So there I was resting me eyes thinking about rocket drives, and what-not. The thought struck me that, perhaps, even before Mr. Einstein interferes with the increasing velocity of the spacecraft Mr. Newton may have something to say.

Please poke me in the rib if my comprehension is wrong -

A rocket, basically a reaction mechanism, must push gases out of it's exhaust to impart velocity to the spacecraft. Assuming adequate fuel is available, what happens when the spacecraft velocity is equal to velocity at the exhaust? At this stage, does the rocket still accelerate the craft?

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    $\begingroup$ Remember that relativity didn't start with Einstein; Galilean relativity handles this condition just fine, and since you can always chose a frame in which this condition applies we see that it is in no way special. $\endgroup$ – dmckee Aug 9 '13 at 20:43

At this stage, does the rocket still accelerate the craft?

If by "velocity of the exhaust" we are talking about its velocity measured in the frame of the rocket, then Yes. Let $\mathbf u$ be the exhaust velocity as measured in the rocket frame, then in free space, the non-relativistic rocket equation is \begin{align} \frac{d\mathbf v}{dt} = \frac{\mathbf u}{M} \frac{dM}{dt} \end{align} where $M(t)$ is the mass of the rocket plus whatever fuel is on board at time $t$ and $\mathbf v$ is the rocket velocity in some inertial frame outside of the rocket. Let's say, for simplicity, that the exhaust velocity is constant, then this equation has solution \begin{align} \mathbf v(t) = \mathbf v(0) - \ln\frac{M(0)}{M(t)}\mathbf u \end{align} The rocket keeps going faster and faster until its fuel is exhausted. In particular, there is nothing preventing the rocket from going faster as measured in the inertial frame than its exhaust as measured in its own frame.

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    $\begingroup$ +1 Think of it as an exchange of momentum. As long as you are pushing mass away from you you will get the opposite force moving your mass. The speed at which you push mass away only helps to give extra acceleration, but you'll always get some acceleration. $\endgroup$ – user6972 Aug 9 '13 at 23:27

The velocity of the rocket is irrelevant. The rocket can be considered as a frame of reference from which the exhaust gases are ejected at a certain speed - therefore the exhaust gases will always move away from the rocket at the same speed. If the rocket keeps producing thrust then it will keep accelerating, and the gases will keep moving away from the rocket at the same speed (relative to the rocket). If the rocket travels faster (relative to an observer) than the exhaust gases are being expelled (relative to the same observer), then both the rocket and its exhaust will be seen (by the observer) to travel in the same direction, albeit the rocket travelling much faster than its exhaust gases.

If the velocity of the rocket is sufficiently close to the speed of light compared to the observer, then special relativity comes into play. Approaching the speed of light means that the rocket would have to be travelling at at least 1,000 miles per second relative to the observer for the classical mechanics to yield relativistic error of just a few thousandths of a percent (if you look at the relativistic component of motion as described by special relativity, you will see the that salient term is the square of the velocity divided by the square of the speed of light, meaning that the velocity has to be very, very large for that term to be of any appreciable size). Of course the rocket might have only just started producing thrust, yet have a velocity close to the speed of light relative to an observer who is travelling close to the speed of light in the opposite direction. That observer would see space-time distorted, causing the the rocket (and its exhaust gas) to have a greater mass, and experience time dilation (time would slow down). Whether those effects were greater for the rocket of the exhaust gas would depend in which direction the observer was travelling (or the rocket travelling relative to them). Meanwhile, to someone aboard the rocket, the exhaust gases would continue to appear to be ejected at the same speed relative to the rocket.


Actually, if you want to see it simply, all the analysis and computation has to be done with respect to the spacecraft. As long as the craft is expelling mass at some speed, hence with some momentum, it gets equivalent momentum in the opposite direction. That is how they can increase their speed (or decrease it if they turn head to tail).

But your questionning is not fully unjustified, in the following sense: this works because the mass expelled from the exhaust was already going at the same speed as the spacecraft. It was carried and accelerated by the spacecraft in the earlier parts of the flight.

This is actually a major problem. Much of the fuel carried by spacecrafts when they leave earth is used only to lift and later accelerate the fuel that will be needed later on, so that the useful payload is actually quite small. The same problem will exist for most reaction crafts: the have first to carry and accelerate with themselves the mass they intend to exhaust later for propulsion.

But there are ways around it. Momentum can be increased either by increasing exhausted mass or by increasing its speed. So the first improvement is to save on mass by increasing speed of exhaust. But designing engines that exhaust matter at very high speed the spacecraft can get momentum with very little mass exhausted. This allows a larger payload as much less reactive mass needs to be used for the same result.

This can be done, for example, with ionic engines that exhaust ions accelerated at very high speed The main drawback of these engines is that they have very low thrust, so the craft gathers speed slowly. But that is often not a problem, except for taking off from Earth or any planet with an atmosphere. When there is no atmosphere, as is the case on the moon, spacecrafts can accelerate to orbital speed almost horizontally (if supported while orbital speed is not attained and centrigugal force is still insufficient). Hence they can be accelerated by ground devices and no longer need powerful engines for takeoff. This however requires large installations that do not exist yet.

Though they are more mass efficient, such spacecrafts still have to carry the reactive mass they will exhaust. And they have to carry a source of energy to accelerate that mass.

  • science-fiction warning -

Solutions have been imagined to avoid these problems. One solution would be to use electromagnetic fields to collect interstellar matter that would then be accelerated as reactive mass for thrust. The problem then is that this matter may not be moving at the same velocity as the spacecraft and collecting it may induce a drag on its velocity. But this is still workable if it is exhausted at a speed that much exceed its speed relative to the craft when collected.

The drag induced by the collecting can be perceived as a head wind. It can probably be reduced or eliminated altogether by simply channelling this wind through the spacecraft exhaust acceleration system so that it simply gets out faster than it came in, without ever moving in the direction of the craft. That is typically the functionning mode of ramjet engines used in some aircraft. The problem with ramjet is that they only work when the craft is already moving at sufficient speed.

The next step is also to find in space the energy source that is use to accelerate the reaction mass. The best know proposal is the Bussard Ramjet.

Strangely, one interesting source of ideas for powering motion might be with bacteria who extract, in many diverse way, the energy and the reactive mass they need from the surrounding medium. The problems are of course quite different technically, but still ... (just my own wild perception of it).


The rocket gases move in opposite direction of the rocket. This means they will never have equal velocities, the gases will have negative velocity and the rocket positive velocity.

The gases will always propel the rocket indepent of their speed in relation to the rocket and, actually, the speed of the gases will almost always remain constant in relation to the rocket, as their fuel is in the rocket and in increasing speed.

The rocket will be propelled until runned out of fuel.


A rocket engine works at its very best when the rocket is travelling at the same speed as the exhaust gases.

The efficiency of the rocket engine varies with the speed of the rocket. When the rocket is standing still, efficiency is minimal as most of the energy is used to keep the exhaust going at high velocity. On the other hand, if the rocket is going much faster than the exhaust speed, the again, the efficiency is low as the exhaust now ends up moving in the same direction as the rocket.

The optimum efficiency is when the rocket is going at the same speed as the exhaust gases. In this case, the gas ends up stationary, and all its energy is transferred to the rocket

  • $\begingroup$ Completely wrong. This answer fails to understand that even in Galilean relativity there is no preferred frame for simple mechanics (it is the introduction of a constant speed of light that requires Einsteinian relativity). All frames are equally valid and the rocket has the same efficiency without reference to how fast it may appear to be going to an outside observer. $\endgroup$ – dmckee Aug 10 '13 at 21:11

This is no rocket science...

You see a frictionless railroad wagon passing by at speed $V$. A passenger on the wagon throws a ball at speed $u$ (relative to the train) in the opposite direction of travel. Will the speed of the wagon increase when $u \le V$?

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To arrive at an answer, all we need is momentum conservation. Let's call the mass of the train plus passenger $M$, and the mass of the ball $m$. Before throwing the ball, total momentum is $(M +m)\ V$. After throwing the ball, the total momentum equals the momentum $m(V-u)$ of the ball, plus the momentum $M(V+v)$ of the wagon with passenger (here, $v$ donates the unknown speed increase). Momentum conservation leads to the condition

$$M(V+v) \ + \ m(V-u) \ = \ (M + m)\ V$$

This can be re-arranged into

$$M\ v \ = \ m \ u$$

The speed $V$ has dropped out of this equation, and we have arrived at the momentum conservation condition as observed from the reference frame moving with the initial velocity of the train.

So the speed $V$ is irrelevant, and the value of ratio $u/V$ doesn't matter: as long as $u$ is nonzero, the wagon will increase its speed.

Change $\ M\ v \ = \ m \ u\ $ into $\ M\ dv \ = \ -dM \ u\ $ and we refer to it as rocket science. Yet, it's nothing more than Newton's law of momentum conservation.


If you mean that the spacecraft velocity and the velocity (of gases) at the exhaust are both measured with respect to, say, Earth and are equal, then the rocket will not accelerate the craft, but this is not easy to imagine: the gases at the exhaust usually have relatively high temperature, so they typically move away from the spacecraft, so their (average) velocity with respect to the spacecraft is not zero, so spacecraft velocity and the velocity (of gases) at the exhaust with respect to Earth are not equal.


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