Is this rocket equation right?

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  • #1
heyhey281
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Homework Statement:
A rocket with the initial mass M = Mp + Mf (payload & fuel) rests on a platform and is
ready for launch. It has an engine which ejects a variable propellant mass always with the same
velocity w > 0 downwards, in the moving reference frame of the rocket. Because the propellant
is ejected, the total mass of the rocket changes with time. The rocket can
can only move vertically and is in the homogeneous gravitational field of the Earth.
Relevant Equations:
https://imgur.com/ARae8EH
I tried to use conservation of momentum

I only want to know if the equation I came up with is correct and if not, why?
 

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  • #2
kuruman
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Can you explain the logic behind the equation using words and not symbols? The left-hand side is obviously the rocket's momentum at time ##t##. What are the terms on the right-hand side and why is their sum equal to the instantaneous momentum of the rocket? Also, is ##\xi## a variable or a constant and what does it stand for?
 
  • #3
haruspex
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The layout of the equation is confusing because it reverses time. More usual is "momentum at time (t+δt) = (momentum at time t) + …."
I think you have a sign error wrt the first g on the right. You can check that by considering the trivial case of no propulsion.
There should be no integral in the equation. You can assume your δt is sufficiently short that the rate of ejection of mass is constant for that interval.
 
  • #4
heyhey281
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The layout of the equation is confusing because it reverses time. More usual is "momentum at time (t+δt) = (momentum at time t) + …."
I think you have a sign error wrt the first g on the right. You can check that by considering the trivial case of no propulsion.
There should be no integral in the equation. You can assume your δt is sufficiently short that the rate of ejection of mass is constant for that interval.
My reasoning was this:

After a certain time ξ, the momentum is the rocket times its velocity and the mass times its velocity, which is ejected downward.

At the time t + ξ, however, one must take into account that mass which was ejected at the time t has a higher velocity than particles which are ejected at t + ξ, since the gravitational force had ξ time to accelerate them downward. So I use an integral to multiply all the tiny particles by their velocity and thus calculate the total momentum of the mass particles.

Why is my formula wrong (apart from my g sign as you said)? I know that it certainly makes more sense for calculations not to write an integral to omit the t + ξ, but I don't really see why it should be wrong.
 
  • #5
heyhey281
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Can you explain the logic behind the equation using words and not symbols? The left-hand side is obviously the rocket's momentum at time ##t##. What are the terms on the right-hand side and why is their sum equal to the instantaneous momentum of the rocket? Also, is ##\xi## a variable or a constant and what does it stand for?
Yes sure.

ξ is a constant in the equation and is an arbitrary time step. Think of it as Δt. (Δt > 0)

After a certain time ξ, the momentum is the rocket times its velocity and the mass times its velocity, which is ejected downward.

At the time t + ξ, however, one must take into account that mass which was ejected at the time t has a higher velocity than particles which are ejected at t + ξ, since the gravitational force had ξ time to accelerate them downward. So I use an integral to multiply all the tiny particles by their velocity and thus calculate the total momentum of the mass particles.
 
  • #6
haruspex
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Why is my formula wrong (apart from my g sign as you said)?
First, it is just unnecessarily complicated.
You can easily get rid of the integral. First, substitute ##t+\xi-\tilde t ## for each occurrence of ##\tilde t##:
##\int_{\tilde t=\xi}^0 (\tilde tg+v+w)\dot m.(-d\tilde t)=\int_{\tilde t=0}^{\xi} (\tilde tg+v+w)\dot m.d\tilde t##.
Since ##\dot m## is effectively constant over the interval ξ,:
##=\dot m\int_{\tilde t=0}^{\xi} (\tilde tg+v+w).d\tilde t##.
The g term integrates to ##\frac 12\xi^2g##, which is a second order small quantity and can be discarded.
v is also constant within the integral.
Look at what remains and consider whether the signs make sense. (I am not saying they do not.)
 
  • #7
heyhey281
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First, it is just unnecessarily complicated.
You can easily get rid of the integral. First, substitute ##t+\xi-\tilde t ## for each occurrence of ##\tilde t##:
##\int_{\tilde t=\xi}^0 (\tilde tg+v+w)\dot m.(-d\tilde t)=\int_{\tilde t=0}^{\xi} (\tilde tg+v+w)\dot m.d\tilde t##.
Since ##\dot m## is effectively constant over the interval ξ,:
##=\dot m\int_{\tilde t=0}^{\xi} (\tilde tg+v+w).d\tilde t##.
The g term integrates to ##\frac 12\xi^2g##, which is a second order small quantity and can be discarded.
v is also constant within the integral.
Look at what remains and consider whether the signs make sense. (I am not saying they do not.)
but why is ##\dot m## effectively constant over the interval ξ? I never said that ξ is an infinitesimal time interval, it could be for example 5 seconds?
 
  • #8
TSny
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It's really important to know your sign conventions. In particular, which of the following are positive and which are negative quantities: ##g##, ##w##, and ##\dot m(\tilde t)##?

Most textbooks take ##g## to be a positive number (the magnitude of the acceleration of gravity).
If ##m(t)## is the mass of the rocket and fuel inside the rocket, then ##\dot m## would be negative.
You call ##w## the speed that the fuel particles get thrown out. Speed is usually considered positive. So, the velocity of a fuel particle relative to the ground just after the time ##\tilde t## that it is ejected would be ##v(\tilde t) - w##.
Anyway, you can see how important it is to clearly define the signs of your symbols.

The integral in your equation looks OK to me except for possible sign issues and I think ##v(t)## inside the integral should be ##v(\tilde t)##.

The general principle is that the final momentum ##P_f## of the system should equal the initial momentum of the system ##P_i## plus the impulse ##J_{ext}## due to external forces: $$P_f = P_i + J_{ext}$$ Rearrange as $$P_i = P_f - J_{ext}$$The impulse in this case is due to the force of gravity, so ##J_{ext} = -M_{sys} g \xi##, where ##\xi## is the time interval and ##g## is taken to be positive. The mass of the system is the mass ##m(t)## of the rocket and fuel at the initial time ##t##. So ##J_{ext} = -m(t)g \xi##. Then, $$P_i = P_f + m(t) g \xi.$$ The initial momentum of the system is ##P_i = m(t)v(t)## and the final momentum of the system is ##P_f = m(t+\xi)v(t+\xi) +\int_t^{t+\xi}## , where ##\int_t^{t+\xi}## is your integral (with possible corrections) that represents the momentum of the ejected fuel.

So, $$m(t)v(t) = m(t+\xi)v(t+\xi) + \left(\int_t^{t+\xi} \right) + m(t) g \xi$$

Now, I'm not sure what the use of this equation would be.
 
  • #9
haruspex
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but why is ##\dot m## effectively constant over the interval ξ? I never said that ξ is an infinitesimal time interval, it could be for example 5 seconds?
Then I misunderstood. I assumed you were trying to obtain a differential equation.
Anyway, treating ##\xi## as infinitesimal will produce a DE which you can sanity check.
 

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