# Is there an equal and opposite force from gas movement due to pressure gradient force?

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Let us say you have a tote full of water and you connect an air pump to draw the water using negative pressure generated by the pump and you used a very flexible hose connected to the side of the tote. Would there be an opposite force on the tote moving away from the pump and perhaps even create tension on the flex hose?

Let us say you have a cup full of air and you use a vacuum to remove the air from the cup, would there be an opposite force on the  bottom of the cup?

When wind starts to blow on the back of your head and you feel the wind, why can't you feel the wind push off your face as the air in front of you starts to move away from you?

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It's atmospheric pressure or gravity that is doing the work in first two examples. Air molecules will try to come in on all sides.

0 psig = 14.7 psia

you can go below zero on the guage but still have a positive pressure in absolute units.

You do see such issues when using a positive displacement pump. Those work the opposite way via cramming rather than use the weight of the atmosphere. They'll happily cause piping to burst or collapse.

Edited by Endy0816

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20 hours ago, Endy0816 said:

It's atmospheric pressure or gravity that is doing the work in first two examples. Air molecules will try to come in on all sides.

0 psig = 14.7 psia

you can go below zero on the guage but still have a positive pressure in absolute units.

You do see such issues when using a positive displacement pump. Those work the opposite way via cramming rather than use the weight of the atmosphere. They'll happily cause piping to burst or collapse.

You didn't really answer the question I was asking. Is there an opposite force on the tote or cup?

Let us assume the drop in pressure in the pump and vacuum is close to -14.7 psig for simplicity.

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On 1/17/2020 at 9:16 PM, Gerrard said:

Let us say you have a tote full of water and you connect an air pump to draw the water using negative pressure generated by the pump and you used a very flexible hose connected to the side of the tote. Would there be an opposite force on the tote moving away from the pump and perhaps even create tension on the flex hose?

"If I have a vacuum pump and a tube going to a container, as I pull a vacuum on the container will there be force that moves the container away from the pump?"

If that is not the question, please try again.

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I don't understand your question ( and examples ) either.

Instead of pulling a vacuum on the exit from the container, let's simplify and pressurize the container.
So, if you consider a pressurized balloon, air escaping from a hole will lead to an imbalance in the pressure through the axis passing through the center and you could say that this imbalance in pressure results in motion away from the hole's direction.
That is the simplistic view of a rocket.

Or you could say the momentum of the air escaping the hole has to be balanced by an equivalent momentum imparted to the balloon in the opposite direction.

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1 hour ago, Bufofrog said:

"If I have a vacuum pump and a tube going to a container, as I pull a vacuum on the container will there be force that moves the container away from the pump?"

If that is not the question, please try again.

Yes that is the question. So since you said no, would there be if the tote was pressurized instead of vacuum pulling on it?  Since space is a vacuum, how come there would be an opposite force then?

47 minutes ago, MigL said:

I don't understand your question ( and examples ) either.

Instead of pulling a vacuum on the exit from the container, let's simplify and pressurize the container.
So, if you consider a pressurized balloon, air escaping from a hole will lead to an imbalance in the pressure through the axis passing through the center and you could say that this imbalance in pressure results in motion away from the hole's direction.
That is the simplistic view of a rocket.

Or you could say the momentum of the air escaping the hole has to be balanced by an equivalent momentum imparted to the balloon in the opposite direction.

When vacuum is "pulling", the container is pressurized in comparison. Either way, the movement is due to pressure differential. Why would there be no opposite force if vacuum is pulling and there would be force when the container is pressurized. Either way the contents is moving toward negative pressure.

Edited by Gerrard

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If the contents are moving in one direction, conservation of momentum demands the container move in the opposite direction.

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9 hours ago, Gerrard said:

Yes that is the question. So since you said no, would there be if the tote was pressurized instead of vacuum pulling on it?  Since space is a vacuum, how come there would be an opposite force then?

If there is a tote that you close with a cap then you would have a tote that is filled with air.  Inside the tote the air pressure would be 14.7 psi (lbs/in^2).  Out side the tote the air pressure would be 14.7 psi.  This means that inside and outside would have 14.7 lbs of force on each square inch of the tote container.  If I pull a vacuum and lower the internal pressure to 12.7 psi, that would mean there would be a 2 lb force on each inch of the tote pushing in on the surface.  If the tote was glass nothing would happen, there would simply be a constant force of 2 psi on the glass.  If the tote was a pliable plastic the force would push in the walls of the tote.  The tote would be compressed in size until the pressure inside the tote was raised to 14.7 psi and then it would be in equilibrium and it would stop compressing.  Since the forces are felt on all sides of the tote there would be no force in a particular direction, so the tote would not move.

The same thing would happen if instead of a vacuum, if I were to pressurize the tote I would see the same sort of thing, the forces would be uniform in all directions on the inside of the tote.

If I were to put a hole in the side of the pressurized tote then the escaping air jet would put a force on the tote that would tend to make it move in the direction away from the streaming air.

Gerrard, I thought about your question a bit more and I think I see where you were going.  If I were to have a extremely high pressure water in a container and a hose going to a tote and I threw a valve so the water rushed to the container, that would cause a force that would make the container shoot away from the hose entrance due to the pressure gradient in the tote.  This is what water hammer is.  I have seen some very large pipes 'dancing' due to someone opening a valve too quickly.

Edited by Bufofrog

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10 hours ago, MigL said:

If the contents are moving in one direction, conservation of momentum demands the container move in the opposite direction.

The contents are moving in one direction regardless of whether vacuum is drawing or the container is pressurized. So there should be movement. But Bufofrog (Senior Member) states that there would be no movement if vacuum were drawing. Is that true?

Also conservation of momentum definition is "Conservation of momentum is a fundamental law of physics which states that the momentum of a system is constant if there are no external forces acting on the system. It is embodied in Newton's first law (the law of inertia)."

So when a pressure differential exists, pressure gradient force exists. Wouldn't this be the force causing change in momentum? If so then force on the container is not necessary in order to conserve momentum?

https://www.shodor.org/os411/courses/_master/tools/calculators/pgf/index.html   Introduction : This calculator calculates the pressure gradient force (PGF) per mass, in units of km-s-2. Mathematically, we use the sign (+ or -) to signify direction: a positive value indicates from location 1 to location 2; a negative sign indicates from location 2 to location 1. The pressure gradient force moves air from areas of high pressure (H) to areas of low pressure (L).

In this calculator, you have three input values:

1. the distance (in km) of the two locations, or centers of high and low pressures
2. the pressure (in kPa) at the first location (the area of low pressure)
3. the pressure (in kPa) at the second location (the area of high pressure)

The algorithm for pressure gradient force per mass of air (in the x-direction) is:

2 hours ago, Bufofrog said:

If there is a tote that you close with a cap then you would have a tote that is filled with air.  Inside the tote the air pressure would be 14.7 psi (lbs/in^2).  Out side the tote the air pressure would be 14.7 psi.  This means that inside and outside would have 14.7 lbs of force on each square inch of the tote container.  If I pull a vacuum and lower the internal pressure to 12.7 psi, that would mean there would be a 2 lb force on each inch of the tote pushing in on the surface.  If the tote was glass nothing would happen, there would simply be a constant force of 2 psi on the glass.  If the tote was a pliable plastic the force would push in the walls of the tote.  The tote would be compressed in size until the pressure inside the tote was raised to 14.7 psi and then it would be in equilibrium and it would stop compressing.  Since the forces are felt on all sides of the tote there would be no force in a particular direction, so the tote would not move.

The same thing would happen if instead of a vacuum, if I were to pressurize the tote I would see the same sort of thing, the forces would be uniform in all directions on the inside of the tote.

If I were to put a hole in the side of the pressurized tote then the escaping air jet would put a force on the tote that would tend to make it move in the direction away from the streaming air.

Gerrard, I thought about your question a bit more and I think I see where you were going.  If I were to have a extremely high pressure water in a container and a hose going to a tote and I threw a valve so the water rushed to the container, that would cause a force that would make the container shoot away from the hose entrance due to the pressure gradient in the tote.  This is what water hammer is.  I have seen some very large pipes 'dancing' due to someone opening a valve too quickly.

If the container was pressurized to lets say 20 psi, the pressure external to the atmosphere would be 14.7 psi and thus negative pressure. (Basically a weak vacuum). So if you open the hole on the container the pressure in the container would also decrease, lets say to 16 psi. Based on your explanation, wouldn't the force on walls of the tote be 4 psi in a glass tote and compression in plastic tote?

Could it be that the release of extremely high pressure into the atmosphere creates a propeller like effect? See Below. Let's say the tote (pressure rated tote of 2000 psi) was pressurized to 1000 psi. You fully open the valve on the back of the tote and now there is higher than atmospheric pressure immediately behind the tote, perhaps that pressure pushes the tote forward as the pressure moves to low pressure in front of the tote?

"Think of a propeller as a spinning wing. Like a wing, it produces lift, but in a forward direction—a force we refer to as thrust. Its rotary motion through the air creates a difference in air pressure between the front and back surfaces of its blades. In order for a propeller blade to spin, it usually needs the help of an engine. "

Also if the extreme pressure was being released into a vast vacuum, would it create pressure behind the tote or would the pressure keep dissipating in one direction as there are no molecules to create resistance?

Joules-Thomson states that no work is done when the external pressure is 0. But if the external pressure is greater than 0, then work is done. Does this apply to the above scenario?

Edited by Gerrard

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12 hours ago, Gerrard said:

When vacuum is "pulling", the container is pressurized in comparison. Either way, the movement is due to pressure differential. Why would there be no opposite force if vacuum is pulling and there would be force when the container is pressurized. Either way the contents is moving toward negative pressure.

That movement requires a similar force (I think you get that) in either case, excess pressure escaping or vacuum pulling ambient pressure.

The difference is that with the vacuum you are creating a pressure less than ambient in the container and over the area where the vacuum is being pulled, so the net force will be toward the vacuum in that case, where it will be net force in the opposite direction in the case of the contents escaping due to excess or above ambient pressure.

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59 minutes ago, J.C.MacSwell said:

That movement requires a similar force (I think you get that) in either case, excess pressure escaping or vacuum pulling ambient pressure.

The difference is that with the vacuum you are creating a pressure less than ambient in the container and over the area where the vacuum is being pulled, so the net force will be toward the vacuum in that case, where it will be net force in the opposite direction in the case of the contents escaping due to excess or above ambient pressure.

When you pressurize the container, you also create a pressure less than the ambient in the container. If the container was pressurized to lets say 20 psi, the pressure external to the atmosphere would be 14.7 psi and thus negative pressure. (Basically a weak vacuum).

Edited by Gerrard

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8 minutes ago, Gerrard said:

When you pressurize the container, you also create a pressure less than the ambient in the containerIf the container was pressurized to lets say 20 psi, the pressure external to the atmosphere would be 14.7 psi and thus negative pressure. (Basically a weak vacuum).

The pressurized container is above ambient. The pressure outside the container would be at ambient (generally speaking 14.7 at sea level. give or take)

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2 hours ago, Gerrard said:

If the container was pressurized to lets say 20 psi, the pressure external to the atmosphere would be 14.7 psi and thus negative pressure. (Basically a weak vacuum). So if you open the hole on the container the pressure in the container would also decrease, lets say to 16 psi. Based on your explanation, wouldn't the force on walls of the tote be 4 psi in a glass tote and compression in plastic tote?

If the pressure was 20 psi then there would be a pressure of 5.3 psi or 5.3 pounds of force on each inch of the container pushing towards the outside.  If you bled off 4 psi of pressure then there would be a pressure of 1.3 psi pushing towards the outside of the tote.  If there was a balloon that was filled with air to 20 psi and you let out air to decrease the pressure to 16 psi, the balloon would compress.

2 hours ago, Gerrard said:

Could it be that the release of extremely high pressure into the atmosphere creates a propeller like effect? See Below. Let's say the tote (pressure rated tote of 2000 psi) was pressurized to 1000 psi. You fully open the valve on the back of the tote and now there is higher than atmospheric pressure immediately behind the tote, perhaps that pressure pushes the tote forward as the pressure moves to low pressure in front of the tote.

The effect is not a propeller, it is a rocket effect.  The tote would move if you opened the valve (assuming the tote wasn't too heavy to move).  The movement is not due to a pressure gradient in the tote, the movement would be due to the reactive force from the mass of air moving out through the valve.

If this was done in the vacuum of space the tote would still move because the movement is due to mass of the air leaving the tote and not due to it pushing against anything.

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5 hours ago, J.C.MacSwell said:

The pressurized container is above ambient. The pressure outside the container would be at ambient (generally speaking 14.7 at sea level. give or take)

true. what I meant was the pressure outside is negative compared to the pressure inside the tote. Therefore, there is a weak vacuum outside the tote as it will draw pressure from the tote.

5 hours ago, Bufofrog said:

If the pressure was 20 psi then there would be a pressure of 5.3 psi or 5.3 pounds of force on each inch of the container pushing towards the outside.  If you bled off 4 psi of pressure then there would be a pressure of 1.3 psi pushing towards the outside of the tote.  If there was a balloon that was filled with air to 20 psi and you let out air to decrease the pressure to 16 psi, the balloon would compress.

The effect is not a propeller, it is a rocket effect.  The tote would move if you opened the valve (assuming the tote wasn't too heavy to move).  The movement is not due to a pressure gradient in the tote, the movement would be due to the reactive force from the mass of air moving out through the valve.

If this was done in the vacuum of space the tote would still move because the movement is due to mass of the air leaving the tote and not due to it pushing against anything.

So if the tote internal pressure is below ambient pressure, then there is no movement, but if the tote is above ambient pressure then there is movement?

A rocket effect is that there is a pressure imbalance on two sides of the of the tote. The hole side of the tote would have less pressure pushing on it than the opposite side with no hole. So let us say that the internal tote pressure is 20 psi, 5.3 psi would be pushing outward on each side except the side with the hole which would be less than 5.3 psi (rocket effect). Similarly if the internal tote pressure is 12 psi, 2.7 psi would push inward on each side except the side with hole which the inward push would be greater than 2.7 psi (rocket effect because less than 12.7 is pushing on the side with the hole). So shouldn't there be movement even if the pump reduces the tote pressure to 12 psi? You said there wouldn't.

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1 hour ago, Gerrard said:

true. what I meant was the pressure outside is negative compared to the pressure inside the tote. Therefore, there is a weak vacuum outside the tote as it will draw pressure from the tote.

That would be a non-traditional interpretation.  Typically ambient pressure is the baseline.  So a standard pressure gage on the tote would read 5.3 psi, since the tote is pressurized.  If the non-traditional interpretation helps you to understand that's fine since it is not incorrect.

1 hour ago, Gerrard said:

So if the tote internal pressure is below ambient pressure, then there is no movement, but if the tote is above ambient pressure then there is movement?

Why do you think that?  I certainly did not say that.

1 hour ago, Gerrard said:

The rocket effect is that there is a pressure imbalance on two sides of the of the tote. The hole side of the tote would have less pressure pushing on it than the opposite side with no hole. So let us say that the internal tote pressure is 20 psi, 5.3 psi would be pushing outward on each side except the side with the hole which would be less than 5.3 psi (rocket effect).

No, that is not the rocket effect, I described how rockets work earlier in the thread.

1 hour ago, Gerrard said:

You said there wouldn't.

No I didn't say that.  I never discussed what would happen if we put a hole in the tote with a vacuum.

Can first agree on how a rocket achieves a force before we move on?

Look at this:  How Rockets work.

Edited by Bufofrog

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5 minutes ago, Bufofrog said:

That would be a non-traditional interpretation.  Typically ambient pressure is the baseline.  So a standard pressure gage on the tote would read 5.3 psi, since the tote is pressurized.  If the non-traditional interpretation helps you to understand that's fine since it is not incorrect.

Why do you think that?  I certainly did not say that.

No, that is not the rocket effect, I described how rockets work earlier in the thread.

No I didn't say that.  I never discussed what would happen if we put a hole in the tote with a vacuum.

Can first agree on how a rocket achieves a force before we move on?

Look at this:  How Rockets work.

"If there is a tote that you close with a cap then you would have a tote that is filled with air.  Inside the tote the air pressure would be 14.7 psi (lbs/in^2).  Out side the tote the air pressure would be 14.7 psi.  This means that inside and outside would have 14.7 lbs of force on each square inch of the tote container.  If I pull a vacuum and lower the internal pressure to 12.7 psi, that would mean there would be a 2 lb force on each inch of the tote pushing in on the surface.  If the tote was glass nothing would happen, there would simply be a constant force of 2 psi on the glass.  If the tote was a pliable plastic the force would push in the walls of the tote.  The tote would be compressed in size until the pressure inside the tote was raised to 14.7 psi and then it would be in equilibrium and it would stop compressing.  Since the forces are felt on all sides of the tote there would be no force in a particular direction, so the tote would not move."

Since my question was about a pump removing contents of the pump using negative pressure,  I presumed that when you pull a vacuum and lower the internal pressure, you are removing contents from the tote with the pump. Opening a valve to the flex hose and pump is putting a hole in the tote.

So when you open a hole in the tote, the force vector on the hole side (Action) is smaller than the force vector on the reaction side (reaction). The force vector being created by the internal pressure of the tote. Hence an imbalanced force causing movement.  https://www.grc.nasa.gov/www/k-12/rocket/TRCRocket/rocket_principles.html

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23 minutes ago, Gerrard said:

So when you open a hole in the tote, the force vector on the hole side (Action) is smaller than the force vector on the reaction side (reaction). The force vector being created by the internal pressure of the tote. Hence an imbalanced force causing movement.

You are misreading the description.  First the action and reaction are the equal and opposite, one is not larger.  The thrust does not come from the internal pressure of the tote or rocket.  The thrust is a reaction of the high velocity fluid exiting the rocket.

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25 minutes ago, Bufofrog said:

You are misreading the description.  First the action and reaction are the equal and opposite, one is not larger.  The thrust does not come from the internal pressure of the tote or rocket.  The thrust is a reaction of the high velocity fluid exiting the rocket.

There's a little more to it than simply action and reaction by throwing mass. In fact the balloon or rocket isn't throwing mass, the mass moves because of pressure differential.

"The pressure exerted by the air molecules on the walls of the rubber balloon is balanced by the tension/stress induced in the material of the balloon, plus the atmospheric pressure acting on the surface of the balloon. The pressure exerted on the interior walls of the balloon by the air in it, Pair, is the same across any given area of the balloon. Pair is the value of the pressure inside the balloon above the atmospheric pressure value. See Figure 1. As a result the pressures acting on geometrically opposite walls of the balloon balance (cancel) each other. Hence the net force acting on the balloon is zero. The air starts escaping through the mouth of the balloon. Since there is no wall now at the mouth of the balloon, no pressure is exerted by the escaping air over that area. Hence less force is exerted on that side of the balloon. This causes imbalance in the forces acting on the interior walls of the balloon. Net unbalanced force is now acting on the area diametrically opposite the mouth of the balloon. This unbalanced force is the thrust Tb , acting on the balloon, which moves/propels the balloon in its direction. The magnitude of the thrust is equal to the force that would have been exerted on that area had there been a wall at the mouth of the balloon."

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2 hours ago, Gerrard said:

There's a little more to it than simply action and reaction by throwing mass. In fact the balloon or rocket isn't throwing mass, the mass moves because of pressure differential.

"The pressure exerted by the air molecules on the walls of the rubber balloon is balanced by the tension/stress induced in the material of the balloon, plus the atmospheric pressure acting on the surface of the balloon. The pressure exerted on the interior walls of the balloon by the air in it, Pair, is the same across any given area of the balloon. Pair is the value of the pressure inside the balloon above the atmospheric pressure value. See Figure 1. As a result the pressures acting on geometrically opposite walls of the balloon balance (cancel) each other. Hence the net force acting on the balloon is zero. The air starts escaping through the mouth of the balloon. Since there is no wall now at the mouth of the balloon, no pressure is exerted by the escaping air over that area. Hence less force is exerted on that side of the balloon. This causes imbalance in the forces acting on the interior walls of the balloon. Net unbalanced force is now acting on the area diametrically opposite the mouth of the balloon. This unbalanced force is the thrust Tb , acting on the balloon, which moves/propels the balloon in its direction. The magnitude of the thrust is equal to the force that would have been exerted on that area had there been a wall at the mouth of the balloon."

Your source does not agree with NASA and the rest of the scientific community, so I have to disagree with your source.  I can see how this article could cause your confusion on the scenarios you introduced.  Hopefully you will look at other, reputable sources and not base everything on this one incorrect source.

Good luck.

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28 minutes ago, Bufofrog said:

Your source does not agree with NASA and the rest of the scientific community, so I have to disagree with your source.  I can see how this article could cause your confusion on the scenarios you introduced.  Hopefully you will look at other, reputable sources and not base everything on this one incorrect source.

Good luck.

Here’s another source

https://howthingsfly.si.edu/propulsion/rocket-propulsion

“If you fill a balloon with air and hold the neck closed, the pressure inside the balloon is slightly higher than the surrounding atmosphere. However, there is no net force on the balloon in any direction because the internal pressure on the balloon is equal in all directions. If you release the neck of the balloon, it acts like a hole, with no surface area for the internal pressure to act on. There is now an imbalanced force on the balloon, and the internal pressure on the front of the balloon is greater than the internal pressure on the back of the balloon.”

NASA sight doesn’t explain it well but the vector diagram I posted from them show exactly this. They can’t all be wrong

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If I have misunderstood the problem, I apologize
And maybe the OP can restate it in a clearer fashion.

Any fluid-dynamic effect can be considered as a pressure difference effect or a conservation of momentum effect.
Does a wing produce lift because of the reduced pressure above the wing as compared to the pressure below, or does it produce lift by accelerating the air mass downwards such that momentum conservation imparts an upward acceleration to the wing ?

Same with the propeller example previously used. Does the propeller reduce pressure ahead of it or does it accelerate a mass of air rearwards ?

Similarly, a rocket can be considered by either the pressure imbalance method, or, momentum conservation method.
( most physicist prefer the momentum method as it seems more 'formal' )

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9 hours ago, MigL said:

If I have misunderstood the problem, I apologize
And maybe the OP can restate it in a clearer fashion.

Any fluid-dynamic effect can be considered as a pressure difference effect or a conservation of momentum effect.
Does a wing produce lift because of the reduced pressure above the wing as compared to the pressure below, or does it produce lift by accelerating the air mass downwards such that momentum conservation imparts an upward acceleration to the wing ?

Same with the propeller example previously used. Does the propeller reduce pressure ahead of it or does it accelerate a mass of air rearwards ?

Similarly, a rocket can be considered by either the pressure imbalance method, or, momentum conservation method.
( most physicist prefer the momentum method as it seems more 'formal' )

I disagree with the conservation of momentum effect for the following reasons.

As I have stated before in this thread, the rocket is not required to "push" or provide a force for the exhaust to overcome it's inertia. The external force to change the exhaust's inertia comes from pressure gradient force. Therefore, conservation is always conserved without the rocket in the picture. Pressure gradient force is a potential force like gravity. When you drop a ball from your hand from a height, the ball does not need your hand to apply a force in order to conserve momentum as gravity provides the force for the ball to overcome it's inertia.

Using the propeller example, the propeller creates high pressure behind to propeller. If you were to use a very strong vacuum to dissipate the pressure being built behind the propeller, pressure cannot build behind the propeller and therefore will not move towards the front of the propeller in order to create movement.

If a person standing on a skateboard were throwing medicine balls and I use a very strong vacuum to dissipate pressure in the area where the balls were being thrown, or I was catching the balls and throwing them aside, the person on the skateboard would still move. It is because the person on the skateboard is providing the force, hence momentum must be conserved. Similarly, if the propeller was actually throwing the air behind it, there would be still movement even in the presence of the strong vacuum.

So therefore, the conservation of momentum effect is not correct.

Edited by Gerrard

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On 1/20/2020 at 8:41 PM, Gerrard said:

So therefore, the conservation of momentum effect is not correct.

It always holds. If you have enough knowns to solve the equations using it you have your answer, and any other way of doing the analysis will give the same results.

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1 hour ago, J.C.MacSwell said:

It always holds. If you have enough knowns to solve the equations using it you have your answer, and any other way of doing the analysis will give the same results.

Yes it holds where it is required. As I explained, momentum is conserved by pressure gradient force, not the rocket. How can you answer my question when something simple like conservation of momentum cannot applied correctly?

Edited by Gerrard

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You have never clarified the original question.
You have allowed us to interpret the OP as we see it, and once we offer a solution to OUR interpretation, you tell us we are wrong ?

How about reposting the original question, as YOU interpret it ( but in a clear manner, not like previously ), and then we can discuss it properly.

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