Propulsion exhaust speed is lower than speed of light, I think and if it is so, than how could spaceship go faster that exhaust speed? Right? So exhaust speed is a maximum speed for spaceship? (If there is no gravity, etc.. just empty space and propulsion)


3 Answers 3


It's not exactly how you put it, exhaust speed is not a maximum speed for a spaceship. While exhaust velocity does define thrust together with the rate of expelled mass and is given by:

$$T=v\frac{\Delta m}{\Delta t}$$

where thrust $T $ is a reaction force that is a product of exhaust velocity $v$ and mass change $\Delta m$ in time $\Delta t$, you're neglecting inertia. In a sense, if you invest some of the propellants to achieve some thrust and you're on your way at a given speed, the remaining rocket's propellants now carry this same rocket's momentum force. This is more usually referred to as conservation of momentum.

So while the exhaust velocity normally doesn't change (assuming same performance engines and propellants) and is directly related to rocket's performance (what we usually refer to as specific impulse), all of the remaining mass of the rocket, including remaining propellants, now carry this same speed. Since thrust is given by exhaust velocity with respect to the rocket itself, you'd constantly add rocket's own speed to new thrust in time, to potentially go beyond the speed of the exhaust velocity. Does this make sense?

$$\Delta v = v_\text{e} \ln \frac {m_0} {m_1}$$

If you want to read more about how to calculate maximum achievable speed (mostly referred to as $\Delta v$, delta-v, or change in velocity), I'd suggest reading the Wikipedia article on Tsiolkovsky rocket equation or other similar questions on our own site.

For example, if we assume wet-to-dry mass ratio of 10:1 (payload and the rocket weigh together 1/10th the weight of propellants), the rocket will reach $\ln(10) \approx 2.3 \cdot v_e$, or delta-v of roughly 2.3 times the propellant exhaust velocity. For more practical example, say our rocket's total mass (wet mass; propellants + rocket's dry mass) is 100 metric tons, and the rocket's dry mass (when it used up all its propellants) is 10 metric tons. Assuming specific impulse $I_{sp}$ of 460 seconds (exhaust velocity of ~ 4,511 m/s, or $I_{sp}$ in seconds multiplied by $g \approx 9.80665\ m/s^2$) that's achievable with LOX/LH (liquid oxygen / liquid hydrogen) cryogenic propellant engines such as Space Shuttle Main Engines in vacuum, the total delta-v using Tsiolkovsky rocket equation above comes out at 10,387 m/s. So again those ~ 2.3 times the exhaust velocity, i.e. natural logarithm $\ln(x)$, where $x$ is $\frac {m_0} {m_1}$, or our wet-to-dry mass ratio. If you want to play with other parameters, you can use this Delta-V Calculator for convenience.

By the way, what you said in the questions would be true for some other methods of propulsion where the source of your thrust is stationary to the point you use to define your speed against, such as, say, solar sails or some forms of beam-powered propulsion.

As for limits, well traditional rocket propulsion methods come with their own limit often referred to as the tyranny of the rocket equation since, simplifying - your total delta-v depends on how much of propellants you carry, but the more of them you carry the heavier you get and it becomes more expensive to accelerate them to any speed (not to be confused with exhaust velocity, that stays the same).

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    $\begingroup$ Also note that staging allows you to effectively have an arbitrarily small dry mass ratio. Then you are limited only by how big of a rocket you can afford, which increases in size geometrically for each additive increment in velocity. $\endgroup$
    – Mark Adler
    Aug 10, 2014 at 5:23
  • $\begingroup$ Thanks for your answer with lots of useful links and good explanation. :) So if I understand correctly, I could achieve the speed close to speed of light if I would have imaginary propellants running out very slowly? $\endgroup$
    – Luckylooke
    Aug 10, 2014 at 12:31
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    $\begingroup$ @Luckylooke Well, it would have to be quite some propellants. Currently, we don't have any such propulsion available to us that would come close to even a fraction of c. Play with that Delta-V Calculator for a while and use Isp of say 12,000 s that's quoted as theoretically possible for VASIMR by some. You're gonna soon notice that the dry mass ratio has to be riddiculously low, making it infeasible for any meaningful mass payloads. $\endgroup$
    – TildalWave
    Aug 10, 2014 at 13:03

TildalWave gave a correct and detailed answer. For a very simple proof, though:

Even hydrogen/oxygen engines (the most powerful fuel that has actually been used) have an exhaust velocity far below orbital velocity.


Actually u r right for every action there is an equal and opposite reaction. So when there is no gravity the max speed will be equal to it. But there is no fixed limit lesser then the speed of light. Only speed of light is constant and unbreakable . according to theory of relativity and that's because time become slow to preserve that constant value.

  • $\begingroup$ Wrong, this would only be true if your wet-to-dry mass ratio is 1:1 and you instantaneously expel all the propellants (instantaneous impulse), or it's e:1 (2.718281828:1) and you take as much of time as you want. I.e. a rocket can go faster than its own propellants' exhaust velocity. $\endgroup$
    – TildalWave
    Aug 8, 2014 at 21:17

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