Why do rocket momentum solutions treat mass as constant when deriving?

[M-Speezy]

The concept of a problem is that if rockets continue to jettison fuel, at some point they will get a very small momentum because their mas will be so small. (Bit of a silly problem, there's a lot of mass in the rocket itself) Anyways, the idea is that you would take the derivative of the momentum and find the time at which you should stop ejecting fuel. There are some solutions floating around, but they all say that momentum is m times the common velocity equation (exit velocity times the natural log of the ratio of initial mass to remaining mass). I said that mass was the (initial mass minus the rate of mass loss times time), which gives me a terrible derivative. The solutions I've seen just left mass as m. Is that a reasonable thing to do?

 

[dauto]

What you wrote above makes little to no sense to me. Try actually writing down the equations instead of describing them. That might help it make some sense.

 

[M-Speezy]

Generally, we would say that the there's an equal and opposite force between ejected fuel and the momentary acceleration of a rocket. (This is how they accelerate in space)

Specifically, d(m)/dt*velocity of jettisoned gas = -m*dv/dt

This is rearranged so that it's integral (dm/m)=integral (dv) *1/v(exit). This gives the equation I initally spoke of, that the change of the velocity =V(exit)*Ln(M(initial)/M(final))
If it starts at rest, then that is the velocity.

Momentum is, of course, mass times velocity. The mass is dependent upon time, as it's ejected at some constant rate (we're assuming). Specifically, the mass at any time is M(initial)-d(m)/dt *T, or initial mass minus the rate of mass ejection times time. This expression for mass at some time ought to be substituted in for M(final) in the velocity I gave earlier.
This gives me, finally, momentum for some time: (M(initial)-dm/dt*T)*v(exit)ln(M(initial)/(M(initial)-dm/dt*T))
The derivative for this expression is pretty terrible. The other solution, which I have linked, just left m in its simplest form. My main question is if this is an appropriate thing to do, or not.

http://physics.oregonstate.edu/~jansenh/COURSES/ph435/Homework/Solution_4_3.pdf (The very first problem)

 

[Dale]

If you leave m as a constant then there will be no maximum. The rocket will continue to gain momentum as it burns fuel. So, given your desire to find the maximum momentum and given the fact that the simplification prevents exactly that then I think it is clear that you cannot use the simplification.

 

[M-Speezy]

Nevermind, I realize now what this is. M is the variable that's being used. Rather than doing what I did, finding the specific time where it's maximized, it's easier to find the mass where it's maximized. At that point, it's trivial to find the time. Sorry, I should have been more observant of the solution I saw.

Thanks for the help!

 

[Dale]

Yes, I agree. It is definitely easier to find the maximum wrt mass than wrt time.

 

[willem2]

Yes, I agree. It is definitely easier to find the maximum wrt mass than wrt time.

It's still easier to find the maximum wrt velocity. Once the speed of the rocket is bigger than the exhaust speed, the exhaust will also have momentum in the same direction as the rocket, so the rocket will lose momentum. While the velocity is still smaller than the exhaust speed, the exhaust will have momentum in the opposite direction, so the rocket will gain momentum. The maximum momentum is reached when the velocity of the rocket is equal to the exhaust speed.

 

[Dale]

D'oh!

That is easy.

 

[M-Speezy]

The exhaust velocity is considered to be with respect to the rocket, and assumed constant. The faster the rocket, the higher the exhaust speed would be to a rest frame. Also, the derivative of velocity is the reason why it was an issue, because the velocity has such a complex relationship with both mass and therefore time. The problem is strange because of the added complexity that mass and time are related.