum, we have

\underbrace{\Delta m \cdot v_e}_\text{Change in exhaust momentum} = \underbrace{M_t \cdot \Delta v_t}_\text{Change in rocket momentum}

at each step, which we can rearrange to

\Delta v_t = v_e \frac{\Delta m}{M_t} = -v_e \frac{\Delta M}{M_t}

where we substitute $\Delta M = -\Delta m$ to indicate the change in the rocket's mass, which is simply the negative mass ejected.

Thus far we've considered changes over a unit-length timestep, implicitly using the equivalence $\Delta v = \frac{\Delta v}{\Delta t}$ for $\Delta t = 1$ . Writing the timestep explicitly and letting $\frac{\Delta v}{\Delta t} \to \frac{dv}{dt}$ (and similarly for $M$ ) we arrive at integrals over time

\int_{t_0}^{t_1} \frac{dv(t)}{dt} dt = -v_e \int_{t_0}^{t_1} \frac{1}{M_t} \frac{dM(t)}{dt} dt

which we can simplify to integrals over velocity and mass, respectively,

\int_{v(t_0)}^{v(t_1)} dv(t) = -v_e \int_{M(t_0)}^{M(t_1)} \frac{1}{M_t} dM(t)

and thus solve for the total change in velocity

v(t_1) - v(t_0) = v_e \log \frac{M(t_0)}{M(t_1)},

which we write in stylized form as the rocket equation

\Delta V = v_e \log\left(\frac{M_\text{fuel} + M_\text{payload}}{M_\text{payload}}\right).

This implies that

\frac{M_\text{fuel}}{M_\text{payload}} = e^{\Delta V / v_e} - 1,

i.e., the required fuel-to-payload ratio grows exponentially in the required $\Delta V$ . Even if we allow extremely high fuel-to-payload ratios in the range of a million even up to a billion (!), this implies a limit on final velocity of any rocket capped at just 14-20 times its exhaust velocity.

This seems like a case against using chemical rockets. In fact, it's even more broadly a case against any rocket system that operates by ejecting a propellant, even if the energy to accelerate that propellant comes from some other source (solar, nuclear, etc). That said, one difference is the role of the exhaust velocity $v_e$ . As the equation shows, increasing exhaust velocity yields exponential returns in $\Delta V$ . Unfortunately this is generally not possible for chemical rockets, since a unit mass of fuel+oxidizer supplies a fixed and finite amount of energy, sufficient to accelerate itself to some fixed velocity but not to arbitrarily high velocities. However, using a separate energy source like a nuclear reactor does work around this and could in principle allow for higher exhaust velocities.

A nice twitter thread on why chemical rockets still have major advantages over the plausible alternatives in the near and medium-term: https://twitter.com/Robotbeat/status/1563749118526332929

Staging. One way to partially 'escape' the rocket equation is to use multi-stage rockets. Of course, fundamentally the multi-stage system is still subject to the rocket equation. But a finer-grained breakdown of the rocket mass,

M = M_\text{fuel} + M_\text{payload} + \sum_i M^{(i)}_\text{structure}

reveals the role of the rocket structure (fuel tanks, engines, etc.), which in the naive calculation above was simply treated as part of the payload. But we don't necessarily need to accelerate the entire rocket up to final velocity; if we can drop pieces of the rocket as we go, then these terms effectively act more like fuel than like payload, which is incredibly valuable in practice.

rocket equation

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