B Finding the mass of gas ejected by a rocket

[marksyncm]

I ran into this problem and need some guidance. Here's the question exactly as it appears in my sheet:

"A 500 kg rocket is set for vertical take-off. If the exhaust speed is 1,000 m/s, the mass of gas ejected per second to supply the thrust needed to overcome the weight of the rocket is:" (multiple choices follow)

The answer to this is supposed to be the following:

F = ma = m * \frac{dv}{dt} = 500 * 9.81
m * \frac{1000-0}{dt} = 500 * 9.81
\frac{m}{dt} = 4.9 Kg /s

What I do not understand is how we are able to do this only knowing the velocity of the exhaust and without knowing the acceleration of the exhaust. For all we know, the exhaust could have been accelerated to 1,000 m/s over the course of 0.0001 second, or 0.00423 seconds. Isn't it the acceleration, not velocity, which determines how much force was imparted to the gas, and hence how much force was imparted to the rocket?

What am I missing?

Thank you.

 

[jbriggs444]

What I do not understand is how we are able to do this only knowing the velocity of the exhaust and without knowing the acceleration of the exhaust. For all we know, the exhaust could have been accelerated to 1,000 m/s over the course of 0.0001 second, or 0.00423 seconds. Isn't it the acceleration, not velocity, which determines how much force was imparted to the gas, and hence how much force was imparted to the rocket?

If the acceleration of the exhaust is lower, that means that there must be more gas in the process of acceleration, somewhere in the nozzle. If the acceleration of the exhaust is higher, that means that the exhaust spends less time in the nozzle.

The net force produced by the acceleration of the gas currently in the nozzle is the same either way.

A better way to look at it is to ignore force entirely and concentrate on the rate at which momentum leaves the business end of the nozzle. Conservation of momentum will then lead to the result.

 

[NFuller]

We know that in some time interval $dt$, the fuel starts at rest and accelerates to $1000\text{m/s}$. The acceleration is then
$$a=\frac{1000\text{m/s}-0\text{m/s}}{dt}$$
This expression for the acceleration is related to the force as shown in your work.
$$F=ma=m\frac{1000\text{m/s}-0\text{m/s}}{dt}$$
If you knew the time interval $dt$ then you could find the acceleration but that is not what the problem asks for. Instead it just wants the rate of the mass ejected from the rocket. This rate can be written as
$$\frac{m}{dt}=\frac{F}{1000\text{m/s}}$$
Since the force must equal the force of gravity, this gives
$$\frac{m}{dt}=\frac{Mg}{1000\text{m/s}}=\frac{500\text{kg}\cdot9.81\text{m/s}^{2}}{1000\text{m/s}}=4.9\text{kg/s}$$

 

[marksyncm]

We know that in some time interval $dt$, the fuel starts at rest and accelerates to $1000\text{m/s}$. The acceleration is then
$$a=\frac{1000\text{m/s}-0\text{m/s}}{dt}$$
This expression for the acceleration is related to the force as shown in your work.
$$F=ma=m\frac{1000\text{m/s}-0\text{m/s}}{dt}$$
If you knew the time interval $dt$ then you could find the acceleration but that is not what the problem asks for. Instead it just wants the rate of the mass ejected from the rocket. This rate can be written as
$$\frac{m}{dt}=\frac{F}{1000\text{m/s}}$$
Since the force must equal the force of gravity, this gives
$$\frac{m}{dt}=\frac{Mg}{1000\text{m/s}}=\frac{500\text{kg}\cdot9.81\text{m/s}^{2}}{1000\text{m/s}}=4.9\text{kg/s}$$

Thank you very much - by including the units in your calculations, you've made this perfectly clear.

However, I'm still not sure I'm getting this on a more intuitive level (although jbriqqs444's explanation helps some). I'm guessing it's an issue with my conceptual understanding of how rockets work.

The way I was imagining a rocket launch is similar to where I would be sitting in a wheeled office chair and throwing (accelerating) a heavy ball forward in order to move myself and the chair backwards.

I was imagining the rocket launch as a similar phenomena, only instead of the rocket "throwing" (accelerating) a ball forward, it was throwing (accelerating) a “ball of gas” of certain mass downward in order to move itself upward.

The rocket would of course do so repeatedly and in quick succession, i.e. it would accelerate one ball of gas of mass X after another, after another, after another.

This would keep the rocket moving upwards, the same way throwing "infinite" balls while sitting in my wheeled chair would keep my body moving backwards.

Because of this mental model, I imagined that the only way to calculate mass is to know the acceleration of the "ball of gas."

As a result, I thought the only way to solve such a problem would be to answer the question: "what is the smallest mass of a ball of gas which, when accelerated at a certain rate, will have the required minimum force acting on it (and hence the force acting on the rocket).

As such, it appeared to me impossible that this problem could be solved without knowing the acceleration. I see that this is not the case now, the algebra proves it.

But I still don't fully understand it on an intuitive level.

That we can solve for the mass without knowing the acceleration means my mental model of a rocket is missing some key component. Or my understanding of forces and acceleration is missing a key component. Or both.

I think I'm basically having a problem seeing how "rate of ejection of mass per second" helps us determine the forces involved. After all, shouldn't these forces depend on the acceleration of that mass ejected during that second? How do we know that at any point during that one second where 4.9 kilograms of mass was ejected, that the mass was being ejected with enough acceleration to generate the required forces?

Anyone able to spot where the lack in my understanding / limiting belief is? I'm certain that if I don't figure it out, I'll make a similar mass/forces-related mistake in the future.

EDIT:

If the acceleration of the exhaust is lower, that means that there must be more gas in the process of acceleration, somewhere in the nozzle. If the acceleration of the exhaust is higher, that means that the exhaust spends less time in the nozzle.

The net force produced by the acceleration of the gas currently in the nozzle is the same either way.

After giving this a bit more thought, I feel like my sticking point might be related to the bolded part.

I was under the impression that the thrust must be larger than the force of gravity at any given instant, not just cumulatively. This seems impossible, however, because it implies that during an infinitesimally short period of time, when an infinitesimally small amount of gas is being expelled, the acceleration of gas would need to be extreme in order to surpass the force of gravity. In my mind, a mass of X is either being accelerated fast enough to generate the required forces, or it isn't. I'm failing to see how the cumulative net force produced by the acceleration of the gas over one second "adds up" to overcome the force of gravity.

 

[jbriggs444]

I was under the impression that the thrust must be larger than the force of gravity at any given instant, not just cumulatively. This seems impossible, however, because it implies that during an infinitesimally short period of time, when an infinitesimally small amount of gas is being expelled, the acceleration of gas would need to be extreme in order to surpass the force of gravity. I think I'm failing to see how the cumulative net force produced by the acceleration of the gas over one second "adds up" to overcome the force of gravity.

Gravity is distracting and adds nothing to the discussion. It is just a performance constraint that the rocket must surpass in order to hover or take off.

If you are doing a force-based analysis then it is not the infinitesimally small amount of gas that is leaving the nozzle that is providing the propulsion. That gas has already done its work. The force is supplied by the nozzle as it accelerates the gas that is in the nozzle while gas is in still the nozzle. There is pressure in the nozzle. That pressure acts upward, pushing against the nozzle and downward, accelerating the gas.

The volume of gas in the nozzle at any given time is not infinitesimal.

 

[NFuller]

Maybe working backwards will give you some intuition. The force of gravity on the rocket is
$$F=Mg=500\text{kg}\cdot9.8\text{m/s}^{2}=4900\text{N}$$
If the rate of mass ejection is $4.9\text{kg/s}$ then every second the rocket must accelerate $4.9\text{kg}$ of material from rest to $1000\text{m/s}$. The change in momentum during this one second interval is then
$$F=\frac{dp}{dt}=\frac{d}{dt}mv=4.9\text{kg}\frac{1000\text{m/s}}{1\text{s}}=4900\text{N}$$
Which is exactly enough to overcome the force of gravity.

I was under the impression that the thrust must be larger than the force of gravity at any given instant, not just cumulatively. This seems impossible, however, because it implies that during an infinitesimally short period of time, when an infinitesimally small amount of gas is being expelled, the acceleration of gas would need to be extreme in order to surpass the force of gravity.

Notice that even if you consider $2.45\text{kg}$ ejected in $0.5\text{s}$ or any other fraction, you will get the same force from the above equation.

 

[marksyncm]

Gravity is distracting and adds nothing to the discussion. It is just a performance constraint that the rocket must surpass in order to hover or take off.

If you are doing a force-based analysis then it is not the infinitesimally small amount of gas that is leaving the nozzle that is providing the propulsion. That gas has already done its work. The force is supplied by the nozzle as it accelerates the gas that is in the nozzle while gas is in still the nozzle. There is pressure in the nozzle. That pressure acts upward, pushing against the nozzle and downward, accelerating the gas.

The volume of gas in the nozzle at any given time is not infinitesimal.

Thank you, this helped me modify my mental model of how a rocket generates thrust.

Maybe working backwards will give you some intuition. The force of gravity on the rocket is
$$F=Mg=500\text{kg}\cdot9.8\text{m/s}^{2}=4900\text{N}$$
If the rate of mass ejection is $4.9\text{kg/s}$ then every second the rocket must accelerate $4.9\text{kg}$ of material from rest to $1000\text{m/s}$. The change in momentum during this one second interval is then
$$F=\frac{dp}{dt}=\frac{d}{dt}mv=4.9\text{kg}\frac{1000\text{m/s}}{1\text{s}}=4900\text{N}$$
Which is exactly enough to overcome the force of gravity.

Notice that even if you consider $2.45\text{kg}$ ejected in $0.5\text{s}$ or any other fraction, you will get the same force from the above equation.

I think this is exactly what I was looking for. For some reason I was stuck thinking that unless acceleration is known, there's no way to determine anything about mass. This biased my thinking.

Now that you've shown me a way to relate force to change in momentum, I can see where I was mistaken.

Also, for some reason it did not occur to me that we were measuring the speed of the gas outside the nozzle.

This is very important because it implies the velocity of the gas cannot increase beyond 1000m/s as there is no force acting on it any longer.

Once this dawned on me, things started making more sense.

So when we say "the answer is 4.9 kg/s", what we're essentially saying is this: "irrespective of the rate of acceleration, if a certain mass was accelerated to 1,000 m/s over the course of one second by a force of 4900 Newtons, then that mass must have been equal to 4.9 kg."

And like you said, we can assume the acceleration to be anything. For instance, if it were 5000 m/s^2, then 4900 = m*5000 \iff m = 0.98 kg. But because the gas was accelerated to 1000m/s only (because it cannot be accelerated any further as it has left the nozzle), the force will have acted on the 0.98 mass for 1000/5000 = 0.2 of a second. So we end up with 5 * 9.8 = 4.9 kg per second one way or another.

I'm writing this out for my own sake and for anyone else who might find this thread in the future, but please do let me know if it all sounds correct.

And thank you again.