# Rocket with variable mass, how do you solve for ratio of weight of fuel and weight of rocket?

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1. Sep 28, 2016

### Sudo

1. The problem statement, all variables and given/known data
Rockets are propelled by the momentum of the exhaust gases expelled from the tail. Since these gases arise from the reaction of the fuels carried in the rocket, the mass of the rocket is not constant, but decreases as the fuel is expended. Show that for a Rocket starting initially from rest, and taking the velocity of the exhaust gases relative to the rocket, $v'$ = 2.1 m/s and a rate of mass loss per second L = 1/60 of the initial mass, to reach the escape velocity of the Earth ($v_e$ = 11.2 km/s), the ratio of the weight of the fuel to the weight of the rocket must be almost 300!

2. Relevant equations
I know that $F=m(\frac {dp} {dt})=ma=m\frac {dv} {dt}$

3. The attempt at a solution
First, I work in the frame of the rocket; so we have two forces, the weight $W=mg$ and the force of the exhaust exiting the rocket $\frac {dm} {dt}v'$. Both forces are pointing downward, thus I set up the following equation: $$ma=m\frac {dv} {dt}=-v'(\frac {dm} {dt})-mg$$ Now I rewrite $m\frac {dv} {dt}$ as $m\frac {dv} {dm}\frac {dm} {dt}$; now I have this expression:$$m\frac {dv} {dm}\frac {dm} {dt}=-v'(\frac {dm} {dt})-mg$$As $\frac {dm} {dt}=L=-\frac {m_0} {60}$ and dividing both sides by $m$ we have:$$\frac {dv} {dm}(-\frac {m_0} {60})=-\frac {v'} {m}(-\frac {m_0} {60})-g$$Dividing both sides by $L$ yields:$$\frac {dv} {dm}=-\frac {v'} {m}+\frac {60g} {m_0}$$Now we're ready to write our differential equation as follows:$$dv=-\frac {v'} {m}~dm+\frac {60g} {m_0}~dm$$We can now integrate, this knowing that we start from $v_0=0$ and from a certain initial mass given by $m_0=m_{rocket}+m_{fuel}$ and a final mass $m_f=m_{rocket}$$$\int_0^v dv=-v'\int_{m_0}^{m_f} \frac {1} {m} \ dm+\frac {60g} {m_0}\int_{m_0}^{m_f} dm$$After integrating and evaluating I get the following expression:$$v=-v'ln(\frac {m_f} {m_0})+\frac {60g} {m_0}(m_f-m_0)$$This can be rewritten as:$$v=-v'ln(\frac {m_{rocket}} {m_{rocket}+m_{fuel}})-\frac {60g} {m_{rocket}+m_{fuel}}(m_{fuel})$$Then, if $m_{fuel}>>m_{rocket}$ we can ignore the term $m_{rocket}$ in $\frac {1} {m_{rocket}+m_{fuel}}$, and the expression can be rewritten as:$$v=-v'ln(\frac {m_{rocket}} {m_{fuel}})-60g=v'ln(\frac {m_{fuel}} {m_{rocket}})-60g$$Solving for the desired ratio I get:$$\frac {m_{fuel}} {m_{rocket}}=e^{\frac {v+60g} {v'}}$$However when plugging in the equation the values $v'=2.1 m/s$, $v=11.2 km/s$, and $g$ the value is much, much bigger than 300.

What did I did wrong? Did I chooses the incorrect frame? To my understanding my equation $ma=m\frac {dv} {dt}=-v'(\frac {dm} {dt})-mg$ should be true for this frame, thus $v'$ indeed is 2.1 m/s

Thank you for your time! As always I appreciate detailed answers, thanks again!

2. Sep 28, 2016

### TSny

The value given for $v'$ looks too small. Could it actually be 2.1 km/s?

You stated that both the thrust force and the gravitational force act downward. But the rocket goes upward.
However, I think your equations are correct.

Last edited: Sep 28, 2016
3. Sep 29, 2016

### Sudo

I'm not completely sure, the problem states m/s.

Do you think that, even when considering that $v'$ is relative to the rocket (i.e., measured by someone in the rocket, not an observer on ground), 2.1 m/s would still, anyway, be too small?

4. Sep 29, 2016

### jbriggs444

Too small by exactly a factor of 1000. An exhaust velocity so laughably low and a coincidence so striking that it is quite obviously an error in the problem statement.

An arthritic cripple could throw bowling balls out the back end of a rocket faster than 2.1 m/s. 300 such throws would achieve a velocity around 25 miles per hour (*) -- somewhat less than escape velocity.

(*) Assuming that the arthritic cripple starts with a pile of 300 bowling balls on his lap and has the same mass as one of the balls.

Last edited: Sep 29, 2016