From Q7 on Pg.22 of "Upgrade Your Physics" by BPhO/Machacek

A rocket of initial mass $M_0$ is being launched vertically in a uniform gravitational field of strength $g$.

(a) Calculate the final velocity of the rocket 90 % of whose launch mass is propellant, with a constant exhaust velocity $u$. Assume that the propellant is consumed evenly over one minute.


Let $\alpha$ denote the fuel consumption in $\mathrm{kg\ s^{-1}}$

Then the constant thrust provided by the exhaust is given by: $$T=\alpha u \tag{1}$$

The acceleration $a(t)$ of the rocket at some time $t$ after the launch: $$T-M(t)g=M(t)a(t) \tag{2}$$

where $$M(t)=M_0-\alpha t \tag{3}$$ is the mass of the rocket at time $t$.

Using $v=\int a(t)\,\mathrm dt $, I got $$v(t)=\int\limits_0^t\left(\frac{\alpha u}{M_0-\alpha t}-g\right)\,\mathrm dt=u\ln\left(\frac{M_0}{M_0-\alpha t}\right)-gt \tag{4}$$ since $v_0=0$.

Can $\alpha$ and $t$ somehow be eliminated or do I need more information to answer the question? Any conceptual errors in my working?

Later on, the question also asks for the velocity at main engine cut-off and the greatest height reached (which I think can be obtained by integrating eq. $(4)$ but the notion of time is again needed here?).


2 Answers 2


As you say :

$$M(t) = M_o - \alpha t$$

But you must know that at time $\tau$ after start the object now has $0.1M_o$ mass (since it consumed all of it's fule).


$$0.1M_o = M_o - \alpha \tau \tag 1$$ $$\Rightarrow \tau = \frac {0.9M_o}{\alpha} \tag 2$$

After substituting $(1)$ and $(2)$ into your equation we get:

$$ v(\tau) = u\ln \left(10 \right)-g\frac {0.9M_o}{\alpha}$$

Now if you know $\alpha$ then you can find $v(\tau)$.

  • $\begingroup$ What about $\alpha$? is there way to eliminate $\alpha$? (Just like in the horizontal case it gets cancelled out, but I am not sure whether it's applicable to a vertical launch?) $\endgroup$ Commented Jan 22, 2020 at 13:42
  • 2
    $\begingroup$ In vertical launch, it does not get cancelled out, as it is needed for fighting gravity. Every second the engine burns, part of its thrust is cancelled by gravity. The larger the thrust, the lesser the fraction. As long as you ignore air resistance and structural loads, shorter burns with higher thrust are better. You can rearrange the equation to be V(tau) = deltaV - gravity losses $\endgroup$
    – AI0867
    Commented Mar 19, 2020 at 9:25

Based on a very nice answer from @Johan Liebert, you can extrapolate several things. First, upper bound of rocket speed is when $\alpha \to \infty$, this gives : $$ v_{max} = u\,\ln(10) $$, because second term approaches zero then.

Second - you can calculate critical fuel consumption, solving for $\alpha$ in : $$ u\,\ln(10) - g \frac{0.9 M_o}{\alpha} = 0 $$ this gives : $$ \alpha_{\text{critical}} = g\frac{0.9\,M_o}{u \ln(10)} $$ If $\alpha < \alpha_{\text{critical}}$, then rocket finally will start to fallback to Earth, i.e. gravity force will win in the end.


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