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I'm fairly new to physics, so please forgive me if I lack basic understanding.

I'm wondering how rocket engines actually work. I know the answer of "actio est reactio", but I feel like this is a very abstract and unclear explanation because it doesnt touch whats actually going on. Here is my issue:

As I understand, the engine generates an explosion which throws out gas particles into one direction (say down) and thus creates an inverted force that pushes the rocket into the other direction (say up). What I dont get is: How do the leaving particles interact with the rocket? They dont necessarily touch the rocket, at least not in a way that would make it go up, so how can they possibly have an effect? As far as I know, a force can either act through

  • contact
  • a field

Now I guess the gas particles dont create some sort of force field, so at some point there has to be a 'meaningful' contact.

I get that mathematically the force of gas particles that are coming out equals the force with which the rocket is accelerated, but I have a hard time understanding how particles that are already on their way out of the rocket are the cause of the acceleration.

My explanation would be that the rocket is flying for the same reason anything is 'flying' near an explosion: The explosives hit other objects and drag them with them (conservation of momentum). The same force applies in a rocket on all walls of the engine, and since the particles can escape "down", the only momentum the rocket gets is "up" (left and right cancel each other out). But I run into issues with that explanation as well, because lets say we make the exit of the engine reaaally long, say a few hundred meters (thought experiment), and then close it up. now that means that at first the effect would be the same, rocket goes up, but once the explosion hits the other end of the engine, the rocket will be 'dragged' back again. seems strange to me.

please help me solve this issue. thanks!

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  • $\begingroup$ A good place to start would be to Google Newton's laws of motion. $\endgroup$
    – BillDOe
    Feb 14, 2019 at 21:35
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    $\begingroup$ I know them all, I even paraphrased the third law $\endgroup$
    – Philipp Murry
    Feb 14, 2019 at 21:41
  • $\begingroup$ So the one could paraphrase your question as, "How does a force of action create a force of reaction?" $\endgroup$
    – BillDOe
    Feb 14, 2019 at 21:44
  • $\begingroup$ kind of, but this type of engine specifically raises my eyebrows because I can't follow the breadcrumbs to how the reactio is generated. $\endgroup$
    – Philipp Murry
    Feb 14, 2019 at 21:46
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    $\begingroup$ Re, "the engine generates an explosion." Most people would think of an explosion as an event in which some thing suddenly disintegrates, and the parts of it fly off in different directions. Might be more accurate to say that the distilled essence of a rocket engine is a nozzle from which a high-pressure fluid is allowed to escape. The nature of that fluid and the mechanism or process that pressurizes it are incidental to the discussion. $\endgroup$
    – Solomon Slow
    Oct 29, 2021 at 19:08

3 Answers 3

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"The same force applies in a rocket on all walls of the engine, and since the particles can escape "down", the only momentum the rocket gets is "up""

That's it: the gas molecules collide with the walls of the combustion chamber, but there is a hole in one wall, so the force on that wall is less!

The net force on the rocket is equal and opposite to the momentum per unit time of the escaping molecules, as these don't hit the wall with the hole in it!

"lets say we make the exit of the engine reaaally long, say a few hundred meters (thought experiment), and then close it up. now that means that at first the effect would be the same, rocket goes up,"

Yes.

"but once the explosion hits the other end of the engine, the rocket will be 'dragged' back again."

No, but it will stop accelerating.

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  • $\begingroup$ thank you! so only my thinking with the "long engine" was wrong. I will change the question a little bit $\endgroup$
    – Philipp Murry
    Feb 14, 2019 at 21:47
  • $\begingroup$ Michio Kaku once said on a Discovery channel show (Science channel) that rockets don't work as well in space as in the atmosphere because the exhaust doesn't have any air particles to push against in space. $\endgroup$
    – BillDOe
    Feb 14, 2019 at 21:51
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    $\begingroup$ in this case, kaku doesn't know what he is talking about. $\endgroup$
    – niels nielsen
    Feb 14, 2019 at 21:54
  • $\begingroup$ I was thinking about changing the question, but I'll ask it here: so my issue is that the explanation of "actio est reactio" makes it seem like the 'outcoming' particles generate the force against the rocket, when in fact its actually the ones that going upwards. whatever comes out of the engine is already 'waste' and has no effect on the rocket anymore, no? I'm a bit frustrated/confused why everyone is using the third law here, when its merely seems like a mathematical model $\endgroup$
    – Philipp Murry
    Feb 14, 2019 at 22:07
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Here is another way to think of this problem which might help:

Imagine you are sitting on a skateboard with a cannon in your lap. This is a special cannon, consisting only of a short tube closed at one end. Inside the tube is a bowling ball that is held tightly in position against a squeezed spring sitting between the ball and the closed end. a trigger mechanism holds the system in its "cocked" state.

you point the cannon away from you horizontally, put your feet up on the skateboard, and pull the trigger.

The spring begins to expand violently, pushing the bowling ball backwards out the tube at great speed. But it is also pushing equally hard on the closed end of the tube, which recoils and kicks you right in the chest and thereby pushes you off in the opposite direction taken by the departing bowling ball.

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  • $\begingroup$ I guess my issue with Newtons third law is that it makes it seem like the flying canon ball is actively 'doing something' to me. but the spring could've pushed off a stable surface as well (a strange beton wall that reaches into the tube), without any canon ball flying out of the tube. The third law might be a good way to calculate the forces, but its terrible at explaining the actual physical cause (imo). or maybe its just my bad habit of thinking $\endgroup$
    – Philipp Murry
    Feb 14, 2019 at 22:20
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    $\begingroup$ @Philipp The key idea is conservation of momentum. Whatever direction the exhaust goes in, the centre of mass of the rocket + exhaust cannot move. $\endgroup$
    – PM 2Ring
    Feb 14, 2019 at 23:02
  • $\begingroup$ Yes, conservation of momentum makes much more sense. why not just say that the rocket goes up because particles hit the rocket at the upper side of the engine? this is backed by everyday experience (A hits B, B moves). I dont understand the need for this detour of newtons third law. Saying that the leaving particles are somehow creating an upward force is ontologically incorrect (imo). their force might match the upward force mathematically, but thats all. Maybe I'll spare this for a new question, since I have my issues with Newtons laws and the concept of force in general. $\endgroup$
    – Philipp Murry
    Feb 14, 2019 at 23:22
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How do the leaving particles interact with the rocket? They dont necessarily touch the rocket, at least not in a way that would make it go up, so how can they possibly have an effect?

Consider the following simple situation.

enter image description here

Initially the mass $m$ is travelling at the same speed as the hollow cylinder which is sealed on the left hand side.
The mass $m$ explodes into two equal mass fragments one moving to the left and the other to the right with speed $v$.
The mass moving to the right does not interact with the tube and carries with it momentum $\frac m 2\,v\, \hat i$ and it escapes out of the tube.

The mass moving to the left carries with it momentum $\frac m 2\,v\, (-\hat i)=-\frac m 2\,v\, \hat i$, hits the end of the tube and rebounds with speed $v$ now having momentum $\frac m 2\,v\, \hat i$.
So the change in momentum of this mass is $\frac m 2\,v\, \hat i- (-\frac m 2\,v\, \hat i)= m\,v\, \hat i$.

The change in momentum of the mass which was moving left is equal to minus the change in momentum of the cylinder, $-m\,v\, \hat i$, ie the cylinder is given an impulse to the left thus increasing the cylinders speed to the left.

What happens in a rocket engine is much more complex than this and here is a diagram from the Wikipedia page Rocket engine showing the forces acting on the rocket combustion chamber on the left and a (de Laval) nozzle on the right both of which contribute to the forward propulsion of a rocket.

enter image description here

But I run into issues with that explanation as well, because lets say we make the exit of the engine reaaally long, say a few hundred meters (thought experiment), and then close it up. now that means that at first the effect would be the same, rocket goes up, but once the explosion hits the other end of the engine, the rocket will be 'dragged' back again. seems strange to me.

Consider the cylinder closed at both ends and the "explosion" occurring close to the left hand wall.
The mass moving left transfers momentum to the cylinder which now moves faster to the left.
Then the mass which was initially moving right rebounds from the right hand end of the cylinder and reduces its speed to what it was originally.

Then the mass which originally was moving left and is now moving right hits the right hand end of the cylinder and the cylinder now moves right.

etc

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  • $\begingroup$ Thank you! One more thing: can you explain why a smaller escape whole (choke) will accelerate the rocket faster? lets say there are three particles moving in each direction in your illustration. on the left side all three hit the rocket, giving it their momentum. on the right side though, only one will escape (small hole), and the other two hit the rocket as well, giving it momentum to the other side. so the rocket only gains the total momentum of 1 particle instead of 3, thus it would move faster if the hole was bigger and all three particles could escape to the right. where am I wrong? $\endgroup$
    – Philipp Murry
    Feb 15, 2019 at 10:44
  • $\begingroup$ I started my answer with the sentence "Consider the following simple situation". A real rocket engine is not a simple system. The main aim is to have the expelled gases moving as fast as possible in a direction opposite to that of the motion of the rocket. To that end the throat between the combustion chamber and the nozzle speeds the gases up - same mass per second flowing through a small cross section - and then adiabatic expansion in the nozzle area does the rest. Note that the speed of the gas in the combustion chamber is less than in the throat. $\endgroup$
    – Farcher
    Feb 15, 2019 at 11:03
  • $\begingroup$ but how is the force of the now faster moving particles transferred to the rocket? this is the crux of my inital question. there has to be some contact, no? $\endgroup$
    – Philipp Murry
    Feb 15, 2019 at 11:43
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    $\begingroup$ The gas in the combustion chamber is at high pressure and the force on the sealed side is greater than the force on the throat side. The force acting on the throat accelerates the gases through the throat. The gas molecules “communicate” with each other via collisions. $\endgroup$
    – Farcher
    Feb 15, 2019 at 11:50
  • $\begingroup$ spot on answer, thank you! I feel like Newtons third law describes the amount of the force against the rocket, but not the cause of the force. This confused me for quite a while now. Idk, maybe as I go one I learn to deal with it better and use it more intuitively. $\endgroup$
    – Philipp Murry
    Feb 15, 2019 at 13:05

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