How do you solve the problem of a rocket with variable mass?

  • #1
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Homework Statement


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!

Homework Equations


I know that ##F=m(\frac {dp} {dt})=ma=m\frac {dv} {dt}##

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!
 

Answers and Replies

  • #2
andrewkirk
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As soon as I read the problem, the figure of 2.1 m/s for the gas velocity struck me as ridiculously low. Rockets accelerate really fast, so the velocity of the gases they emit must be several times faster than that. I suspect the figure is supposed to be 2.1 km/s.

A quick check with wiki seems to corroborate this. This article says the exhaust gases travel at hypersonic velocities, which apparently means velocities of Mach 5 and above. 2.1 km/s is about Mach 7.

2.1m/s is a slow walk, which is not going to send a big cylinder of metal anywhere!

EDIT: I forgot to put the link to the article on rocket engines. I've put it in now.
 
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  • #3
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As soon as I read the problem, the figure of 2.1 m/s for the gas velocity struck me as ridiculously low. Rockets accelerate really fast, so the velocity of the gases they emit must be several times faster than that. I suspect the figure is supposed to be 2.1 km/s.

A quick check with wiki seems to corroborate this. This article says the exhaust gases travel at hypersonic velocities, which apparently means velocities of Mach 5 and above. 2.1 km/s is about Mach 7.

2.1m/s is a slow walk, which is not going to send a big cylinder of metal anywhere!
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?

Thanks for the reply!
 
  • #4
andrewkirk
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Do you think that, even when considering that v' is relative to the rocket
Yes, and that co-moving frame of the rocket is the natural one to use. Note that that is the same as the Launchpad frame when the countdown hits zero.

2.1 m/s is a waft, not a rocket blast. It would be overstating it to call it a gentle breeze.
 
  • #5
Ray Vickson
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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?

Thanks for the reply!
Around where I now live there a lot of elderly people who zip along in their motorized wheelchairs faster than 2.1 m/s.
 
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