This question was initially inspired through playing Kerbal Space Program, however I am seeking a more general answer.

I've been playing the excellent Near Future Propulsion mod for KSP, using theVASIMR engine. For those not in the know, a VASIMR engine has variable thrust - A low thrust-high isp option which is suitable for deep space travel, and a high thrust-low isp option which is powerful enough to lift off from smaller solar system bodies.

At what point in the process of taking off does it become optimal (in terms of delta-V) to make the switch over from high thrust to high isp? Would the optimum change be gradual or sudden?

My intuition is that for as long as you're on the ground or on a sub-orbital trajectory, you must accelerate as fast as possible to enter orbit, and once in orbit, switch to low thrust-high isp. It is not clear to me, once you are on a trajectory that is not quite an orbit but is only a few m/s off, is whether the changeover should be sudden or gradual. I suspect there would be a curve describing the optimal delta-V usage depending on the specifics of the engine.

  • $\begingroup$ This is quite a complex problem, especially if you want to include aerodynamic effects? The answer will also depend on the gravity of the body in question. Extreme examples: with very low surface gravity you can perform the entire manoeuvre at your low thrust setting. At very high gravity you can burn all your fuel without ever moving so it wouldn't matter which setting you chose. $\endgroup$
    – Jack
    Jun 13, 2018 at 10:42
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    $\begingroup$ The problem is - you need more energy for the same deltaV with higher Isp. So in real world power of energy source limits your acceleration at given Isp. Acceleration rates by VASIMR and other types of ion engines are actually far less than needs to liftoff ever from big asteroid. So, to liftoff you should use chemical engines. $\endgroup$
    – Heopps
    Jun 13, 2018 at 10:47
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    $\begingroup$ And yes, as @Heopps points out, the real world VASIMR produces far less thrust than in KSP. Your question is still a good one for a hypothetical variable ISP engine though. $\endgroup$
    – Jack
    Jun 13, 2018 at 10:52
  • $\begingroup$ @Jack the question is strictly limited to airless bodies. High-Isp engines like VASIMR would behave similar to ion engines in atmosphere - that is to say badly. It seems that KSP has dramatically exaggerated its thrust capabilities in the cause of gameplay. $\endgroup$
    – Ingolifs
    Jun 13, 2018 at 10:57
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    $\begingroup$ A good related question on aviation about afterburners aviation.stackexchange.com/questions/17289/… $\endgroup$
    – Antzi
    Apr 12, 2019 at 5:42

1 Answer 1


I've done the requisite maths to be able to answer my own question. The answer turned out to be surprisingly simple and amenable to calculation.


Your spacecraft is standing on a small, airless, perfectly spherical planetary body.

Your spacecraft has a variable $I_{sp}$ engine which operates at a constant power $P$. The thrust force and exhaust velocity are governed by the following equation $$ P =Fv_e $$ where $F$ is the force (thrust) of your engine and $v_e$ is the variable exhaust velocity. Increasing $v_e$ increases the fuel efficiency, but comes at the cost of thrust. Thrust is inversely proportional to exhaust velocity. $$ F \quad \alpha \quad \frac{1}{v_e}$$

You want to get to orbit as efficiently as possible. Once you are in orbit, you can take your time with maneuvers, and you can set your $I_{sp}$ to be as efficient as possible. While in suborbital flight, you have no such luxury, as time is not on your side. You must accelerate to orbital velocity $v_o$ as quickly as you can or else you'll crash.


It is generally agreed (at least in the KSP community) that the best way to get off an airless body is to boost upwards till you've cleared local topography, then boost as hard as you can for the horizon. The orbit should be as low as possible without crashing into terrain. In my perfectly smooth sphere solution, this orbit is placed a negligible distance above the surface.

This makes for a simple vector diagram. After a negligible boost into the air, the spacecraft is gliding over the surface. The total acceleration $a_t$ is separated into a vertical component $a_v$, which is exactly counteracting the acceleration from gravity, $g$, and a horizontal component $a_h$ which is the one contributing to the overall orbital velocity.

enter image description here $$ a_t = \sqrt{a_h^2 + a_v^2} $$ $$ a_v = g $$ $$ a_h = \sqrt{a_t^2-g^2}$$ The total acceleration $a_t$ can be derived from the thrust of our engine and the mass of the craft $m_t$ $$a_t = \frac{F}{m_t} = \frac{P}{v_e m_t}$$ and the time taken to reach orbital velocity $v_o$ is: $$ t = \frac{v_o}{\sqrt{a_t^2-g^2}} = \frac{v_o}{\sqrt{(\frac{P}{v_e m_t})^2-g^2}}$$ At this point, you can sort of see what the problem is. At one extreme, the acceleration is very high and so therefore most of it goes towards reaching orbital velocity. This is an efficient use of propellant. However the force required to make this acceleration is so high that the engine needs to operate very inefficiently, nullifying your efficiency gains. At the other extreme, even though the engine is running efficiently, the thrust is so small that most of it is spent fighting gravity, and the residual horizontal acceleration occurs at a snails pace. It takes a very long time to accelerate to orbital velocity, and in the meantime you've actually wasted most of your fuel just fighting gravity. There is a hard limit where $a_t$ is equal to or less than $g$, and in that case, your spacecraft just isn't going anywhere. It can't even lift itself off the surface.

enter image description here


We are minimising mass loss from propellant, not delta-v. With a variable $I_{sp}$ engine, the total delta-v budget is not so precisely defined. So first, we need to find the mass loss rate $ \frac{dm}{dt} $as a function of exhaust velocity. A rocket's thrust is defined by $$F = \frac{dm}{dt} v_e$$ and based on our first equation $P =Fv_e$ $$ \frac{dm}{dt} = \frac{P}{v_e^2} $$ and so the mass of propellant required to reach orbital velocity as a function of $v_e$ is: $$ m =t \frac{dm}{dt} = \frac{P v_o}{v_e^2 \sqrt{(\frac{P}{v_e m_t})^2-g^2}}$$

mass loss vs exhaust velocity This graph clearly illustrates the behaviour of the function and the location of the minimum. The 'max exhaust velocity' is $\frac{P}{m_t g}$, and if you plug that into the above equation you'll see why. (it makes the denominator zero)

If we do some quick calculus (left as an exercise to the reader), and make the naughty assumption that the overall mass $m_t$ doesn't change, we find the exhaust velocity that provides the minimum mass loss is $$ v_e(min) = \frac{P}{\sqrt{2} m_t g} $$ which is $\frac{1}{\sqrt{2}}$ times the value of the maximum exhaust velocity.

When plugging this exhaust velocity back into the acceleration vector diagram, we find that the optimal angle is $45^o$.

It's also important to note, that while the above optimum was found assuming no mass loss to orbit, it also holds for any instantaneous velocity - at any point in time, given a certain mass, it will give you the most efficient exhaust velocity.

Accounting for mass loss Since the optimum ascent profile seems to be to maintain an angle of 45 degrees while gliding along the surface (effectively accelerating forward with acceleration $g$), any decrease in mass will result in an increase in exhaust velocity, as $M_t v_e$ will be kept constant. This results in a mercifully easy DE to solve. $$ \frac{dm}{dt} = \frac{-P}{v_e^2}$$ and $$ m v_e = \frac{P}{\sqrt{2}g} $$ whose solution for $m$ is $$m = \frac{Pm_0}{2g^2m_0t+P}$$ where $m_0$ is the initial mass at takeoff

and $$v_e = \sqrt{2}gt+\frac{P}{\sqrt{2}gm_0} $$ Which looks like this when plotted over time (arbitrary units): enter image description here


If you're playing KSP and wondering how to ascend most efficiently, burn upwards to clear local topography, then point your nose at 45 degrees and try to adjust your $I_{sp}$ at a constant rate so that you are neither gaining nor losing altitude. You want to be following the red line in the graph above.

And if you're playing real life, either find yourself a very tiny asteroid, or don't use a VASIMR engine, as ~5 N of thrust is pretty abysmal.

  • $\begingroup$ This is really interesting, but I got stuck as soon as I hit $P = F v_e$. How does arise? $\endgroup$
    – uhoh
    Apr 12, 2019 at 5:57
  • $\begingroup$ It's one of the assumptions. Constant power, variable force and exhaust velocity. I'm no engineer, but to a first approximation it seems reasonable that a VASIMR engine would behave like this. Try dimensional analysis on it: Power is $kgm^2s^{-3}$ force is $kgms^{-2}$ and velocity is $ms^{-1}$ $\endgroup$
    – Ingolifs
    Apr 12, 2019 at 6:22
  • $\begingroup$ I'd run it at maximum acceleration voltage (and therefore $v_e$) at all times to make best use of limited propellant. The primary use of electrical power is the production of the high density plasma, not acceleration, so I'd assume the ion current ($\dot{m}$) would be proportional to Power. That gives $F=dp/dt=\dot{m}v_e \approx Pv_e$ with a physics based proportionality constant to make the units work. But that's just me. $\endgroup$
    – uhoh
    Apr 12, 2019 at 6:41
  • $\begingroup$ Ugh. Way to ruin my nice equations... $\endgroup$
    – Ingolifs
    Apr 12, 2019 at 6:45
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    $\begingroup$ I joke. Also, the other reason I made that assumption is because that's how it works in KSP. $\endgroup$
    – Ingolifs
    Apr 12, 2019 at 6:48

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