The Vertical Launch of a Rocket

Question

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.

Attempt:

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?).

Answer 1

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).

Therfore

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

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 $v(\tau)$.

Answer 2

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.