# How rocket engines work? Newton's Third Law of Motion:

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#### nec208

##### Guest
I'm trying to understand how rocket engines work.

Quote And Newton's Third Law of Motion: for every action there is an equal and opposite reaction Quote

I really do not understand this.And doing a search of how rocket engines work I get different views pressure ,pushing on aire ,or a force of force.

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Newton's third says that for every action their is an equal yet opposite reaction. Rockets work by pushing air molecules (basically applying a huge force on them) the air molecules respond by propelling the rocket upward.
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A rocket engine exerts force all around it.
since the back is open that force blows out the back and most of it's force is absorbed by the air. Meanwhile in the opposite direction that force has no opposite force and moves like a balloon you blow up and let go.
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Combustion chamber has large internal pressure. Hole in back of combustion chamber, with de Laval nozzle, lets gas escape at high velocity, going supersonic. Rocket pushes on gas, accelerating it backward. By Newton's third law, gas must therefore push forward on rocket, accelerating it forward.

Hence rocket goes forward.
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Newton's third law is very simple. Fill a shopping cart. Now put on roller skates. Stand behind the cart and push. The cart will go one way and you will go backwards. If you filled the cart well, it will move slowly forward while you move quickly backward. The key concept is that while you exerted a force on the cart, it also exerted a force on you in return.

Rockets do this. As the exhaust gas leaves, heading backward at tremendous velocity, the rocket (considered part of the same momentum system) moves forward to "conserve" the momentum.

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Rocket exhaust does not push against anything in order to achieve thrust??
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Is all 5 wrong?

M

#### Mee_n_Mac

##### Guest
nec208":13ky1ea3 said:
I'm trying to understand how rocket engines work.

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Rocket exhaust does not push against anything in order to achieve thrust??
==========

Is all 5 wrong?

If by #5 you mean the part I've left above then no. Rocket exhaust does not have to push or hit anything to make the rocket go. It's enough for it to be exhausted out of the rocket motor. A better analogy than the ones given would be to imagine you're sitting on some very very very slippery ice. You have stuck in your pockets a lot of heavy balls. What happens when you throw a ball, as fast as you can, towards the South ? Well you slide / move towards the North. If you can imagine throwing a lot of balls, very fast and one right after another, then the balls are like the rocket's exhaust and you are the rocket motor. The balls don't have to hit anything for you to move just like the rocket exhaust doesn't.

K

#### kyle_baron

##### Guest
Put a brass nozzle at the end of a garden hose, and turn on the water. This is as about as simple an explanation, as I could think of.

M

#### MeteorWayne

##### Guest
kyle_baron":2lacze1n said:
Put a brass nozzle at the end of a garden hose, and turn on the water. This is as about as simple an explanation, as I could think of.

Hidden in there is a great demonstration as well. Without the nozzle, the water comes out slowly, and there is almost no back force on your hand. Put the nozzle on, and even though the amount of water is the same (or even less, due to the restriction) the squirting forces your hand backwards. So with the same amount of water, the encreased velocity of the ejection creates more force. This explains the efficiency of ion and other electric engines.

D

#### DrRocket

##### Guest
nec208":28ar4e3n said:
I'm trying to understand how rocket engines work.

Quote And Newton's Third Law of Motion: for every action there is an equal and opposite reaction Quote

I really do not understand this.And doing a search of how rocket engines work I get different views pressure ,pushing on aire ,or a force of force.

==========
Newton's third says that for every action their is an equal yet opposite reaction. Rockets work by pushing air molecules (basically applying a huge force on them) the air molecules respond by propelling the rocket upward.
==========
A rocket engine exerts force all around it.
since the back is open that force blows out the back and most of it's force is absorbed by the air. Meanwhile in the opposite direction that force has no opposite force and moves like a balloon you blow up and let go.
==========
Combustion chamber has large internal pressure. Hole in back of combustion chamber, with de Laval nozzle, lets gas escape at high velocity, going supersonic. Rocket pushes on gas, accelerating it backward. By Newton's third law, gas must therefore push forward on rocket, accelerating it forward.

Hence rocket goes forward.
==========

Newton's third law is very simple. Fill a shopping cart. Now put on roller skates. Stand behind the cart and push. The cart will go one way and you will go backwards. If you filled the cart well, it will move slowly forward while you move quickly backward. The key concept is that while you exerted a force on the cart, it also exerted a force on you in return.

Rockets do this. As the exhaust gas leaves, heading backward at tremendous velocity, the rocket (considered part of the same momentum system) moves forward to "conserve" the momentum.

==========

Rocket exhaust does not push against anything in order to achieve thrust??
==========

Is all 5 wrong?

Force does not come out of a rocket, mass comes out of a rocket. The mass comes out with a velocity and a direction and mass moving at a given speed in a specific direction has momentum. Rockets work by conservation of momentum, it is as simple as that.

The figure of merit for a rocket engine is called specific impulse. It is the momentum obtained per unit of mass expelled, which is simply the speed of the expelled mass when the rocket operates in a vacuum. Specific impulse, or Isp, is often quoted in units of "seconds" which is technically not correct but is the result of the use of English units and the cancellation of pounds force and pounds mass in the derivation.

You can calculate the speed of a rocket in a vacuum (without gravity) via the "rocket equation", which is, in english units
delta V = gc*Isp*ln(final mass/initial mass).

M

#### Mee_n_Mac

##### Guest
Looking back at the OP I get the sense that 'nec' may be confused by the multiple ways that can be used to analyze the function of a rocket motor. Certain conservation of momentum is the easiest, and to me, the most intuitive of the ways. But he talks about force as well and may be trying to understand how a rocket works from a F=mA standpoint. This, at least in the basics, can also be done.

Imagine a sealed hollow metal sphere or ball. Inside this sphere let's place some combustible material like gunpowder. We ignite and it burns. What happens ? Well for one thing whatever gases inside the sphere get hotter. Also we turn (most) of the powder into gas. The result is that there's more and hotter gas than before the ignition. Naturally the pressure rises as a result. But unless we do something else (and assuming the sphere is strong enough to withstand the presssure) nothing much happens. We have a hot pressured ball. However if we now put a hole in the ball the gas will come out and, like a balloon, the ball will (try to) squirt away. Can this be explained using Newton's 2'nd law ? Yes.

Just prior to puncturing the ball the pressure inside was equal through the interior surface. If one measured the force exerted on any square inch, they would all measure the same. Just as importantly they would all cancel. The force on the interior of the sphere pointing in any one direction is balanced by a force on the opposite side on the interior pointing in the opposite direction. As a result the is no net force and thus no acceleration. This is similar to a tug of war where both sides pull equally hard on the rope. Despite the large forces, they cancel each other and no movement happens. Now we puncture the ball. We remove part of the interior surface, there's a hole. Assuming the pressure inside the ball was greater than the pressure outside the ball the interior gas moves out through the hole. But think of the balance of forces we had just a moment ago. Where the hole is there is a force is pushing out the gases and not on the interior wall of the sphere. Opposite the hole the pressure is still exerting the same force is was just a moment ago (perhaps a bit smaller) but this is no longer balanced. Now there's a net force pushing on the sphere directly opposite the hole. This force causes the sphere to move in that direction which is of course just what we see. We can also realize that a small hole means only a small differential in force, a larger hole = more force = larger acceleration (though we run out of gas sooner) We can also realize that more pressure means more force which means larger acceleration.

So 'nec' there's more than 1 way to "see" how a rocket motor works. At least I hope I've explained well enough for you to see.

D

#### DrRocket

##### Guest
Mee_n_Mac":9wq7be8r said:
Looking back at the OP I get the sense that 'nec' may be confused by the multiple ways that can be used to analyze the function of a rocket motor. Certain conservation of momentum is the easiest, and to me, the most intuitive of the ways. But he talks about force as well and may be trying to understand how a rocket works from a F=mA standpoint. This, at least in the basics, can also be done.

Imagine a sealed hollow metal sphere or ball. Inside this sphere let's place some combustible material like gunpowder. We ignite and it burns. What happens ? Well for one thing whatever gases inside the sphere get hotter. Also we turn (most) of the powder into gas. The result is that there's more and hotter gas than before the ignition. Naturally the pressure rises as a result. But unless we do something else (and assuming the sphere is strong enough to withstand the presssure) nothing much happens. We have a hot pressured ball. However if we now put a hole in the ball the gas will come out and, like a balloon, the ball will (try to) squirt away. Can this be explained using Newton's 2'nd law ? Yes.

Just prior to puncturing the ball the pressure inside was equal through the interior surface. If one measured the force exerted on any square inch, they would all measure the same. Just as importantly they would all cancel. The force on the interior of the sphere pointing in any one direction is balanced by a force on the opposite side on the interior pointing in the opposite direction. As a result the is no net force and thus no acceleration. This is similar to a tug of war where both sides pull equally hard on the rope. Despite the large forces, they cancel each other and no movement happens. Now we puncture the ball. We remove part of the interior surface, there's a hole. Assuming the pressure inside the ball was greater than the pressure outside the ball the interior gas moves out through the hole. But think of the balance of forces we had just a moment ago. Where the hole is there is a force is pushing out the gases and not on the interior wall of the sphere. Opposite the hole the pressure is still exerting the same force is was just a moment ago (perhaps a bit smaller) but this is no longer balanced. Now there's a net force pushing on the sphere directly opposite the hole. This force causes the sphere to move in that direction which is of course just what we see. We can also realize that a small hole means only a small differential in force, a larger hole = more force = larger acceleration (though we run out of gas sooner) We can also realize that more pressure means more force which means larger acceleration.

So 'nec' there's more than 1 way to "see" how a rocket motor works. At least I hope I've explained well enough for you to see.

What you say here is quite true. You can in principle determine the force on a rocket by evaluating the surface integral of pressure over the surface of the rocket. In your example of the closed vessel the integral is zero. In a normal rocket the integral would be the chambe pressure over the projected area of the nozzle throat plus the integral of the spatially varying pressure over the projected area of the nozzle exit cone.

What one misses by taking that perspective, even though itis perfectly valid, is insight into the thermodynamics of the rocket. In particular you miss the importance of specific impulse Isp which can be examined further to determine the thermodynamics of rocket propellant and fluid dynamics of nozzle design that result in maximum performance.

One thing that is important however, is that F=ma DOES NOT hold. Newton's second law is actually F=dp/dt where p is momentum. For a system of constant mass this reduces to F=ma since p=mv, where m is mass and v is velocity. Rockets are not constant mass systems, and F=ma simply does not apply. You also have to be careful about the reference frame in which you work, since rockets are usually accelerating and a reference frame attached the rocket is not a Newtonian reference frame. I have had professionals show confusion on this point. But if you do the math in detail, apply Newton's second law in its original form and use a valid inertial reference frame then everything works out.

D

#### drwayne

##### Guest
"One thing that is important however, is that F=ma DOES NOT hold. Newton's second law is actually F=dp/dt where p is momentum. For a system of constant mass this reduces to F=ma since p=mv, where m is mass and v is velocity. Rockets are not constant mass systems, and F=ma simply does not apply."

Bingo! Important point to be sure. I even saw an ad in a space trade many years ago that that said something
like:

"There are two things you can count on in rocketry"

and the first one was "F = MA"

I used to drive my advisor crazy when I was prepping for my orals by doing the "Car on the banked track"
problem in a frame at rest with respect to the car - which is of course a non-interial frame - and one got into
pseudo-forces...

Wayne

D

#### DrRocket

##### Guest
drwayne":iyxh7nis said:
"One thing that is important however, is that F=ma DOES NOT hold. Newton's second law is actually F=dp/dt where p is momentum. For a system of constant mass this reduces to F=ma since p=mv, where m is mass and v is velocity. Rockets are not constant mass systems, and F=ma simply does not apply."

Bingo! Important point to be sure. I even saw an ad in a space trade many years ago that that said something
like:

"There are two things you can count on in rocketry"

and the first one was "F = MA"

I used to drive my advisor crazy when I was prepping for my orals by doing the "Car on the banked track"
problem in a frame at rest with respect to the car - which is of course a non-interial frame - and one got into
pseudo-forces...

Wayne

Now that you mention it I remember that ad, and I think I even saw it as a poster. I think it was from Rocketdyne. It drove me nuts. Rockets to which F=ma applies, rockets of constant mass and zero acceleration, are called FAILURES.

U

#### undidly

##### Guest
DrRocket":2w6h7zys said:
drwayne":2w6h7zys said:
"One thing that is important however, is that F=ma DOES NOT hold. Newton's second law is actually F=dp/dt where p is momentum. For a system of constant mass this reduces to F=ma since p=mv, where m is mass and v is velocity. Rockets are not constant mass systems, and F=ma simply does not apply."

Bingo! Important point to be sure. I even saw an ad in a space trade many years ago that that said something
like:

"There are two things you can count on in rocketry"

and the first one was "F = MA"

I used to drive my advisor crazy when I was prepping for my orals by doing the "Car on the banked track"
problem in a frame at rest with respect to the car - which is of course a non-interial frame - and one got into
pseudo-forces...

Wayne

Now that you mention it I remember that ad, and I think I even saw it as a poster. I think it was from Rocketdyne. It drove me nuts. Rockets to which F=ma applies, rockets of constant mass and zero acceleration, are called FAILURES.
DrRocket":2w6h7zys said:
drwayne":2w6h7zys said:
"One thing that is important however, is that F=ma DOES NOT hold. Newton's second law is actually F=dp/dt where p is momentum. For a system of constant mass this reduces to F=ma since p=mv, where m is mass and v is velocity. Rockets are not constant mass systems, and F=ma simply does not apply."

Bingo! Important point to be sure. I even saw an ad in a space trade many years ago that that said something
like:

"There are two things you can count on in rocketry"

and the first one was "F = MA"

I used to drive my advisor crazy when I was prepping for my orals by doing the "Car on the banked track"
problem in a frame at rest with respect to the car - which is of course a non-interial frame - and one got into
pseudo-forces...

Wayne

Now that you mention it I remember that ad, and I think I even saw it as a poster. I think it was from Rocketdyne. It drove me nuts. Rockets to which F=ma applies, rockets of constant mass and zero acceleration, are called FAILURES.

F=MA is correct.
So what if M changes?.
Does it not apply to a large rocket with large mass and also to a small rocket with small mass and all masses between?.

D

#### DrRocket

##### Guest
undidly":1ny6i6ct said:
F=MA is correct.
So what if M changes?.
Does it not apply to a large rocket with large mass and also to a small rocket with small mass and all masses between?.

No, sonny, F=ma is not correct, and never was. Newton did not state his second law as F=ma but rather as F=dp/dt (he actually used somewhat different notation and different words, calling what we call momentum the "quantity of motion" but that was the meaning as you can read for yourself in the Principia if you wish).

F=dp/dt = d(mv)/dt=m dv/dt + v dm/dt which if m is constant reduces to F=m dv/dt =ma. But it is applies to neither a large nor a small rocket motor so long as the mass is changing, and with rockets as fuel is expelled the mass always changes.

You ought not be making such positive assertions when you don't know what in the hell you are talking about.

D

#### drwayne

##### Guest
Not to mention that for many rockets, the dM/dt is in fact quite significant. (The shuttle for example empties
its tank with extreme rapidity)

Wayne

D

#### DrRocket

##### Guest
drwayne":n6bz4htz said:
Not to mention that for many rockets, the dM/dt is in fact quite significant. (The shuttle for example empties
its tank with extreme rapidity)

Wayne

The solids on the shuttle each have about 1 million pounds of propellant. That propellant is burned up in about 2 minutes.

The rocket equation might lead one to think that the velocity increment is independent of burn rate. It is, in the absence of gravity and air resistance, which is neglected in the derivation of that equation. However, if you consider the effects of gravity and if the trajectory is essentially vertical (and if you continue to neglect air resistance) it is easy to see that the most efficient burn is an impulsive burn -- burn all of the propellant immediately. This is because by burning the propellant quickly you don't have to take the losses associated with lifting the propellant against the force of gravity. Of course that results in high g loads, and if taken to extremes for manned flight, would result in some rather short squat astronauts.

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