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Ultimately, the factor limiting the maximum speed of a rocket is:

  1. the amount of fuel it carries
  2. the speed of ejection of the gases
  3. the mass of the rocket
  4. the length of the rocket

This was a multiple-choice question in a test I've recently taken. The answer was (1), however, is this disputable, for if we assume that this rocket can potentially achieve relativistic speeds, what implications would this present to the limiting factor on maximum speed?

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Where did that quote come from? It doesn't look like a multiple choice question, since all of those factors except the length affect the rocket's speed. –  David Z May 1 '11 at 16:02
@David: It's a multiple-choice question from an exam we've already taken. In what way would option b) and c) affect the rocket's maximum speed? –  Phillip Wang May 1 '11 at 16:08
A rocket which ejects its exhaust faster will acquire more momentum per unit of fuel, and a rocket which is less massive (not counting the fuel to be used) will move faster for a given amount of momentum. This should make sense intuitively but it's backed up by the "rocket equation" $v(t) = v(0) + v_\text{exhaust}\ln\frac{m(0)}{m(t)}$ which can be derived from the law of momentum conservation. Notice that both the exhaust speed and the rocket mass enter the equation. –  David Z May 1 '11 at 16:23
@David, the length affects the volume and hence the mass of the rocket, as well the amount of fuel it can carry. I'd say all 4 affect the maximum speed. –  John McVirgo May 1 '11 at 17:35
@John: typically yes, but one can build two rockets of different length but carrying the same amount of fuel and having the same mass, and in that case the difference in length would be irrelevant. So I wouldn't say that the length (directly) affects the maximum speed. On the other hand, a difference in mass or fuel payload will change the maximum speed, even if other factors are fixed. Basically I'm saying that $\partial v_\text{max}/\partial(\text{length}) = 0$, but that is not the case for the other factors. –  David Z May 1 '11 at 17:40
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5 Answers

Let us take the rocket to start with an initial mass of $M_{0}$ and to eject mass at a rate $\frac{dm}{dt}$, so after a time $t$, it has ejected a total mass $m$. Furthermore, let the ejection speed be given by $v_{e}$ for simplicity. Then, using the formula for relativistic momentum, we have the momentum gained by the rocket equalling the momentum of the expelled rocket fuel (we also assume the rocket starts from rest):

$$\begin{align*} \frac{dm}{dt}v_{e} &=\frac{d}{dt}\left(\frac{mv}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\right)\\ m\,v_{e} &=\frac{(M_{0}-m)v}{\sqrt{1-\left(\frac{v}{c}\right)^{2}}}\\ \frac{m^{2}v_{e}^{2}}{(M_{0}-m)^{2}}&= \frac{v^{2}}{1-\left(\frac{v}{c}\right)^{2}} \end{align*}$$

After some algebra, you get the final result,


The final speed of the rocket will then be given as a function of the mass $M_{0}-m$ of the rocket itself, the mass $m$ of the fuel expelled, and the velocity $v_{e}$ at which the fuel was expelled. Therefore, the correct answer is definitely everything but the length of the rocket.

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Looking at this again, I should be using $v_{e}\rightarrow \frac{v_{e}}{\sqrt{1-\left(\frac{v_{e}}{c}\right)^{2}}}$, but the general form of everything doesn't change. –  Jerry Schirmer May 1 '11 at 22:29
I'm quite curious about this answer and I don't have a good grasp on your math. Why would you integrate over $dt$? That doesn't make any sense to me. The rate of propellant ejection doesn't matter, so why wouldn't you use $dv$ in it's place ($v$ being rocket velocity). On top of that, does this consider the change in mass of the propellant upon ejection? If not it would be possible to travel with finite propellant mass but infinite energy to expel it with, which is also prohibited by the laws of relativity. I would like to see a reliable solution for $v_e<<c$ and $m<<M$. Tricky tricky. –  AlanSE Jun 3 '11 at 1:21
@Zassounotsukushi: you integrate over dt because you have two time derivatives related to each other and want to solve for velocity. And the rate of propellant ejection is important, because it tells you the rate at which you expel momentum from the system. The whole thing is in terms of rest mass, so don't worry about the "relativstic mass" of the propellant. finally, check the final result--the final velocity is bounded to be less than $c$, with equality reached only if $M_{0}-m=0$, meaning that the rocket's final state is massless. You can just expand the above solution in a series –  Jerry Schirmer Jun 3 '11 at 19:31
in $\frac{v_{e}}{c}$ if you want to see the case of $v_{e}\ll c$ –  Jerry Schirmer Jun 3 '11 at 20:18
I don't think you understand the scope of your equation. It almost exactly matches the classical solution over the entire range. That would be $v=v_e \ln{(M-m)/m}$, yes your equation is different, but it's not relativistic at all (starting at like $v>0.2c$) and misses the mark from any decent relativistic solution, see my answer for what I think those are. –  AlanSE Jun 3 '11 at 22:17
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Take the initial rocket mass to be $M$ and the cumulative quantity of propellant ejected to be $m$, which will be used interchangeably as a dynamic variable and for the final answer. Unit-wise, I will use exclusively $\beta=v/c$ for representation of velocities of all types. I am intentionally avoiding using time in my equations. The essence of the problem is that the velocity depends on the quantity of propellant ejected up until that point (and problem parameters), so I'm going to be finding a solution for $\beta(m)$, that is, the speed of the craft as a function of the mass ejected up until that point. I'm going to consider the problem in the "rest" reference frame, which is more formally defined by the inertial reference for which it is true that $v(0)=0$. I'll use the shorthand of the $\gamma$ function, $\gamma(\beta) = (1-\beta^2)^{-1/2}$.

The first real physics I'll do here is to dissect a specific reaction. I define the specific point-in-time I'm talking about as the point when the moving rocket has mass of $M-m+dm$. It then ejects $dm$ at speed $\beta_e$ relative to itself in the direction opposite of motion. This ejection increases the rocket's momentum by $dp$ as the ejected mass' momentum changes from $p_{dm}$ to $p_{dm}'$ at speeds of $\beta$ and $\beta_{dm}'$. Relativistic velocity addition must be done to obtain the latter.

$$\beta_{dm}' = \frac{\beta-\beta_{e}}{1-\beta \beta_e}$$

Now the reaction momentum balance is written.

$$dp = p_{dm} - p_{dm}' = dm \gamma(\beta) \beta - dm \gamma(\beta_{dm}') \beta_{dm}'$$ $$\frac{dp}{dm} = \gamma(\beta) \beta - \gamma(\beta_{dm}') \beta_{dm}' = \frac{\beta}{\sqrt{1-\beta^2}} - \frac{\beta-\beta_e}{1-\beta \beta_e} \frac{1}{\sqrt{1-\left(\frac{\beta-\beta_e}{1-\beta \beta_e}\right)^2}}$$

Now I hope that by this point it's blatantly obvious what I'm doing. I'm attempting to formulate a differential equation where $m$ is the independent variable and we solve for $\beta(m)$. But we still need the left hand side of that equation. In order to find this we must continue thinking about the balances of the stated interaction and find the change in momentum of the non-ejected part of the rocket, $dp$, after that approximations can be made. The momentum of the non-ejected part of the rocket is $p$ before the ejection and $p'$ after the ejection.

$$dp = p' - p = (M-m) (\beta' \gamma(\beta') - \beta \gamma(\beta))$$ $$\beta'-\beta = d\beta << \beta_e$$ $$dp = (M-m) ((\beta+d\beta) \gamma(\beta+d\beta) - \beta \gamma(\beta))$$

2nd order series expansion about $d\beta=0$.

$$dp = (M-m) d\beta \gamma(\beta)^3$$

Alternatively, this can be found by differentiating. The reason simple differentiating is so hard is that you have to identify what it is you take the derivative of. In order to be consistent with the physics of the situation I had to introduce a special $m''$ variable, which is an invariant form of $m$, although having the same value. Basically, $m''$ is not affected by the loss of $dm$ but $m$ is. Starting from here I have to start writing $\beta$ in terms of $\beta(m)$ as well, which is the objective. Pardon the sudden change in notation. Here is the calculus approach to $dp$.

$$\beta = \beta(m)$$ $$\frac{dp}{dm} = \frac{d}{dm} \left( (M-m'') \beta(m) \gamma(\beta(m)) \right) = (M-m'') \frac{d\beta}{dm} \gamma(\beta)^3 $$

Either approach gives the needed expression for the next step, which is to simply write the differential equation that is the solution to the problem. Pardon the switch back to suppressing the $m$ dependence again (so it fits on the line), just know that it's really $\beta(m)$ and that $\beta_e$ is constant.

$$(M-m) \frac{d\beta}{dm} \gamma(\beta)^3 = \frac{\beta}{\sqrt{1-\beta^2}} - \frac{\beta-\beta_e}{1-\beta \beta_e} \frac{1}{\sqrt{1-\left(\frac{\beta-\beta_e}{1-\beta \beta_e}\right)^2}}$$ $$\beta(0)=0$$

And we're done. This is your answer. With $M$ and $\beta_e$ specified you can find $\beta(m)$ which is the speed of the rocket as a function of the mass ejected, but remember that $m<M$. I'll give a sample plot. This is showing the function of $\beta$, which again is the fraction of the speed of light the rocket is going.

$\beta_e=0.1$ and $M=1$ Rocket Beta as function of propellent

There are some approximations you can get, of course. Doing a 2nd order taylor expand of the RHS of the diff equation above will result in the following solution.

$$\beta = tanh \left( \beta_e ln(\frac{M}{M-m}) \right)$$

And if you simplify even further ($tanh(x)=x$) you'll get the classical version.

$$\beta = \beta_e ln(\frac{M}{M-m}) $$

The first of these seems to be a fairly good approximation, but only for $\beta_e<<1$. Obviously $\beta_e$ and $m$ are relevant to the answer, unless you made some different assumptions I did. I find it most likely, however, that whoever argued the question only has 1 answer out of a, b, c does not have a coherent argument for it.

Edit: had $\sqrt{2}$ factor incorrectly because I typed in an equation wrong. Fixed it now, consistent with Wikipedia for 1st approximation.

When $\beta=1$

In this case, even the very first equation written for $\beta_{dm}'$ is invalid, and we must return to the drawing board for calculating $dp$. I'll have to approach this considering the emission of a single photon, so my notation will be that $dp$ and $dm$ refer to the change in momentum and mass of the rocket according to the stationary observer due to the emission of a single photon. The frequency of the light according to the spacecraft will be $\lambda_e$ and in stationary frame, $\lambda_o$.

$$dp = \frac{h}{\lambda_o}$$

$$p c = \frac{h c}{\lambda_o} = E = c^2 dm \rightarrow dm = \frac{h}{c \lambda_o}$$

$$\frac{dp}{dm} = c$$

Turns out we don't need to do anything with redshift, or even need to know the light frequency! I actually expected that, so it's ok. My previous $dp$ by the way, was the wrong units. I should have written $p=c \beta \gamma(\beta)$, but if you're find with the answer being in terms of $\beta$, then why bother? So I'll just fudge it so that $dp/dm=1$. Now we can take the previous expression for that quantity and set it equal to 1 to find the DE for $\beta(m)$.

$$\frac{dp}{dm} = (M-m) \frac{d\beta}{dm} \gamma(\beta)^3 = 1$$

$$\beta(0) = 0$$

The solution:

$$ \beta = ln{ \frac{M}{M-m} } \sqrt{ \frac{1}{ 1+ ln{ \frac{M}{M-m} }^2 } }$$

I'll plot this with all the others discussed.

M = 1.0, beta_e = 0.5

All plotted together

  • Zassou - My answer in the form of the complete diff equation. Numerically I was only able to evaluate up to $m=0.8$
  • Wikipedia - the tanh( ln( .. )) solution that is also in the wikipedia article for this, it's valid for $\beta_e<<1$, and shows some deviation from the actual here because of it, and when $\beta_e$ gets close to 1 you can see a lot more bending of the curve
  • Jerry - formula he gave in his answer
  • Newton - obviously the classical case, flies off to super-luminary speeds of course
  • Photon - equation I just gave - note this DOESN'T solve the same problem since $\beta_e$ is different, which is evidenced by the initial slope
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@Zassounotsukushi: What you get at $\beta_e=1$? –  Martin Gales Jun 6 '11 at 8:59
@Martin Obviously you get division by zero in the forms here. I added in a solution for photon emission, which is fairly trivial. However, I categorically failed in every attempt to limit take a limit for $\beta_e$ going to 1. I don't know why, I just couldn't get that to work. –  AlanSE Jun 6 '11 at 20:18
@Zassounotsukushi: This means both of your solutions are flawed, i think. –  Martin Gales Jun 7 '11 at 9:44
@Martin It means that I couldn't find a good enough series expansion of $dp/dm$ about $\beta_e=1$ to finish solving the problem. Your statement is implying that the $tan(ln(..))$ solution is also wrong, which is what Wikipedia, as well as literature on the subject reports. And that equation is only valid for $m/M\ll1$ for the case of photons, just like any case where $\beta_e<<1$ is not true. So please, share with us specifically which equations you think are wrong. –  AlanSE Jun 7 '11 at 16:25
@Zassounotsukushi: Consider RSH of your diff.eq: $\frac{\beta}{\sqrt{1-\beta^2}} - \frac{\beta-\beta_e}{1-\beta \beta_e} \frac{1}{\sqrt{1-\left(\frac{\beta-\beta_e}{1-\beta \beta_e}\right)^2}}$ At the point $\beta=\beta_e$ the structure of RHS becomes totally different, equal to:$\frac{\beta}{\sqrt{1-\beta^2}}$. How do you explain this discontinuity? –  Martin Gales Jun 8 '11 at 8:12
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I am not sure whether I have understood the question clearly.

Firstly you need a HUGE fuel resource in order to attain anything even remotely comparable to the speed of light which is practically almost an impossibility for a practical size rocket.

We may calculate the velocity of the rocket assuming that the gas is ejected at a constant rate and at constant velocity.

If we consider the rate of ejection is constant = $\alpha$ at constant velocity then we have $(m_0 - \alpha \times t) \times \frac {dv}{dt} - \alpha{v_0} = - (m_0 - \alpha{t})g$ Solving it we find,

$v = -gt + v_0ln(\frac{m_0}{m_0 - \alpha t})$

For a general case where rate of ejection is not constant we have to know the exact dependence of mass of gas with time and solve the equation,

$m\frac{dv}{dt} + v_0 \frac{dm}{dt} = F$

Second, as usual in S.R., as the rocket approaches the speed of light w.r.t. a stationary inertial observer, it needs more and more energy to accelerate further. It requires an infinite amount of energy to attain the exact speed of light which is the limiting speed for any physical object.

$E = \frac {E_0}{\sqrt {1- v^2/c^2}}$

$E_0$ is the rest mass energy of the rocket and $E$ is the required energy to attain the speed $v$, $c$ is the speed of light. It is clear that as $v$ approaches $c$, $E$ approaches $\infty$

So, ultimately it is the law of nature which will limit the maximum speed attainable by the rocket (or any other physical object).

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Thanks for writing. I agree that this would require a massive fuel resource, but out of the multiple-choice options listed, which feature of the rocket itself (mass, speed of ejection etc.) would limit the maximum speed attainable? –  Phillip Wang May 1 '11 at 16:15
Except the length of the rocket all options are valid to different degrees. –  user1355 May 1 '11 at 16:30
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Both answers 1 and 2 are the correct answers. It appears your teacher made some mental leaps (and expected you to come along) to narrow it down to answer #1.

Ignoring gravitational and relativistic considerations (and for practical rockets in deep space, this is reasonable), and assuming an ideal rocket engine, the final rocket velocity can be expressed as:

Pv = Ev x ln((Pm+Em)/Pm)


Pv = Empty rocket (Payload) Velocity

Pm = Empty rocket (Payload) mass

Ev = Exhaust velocity of rocket engine gases.

Em = Total Ejected propellant (fuel + oxidizer) mass

"ln(x)" means "take the natural logarithm of x."

I think I saw something approximating this form in sb1's answer.

From the eqn, you can see that final payload velocity is linearly proportional to exhaust velocity, and very non-linearly proportional to Em/Pm.

The first mental leap you teacher wanted was for you to realize that there are practical limitations on Ev. If you want to increase the speed of a modern rocket that is pushing the envelope of state of the art science, by 45%, you can't just go out and grab a propellent/rocket combo that increases Ev by 45%. We don't have that.

The second mental leap you teacher wanted was for you to realize that (s)he meant "amount of fuel" as a percentage of total rocket mass. As you increase this percentage you have more leverage to increase final payload velocity.

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Welcome to Physics.SE. The Stack Exchange engine is a question and answer system in which the answers do not have an intrinsic order. Writing "answers 1 and 2" is not particularly helpful. You might consider referring them by their authorship and/or adding links. –  dmckee May 5 '11 at 0:33
@dmckee Do you mean add links to another answer? I would like to hear how to do that. –  AlanSE Jun 3 '11 at 22:33
@Zass: The same way you add links to anything else. To get the url you need, note the "link" button at the bottom of each answer. –  dmckee Jun 4 '11 at 0:02
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Because there is no correct answer for a relativistic case i decided to submit an answer which is correct, in my opinion.

We start from the rocket motion relation which is correct for both, relativistic and non-relativistic, cases:

$$\frac{d}{dt}M\overrightarrow v=\overrightarrow u\frac{dM}{dt};\overrightarrow u= \overrightarrow {u'}+\overrightarrow v$$

where $M$ is the total mass of a rocket(including fuel), $\overrightarrow v$ is the velocity vector of the rocket and $\overrightarrow u$ is the velocity vector of gas jet. $\overrightarrow {u'}$ is velocity vector of the ejected mass with respect to the missile.

$\overrightarrow v$ and $\overrightarrow u$ are taken relative to the inertial coordinate system, which deals with the motion (rather than relative with respect to the missile).

In a relativistic case we have:

$$M=\frac{M'}{\sqrt{1-\frac{v^2}{c^2}}}$$ where $M'$ is a variable rest mass of the rocket in the the moving coordinate system attached to the rocket. After substituting and skipping the math, the reativistic motion equation displays as

$$\frac{M'}{\sqrt{1-\frac{v^2}{c^2}}}\frac{d\overrightarrow v}{dt}=(\overrightarrow u- \overrightarrow v)\frac{d}{dt}\left (\frac{M'}{\sqrt{1-\frac{v^2}{c^2}}}\right )$$

Let assume that the acceleration occurs in the positive direction of the $x$-axis. Then the last equation becomes:

$$\frac{M'}{\sqrt{1-\frac{v^2}{c^2}}}\frac{dv}{dt}=(u_x-v)\frac{d}{dt}\left (\frac{M'}{\sqrt{1-\frac{v^2}{c^2}}}\right )$$ By the reativistic law of velocity-addition we have:

$$u'_x=\frac{u_x-v}{1-\frac{vu_x}{c^2}}$$ where $u'_x$ is velocity of the ejected mass with respect to the missile.

After substituting and skipping the math we have:

$$\frac{dM'}{M'}=-\frac{1}{u'}\frac{dv}{1-\frac{v^2}{c^2}}$$ Here we took $u'_x=-u'$

After integrating we have finally:

$$\frac{M'}{M'_0}=\left(\frac{c-v}{c+v}\right)^{\frac{c}{2u'}}$$ where $M'_0$ is total mass of the rocket at rest ($v=0$).

For a photon rocket case it is sufficient the substitution $u'=c$ here.

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