# Conservation of momentum, rocket

When we analyze a rocket using conservation of momentum we neglect gravity and air drag. We then consider rate of fuel consumption or exhaust rate and by applying the law of conservation of momentum we find this equation - $RV = Ma$, where $R$ is the mass rate of fuel consumption, $V$ is velocity of exhaust w.r.t rocket.

So my question is as follows: Is the mass rate considered here for only that mass which contributes solely in the velocity of the rocket and not in overcoming gravity and air drag? (Because they are neglected, to apply the conservation of momentum, so that system becomes isolated.)

I've also uploaded a screenshot of the derivation of above equation from Halliday, Resnick, Walker - fundamentals of physics: • -1. The text which you provide states explicitly in the first sentence of "Finding the Acceleration" that the rocket accelerate(s) through deep space with no gravitational or atmospheric drag forces acting on it. So your question appears to be answered by the textbook. Aug 1 '17 at 14:55

They are ignoring air and drag. This means it doesn't factor into the calculations at all.

There is no force required to overcome the drag and gravity if they were not considered when forming the relationships. This means that in a real scenario that velocity or fuel consumption rate would have to increase to gain the same acceleration as in this idealized scenario (if the real scenario had gravity and drag, your edit shows they are talking about deep space, where the effects of either would be insignificant).

So my question is as follows; Is the mass rate considered here for only that mass which contributes solely in the velocity of the rocket and not in overcoming gravity and air drag?

The answer is yes as the system is closed and isolated. It becomes more complicated when you take into account air resistance and the gravitational force acting on the rocket from the earth.

This simplification is considered only for rockets in a vacuum (a theoretical ideal).

By the way I have the 8th edition of "Fundamentals of Physics" and have read the derivation in my book. It looks like you have a later version of the book but the derivations seem pretty much identical.

• Yeah, the screenshot is from 9th edition but I also have a 10th edition book. And the derivation is same in all of them, because I guess it's a pretty elementary thing. I think they differ in some questions and sample problems ! Feb 18 '17 at 1:06
• And yes there are changes in quantum physics chapters, Gauss' law and electric potential. Feb 18 '17 at 1:33
• Yeah I thought there would be. For the quantum physics part I had buy the 'extended edition' just to view that section :-( Feb 18 '17 at 1:38
• In quantum chapters, the discussion on Schrodinger's equation is expanded including reflection of matter waves from step potential. And also a module of Planck's blackbody radiation is added! Feb 18 '17 at 2:48

The important thing to realise is that in the rocket equation derivation a system has been defined as the rocket, its fuel and the combustion products.
The system has a constant mass but part of the mass of the system starts from being fuel and ends up as combustion products.

The rate at which fuel is being converted into combustion products is $\dfrac {dM}{dt}$.
It is that process which accelerates the rocket forwards and the fuel as combustion products backwards.

The derivation uses Newton's second law: sum of external force acting on the system = rate of change of momentum of the system.

In the derivation that you have shown there are no external force and so the rate of change momentum of the system is zero although parts of the system do change their velocity, $F_{\rm external} = 0 = \dfrac{\Delta p}{\Delta t}$.
From this you can get an equation for the acceleration of the rocket.

This is probably the most difficult thing to comprehend but in some ways it is no different from the example of a gun with a bullet inside it starting from rest (initial momentum = 0) and then when the gun is fired if there are no external forces the gun and the bullet move in opposite directions and the final momentum of the gun and the bullet is still zero.

If there are external forces all you do is add them to the left hand side of the equation $F_{\rm external} = 0 = \dfrac{\Delta p}{\Delta t}$ and solve it as appropriate.
As you pointed out those forces could be the weight of the rocket and/or the viscous drag on the rocket.
This is what will probably be done in the next few pages of your textbook and you will find that keeping the same rate of conversion of fuel to combustion products will result in a smaller acceleration of the rocket.