Previously we calculated the fuel requirements of rockets to reach escape velocity. In that calculation, we did not take into account the effect of gravity during the fuel burn and underestimated the fuel requirements. We improved on this through a simulation better approximating reality. This simulation takes into account the force of gravity during the burn, as well as that force weakening as the rocket increases its distance from Earth, and takes into account the necessary escape velocity decreasing as the rocket’s distance to Earth grows. On this second post, Basker V commented that a closed-form solution can be found when the burn time is sufficiently small, allowing the rocket’s distance to Earth to be considered constant during the burn. In effect, this enables us to ignore both the diminishing force of gravity and decreasing escape velocity. Let’s find this closed-form.

In the rocket simulation post, we arrived at the following formula:

\frac{dv}{dt} = \frac{-v_e dM / dt}{M} – G\frac{M_\oplus}{r^2}with v the rocket velocity, t the time, v_e the fuel exhaust velocity, M the rocket mass, G the gravitational constant, M_\oplus the mass of Earth, and r the distance of the rocket to the center of Earth. Rewriting, we find the instantaneous change in velocity:

dv = \frac{-v_e}{M} dM – G\frac{M_\oplus}{r^2} dtAs in our original derivation of the rocket equation, we take the definite integral of all change terms from the initial to the final situations: dv is taken from the initial to the final velocity, dM from the initial to the final mass, and dt from the initial to the final time:

\begin{aligned} \int_{v_i}^{v_f} dv &= \int_{M_i}^{M_f} \frac{-v_e}{M} dM – \int_{t_i}^{t_f} G\frac{M_\oplus}{r^2} dt \\ \int_{v_i}^{v_f} 1 dv &= -v_e \int_{M_i}^{M_f} \frac{1}{M} dM – G\frac{M_\oplus}{r^2} \int_{t_i}^{t_f} 1 dt \\ v \Bigr\rvert_{v_i}^{v_f} &= -v_e \left( \ln{M} \Bigr\rvert_{M_i}^{M_f} \right) – G\frac{M_\oplus}{r^2} \left( t \Bigr\rvert_{t_i}^{t_f} \right) \\ v_f – v_i &= -v_e \left( \ln{M_f} – \ln{M_i} \right) – G\frac{M_\oplus}{r^2} \left( t_f – t_i \right) \\ \Delta v &= v_e \left( \ln{M_i} – \ln{M_f} \right) – G\frac{M_\oplus}{r^2} \left( t_f – t_i \right) \\ \Delta v &= v_e \ln{\left(\frac{M_i}{M_f}\right)} – G\frac{M_\oplus}{r^2} t_\text{burn} \end{aligned}here t_\text{burn} = t_f â€“ t_i is the total burn time. To find the fuel requirement, we want the formula for M_i as a function of \Delta v. Symbolic equation solvers happily rewrite the formula for us:

M_i = -v_e b \cdot \left. \text{W}{ \left( – \frac{\exp{\left( \Delta v/v_e – (G\frac{M_\oplus}{r^2} M_f)/(v_e b) \right)} G\frac{M_\oplus}{r^2} M_f }{v_e b} \right) } \middle/ \left(G\frac{M_\oplus}{r^2}\right) \right.with \text{W}{\left(\cdot\right)} the Lambert W function, and b = -dM/dt such that t_\text{burn} = \left( M_i â€“ M_f \right) / b.

We use the same values as before: the dry mass of the rocket is M_f = 2\,000 \text{ kg}, required escape velocity \Delta v = 1.118 \cdot 10^4 \text{ m/s}, and fuel exhaust v_e = 5\,000 \text{ m/s}. Plugging this into the equation, We find that the initial mass is 37\,667 \text{ kg}, and so the fuel required is 35\,667 \text{ kg}.

This number is considerably higher than the 24\,500 \text{ kg} found through simulation. And this makes sense: at the end of the simulated burn, the rocket has achieved a height of 600 \text{ km} over sea level. This is 10% of the radius of the Earth. Deceleration due to gravity at that point is 8.2 \text{ m}/\text{s}^2, rather than 9.8 \text{ m}/\text{s}^2 at sea level. Furthermore, the necessary escape velocity will have decreased to 10.69 \text{ km/s} from 11.18 \text{ km/s} at sea level. As such, this closed-form approximation is overestimating the fuel requirements by taking those values as constant from sea level at Earth.

This approximation gets more accurate as the rocket’s acceleration increases (e.g. through higher burn rates), resulting in lower burn times, thereby making the assumption of constant gravity and escape velocity more valid. Once the burn rate is taken as instantaneous, this approximation yields the same result as the naive rocket equation.

You can try out the equation interactively here. Note that not all parameter combinations make escape velocity achievable.