How to calculate velocity of infinite-stage rocket?

[zenterix]

Homework Statement

Assume that it is impossible to build a structurally sound container that can hold fuel of more than, say, nine times its mass. It would then seem like the limit for the speed of a rocket is $u\ln{10}$. How can you build a rocket faster than this?

Relevant Equations

$F_{ext}=0=m_r(t)v_r'(t)-m_r'(t)u$

My answer to the question is: build a two-stage rocket. Or a $k$-stage rocket. Then I thought: what happens if we try to make $k=\infty$?

To cut to the chase, my question is how to calculate the infinite series

$$\lim\limits_{k\to\infty} \sum\limits_{i=1}^k \ln{\left ( \frac{10}{10-\frac{9}{k-(i-1)}} \right )}$$

I will explain in what follows how this series comes about.

To set this up, let's think about the base case which has a single rocket and no external forces at play.

The rocket has mass $m_r(t)=m_{r,dry}(t)+m_f(t)$ where $m_{r,dry}$ is the mass of the dry rocket (with no fuel in it) and $m_f(t)$ is the mass of the fuel. Fuel is burned and exhausted with a velocity of $\vec{u}=-u\hat{i}$ relative to the rocket's frame. The rocket has velocity $\vec{v}_r(t)=v_r(t)\hat{i}$ relative to a "fixed frame".

Since everything happens in one dimension we drop the vector notation.

By assumption, $m_{r,dry}=\frac{1}{10}m_r(0)=\frac{m_{r,dry}+m_f(0)}{10}\implies m_f(0)=9m_{r,dry}$.

That is, the rocket is composed of one part dry mass and nine parts of fuel. This is the same as just ten times dry mass.

The rocket equation is

$$F_{ext}=0=m_r(t)v_r'(t)-m_r'(t)u$$

which has solution

$$v_r(t)=v_r(0)+u\ln{\frac{m_r(0)}{m_r(t)}}$$

Thus, for a single rocket, when all the fuel has been used we have

$$v_r(t_f)=u\ln{\frac{10m_{r,dry}}{m_{r,dry}}}=u\ln{10}$$

I did the calculations for a two-stage rocket and a k-stage rocket. I will mostly provide the results of the calculations not the calculations themselves.

Let $m_{r,dry}$ be the dry weight in the case of a single-stage rocket. When we increase the number of rockets, we will keep the total weight equal to the case of the single-stage rocket (ie, $10m_{r,dry}$) so we can make a fair comparison between final velocities.

For a 2-stage rocket we have two rockets each with mass $5m_{r,dry}$ and composed of $0.5m_{r,dry}$ of dry mass and $4.5m_{r,dry}$ of fuel.

After the first stage is over (that is, after we've burned all the fuel from the first rocket), we have

$$v_r(t_1)=u \ln{\frac{10}{10-\frac{9}{2-0}}}=u\ln{\frac{10}{5.5}}$$

and after the second stage we have

$$v_r(t_2)=u\ln{\frac{10}{10-\frac{9}{2-0}}}+u\ln{\frac{10}{10-\frac{9}{2-1}}}=u\ln{\frac{10}{5.5}}+u\ln{10}$$

For a k-stage rocket we have a generalization of the pattern we could have imagined from the two-stage case. The final speed after $m$ stages have been used is

$$v_r(t_m)=\sum\limits_{i=1}^m u\ln{\left ( \frac{10}{10-\frac{9}{k-(i-1)}} \right )}$$

After all the stages are done then $m=k$ so

$$v_r(t_k)=\sum\limits_{i=1}^k u\ln{\left ( \frac{10}{10-\frac{9}{k-(i-1)}} \right )}$$

Note that the $k$-th term is $u\ln{10}$. This is the additional speed gained by the final part of the rocket (which has dry mass of only $\frac{m_{r,dry}}{k}$.

My question is how do we calculate

$$\lim\limits_{k\to\infty} v_r(t_k)$$

$$=\lim\limits_{k\to\infty} \sum\limits_{i=1}^k u\ln{\left ( \frac{10}{10-\frac{9}{k-(i-1)}} \right )}$$

 

[haruspex]

Doesn't it at least simplify to $-u\Sigma_{i=1}^{\infty}\ln(1-\frac{0.9}{i})$?
Then I think that could be written as the log of the limit of the ratio of two gamma functions.

 

[Filip Larsen]

I haven't read your derivation, but would just like to comment that for an infinite-stage rocket it seems that some fixed fraction of the propellant mass is also ejected with zero relative speed as spend staging all the time, thus if the the ejected impulse is the same as for the unstaged rocket but the total mass loss per time is higher then the effective ejection speed must be less, i.e. $\dot m v_e = (\dot m + \dot m_s) v_s$, where $v_e$ and $v_s$ is the unstaged and infinitely-staged ejection speed respectively, $m$ is the propellant mass rate and $m_s$ is the "staging" mass rate. Thus, using $v_s$ instead of $v_e$ in the standard rocket equation should give the delta-V. To compare the two delta-Vs's the final payload mass in the unstaged rocket equation should of course include all the mass that is considered staging for the infinitely-staged rocket.