A lucky discovery of a Kerbal engineer's writing gave this deep insight into the mechanics of climbing to orbit:

the best way to minimize d/v losses through the atmosphere is to always go at exactly the terminal velocity at any given altitude (because this is the point of intersection between the graphs of drag losses vs gravity losses).

I had always found the mathematics of the gravity turn to be daunting. This would really toss a lot of that detail away, allowing many quick and useful back-of-the-envelope calculations.

But why should I believe it? Maybe it's just a Kerbal old wives tale. Is there any real theoretical model that suggests something like this?

Furthermore, it's not clear what terminal velocity should be used anyway. The mass of the rocket is changing throughout the climb, and with mass, the terminal velocity changes. Would this rule use the dynamic mass value, or the final mass, or next mass before staging, or something else?

  • $\begingroup$ Terminal velocity and mass - isn't that more about falling? I.e. F=ma so mass changing matters. a=gravity, so force down is mass X 9.8 m/s/s. Balanced against air resistance's force slowing you down. TermVeloc upwards though seems not to be the same thing? $\endgroup$
    – geoffc
    Sep 23, 2014 at 16:36
  • $\begingroup$ The idea is to be at terminal velocity (or near it) at any point during flight, which means to accelerate exponentially through the atmosphere. It may only apply to KSP's atmosphere model, though. Nothing of the sort is mentioned in my trusted orbital mechanics book. $\endgroup$ Sep 25, 2014 at 0:57
  • $\begingroup$ KSP's guides contain a lot on information that works well, if you don't want to nerd out and plan your space missions with a calculator (which I would't do. Ever... not me). But is not technically optimal, or only applies to the limited physics model. $\endgroup$ Sep 25, 2014 at 1:04
  • $\begingroup$ The Basic maneuvers wiki has a cheat sheet for terminal velocity on Kerbal. $\endgroup$ Apr 24, 2015 at 15:12

1 Answer 1


Climbing at terminal velocity is the most fuel efficient speed to gain (vertical) height while in steady state, such that the density, velocity, gravitational acceleration and other parameters stay constant during ascent.

To proof this I will first have to define the required thrust, $F$, which has to overcome gravity and drag (such that the sum of all forces would add up to zero) and can be calculated as follows,

$$ F = mg + \frac{1}{2} \rho v^2 C_D A, $$

where $m$ is the mass of the rocket, $\rho$ the density of the atmosphere, $v$ the velocity, $C_D$ the drag coefficient and $A$ the cross section area of the rocket.

This thrust can also be related to mass flow of the expelled fuel, like so

$$ F = v_e \dot{m}, $$

where $v_e$ is equal to the effective exhaust velocity of the rocket engine and $\dot{m}$ the mass flow ($\frac{dm}{dt}$).

To optimize fuel efficiency you would have to minimize the amount of fuel needed to gain height, thus minimize

$$ \frac{d m}{d y} = \frac{dm/dt}{dy/dt} = \frac{\dot{m}}{v} = \frac{mg + \frac{1}{2} \rho v^2 C_D A}{v_e v}. $$

This can be done by taking the derivative of this expression with respect to $v$ and set this equal to zero,

$$ \frac{d}{dv} \frac{mg + \frac{1}{2} \rho v^2 C_D A}{v_e v} = \frac{v_e \rho v^2 C_D A - v_e (mg + \frac{1}{2} \rho v^2 C_D A)}{v_e^2 v^2}=0, $$

$$ \rho v^2 C_D A - mg - \frac{1}{2} \rho v^2 C_D A = \frac{1}{2} \rho v^2 C_D A - mg = 0, $$

$$ \frac{1}{2} \rho v^2 C_D A = mg. $$

So the best velocity indeed turns out to be the same as the magnitude on terminal velocity (drag equal to gravity), but in the opposite direction of course.

However during the ascent only a small initial portion of the trajectory will be vertical. Once the rocket would start to pitch sideways then terminal velocity will probably not be the most efficient anymore.

  • $\begingroup$ I don't suppose you care to derive the optimum for the non-vertical case now? :) $\endgroup$ Oct 1, 2014 at 19:58
  • 1
    $\begingroup$ @RussellBorogove as far as I know there is no general solution. However optimal trajectories can be found numerically. For KSP there is also a forum thread about this. $\endgroup$
    – fibonatic
    Oct 1, 2014 at 20:39
  • $\begingroup$ The link is broken 😢 $\endgroup$ Dec 6, 2020 at 2:17

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