Ultimately, the factor limiting the maximum speed of a rocket is:
- the amount of fuel it carries
- the speed of ejection of the gases
- the mass of the rocket
- 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?
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,
$$v=\frac{v_{e}}{\sqrt{\left(\frac{M_{0}-m}{m}\right)^2+\left(\frac{v_{e}}{c}\right)^{2}}}$$
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.
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 $M=1$
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
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).
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)
where:
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.
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.
