Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: clevernever on 03/02/2020 00:16:31
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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. The force that causes high pressure to move toward low pressure. 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.
A rocket is said to work using this example: 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.
So therefore, the conservation of momentum effect is not correct. A person standing on a skateboard throwing medicine balls to create movement is a false analogy. Why do schools and most sources teach this false analogy?
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They are both correct.
Microscopically, the difference in pressure is effectively a difference in the number (or speed) of gas particles hitting the walls of the balloon. Each gas particle is a tiny medicine ball. And each one effectively bounces off the inside of the balloon, pushing on the whole balloon. If the balloon i sealed, then the particles will hit all parts evenly, with zero net force. As soon as a hole opens up, then there is a portion of particles that will leave through that hole, without bouncing off of the balloon wall and compensating for their last impact. (Note: this picture is more complex if the mean free path of the particles is less than the radius of the balloon, but the end . result is the same)
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They are both correct.
Microscopically, the difference in pressure is effectively a difference in the number (or speed) of gas particles hitting the walls of the balloon. Each gas particle is a tiny medicine ball. And each one effectively bounces off the inside of the balloon, pushing on the whole balloon. If the balloon i sealed, then the particles will hit all parts evenly, with zero net force. As soon as a hole opens up, then there is a portion of particles that will leave through that hole, without bouncing off of the balloon wall and compensating for their last impact. (Note: this picture is more complex if the mean free path of the particles is less than the radius of the balloon, but the end . result is the same)
You didn't say how the conservation of momentum explanation is correct. How is a person throwing a ball while standing on a skateboard a correct analogy when the exhaust from the rocket moves out due to pressure differential?
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The gas that exits has momentum.
So you can either think of it as the gas leaving and pushing the rocket forward. Or you can thin of the rocket throwing the gas out the back, and thereby moving forward... it's the same thing.
You can also either think of the gas as a uniform continuous fluid with mass and net velocity and heat capacity, or as a collection of particles (ideal gas or more complex models). Again, you will still have a reasonable description of how/why a rocket moves, or how a balloon moves when the end is opened.
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The gas that exits has momentum.
So you can either think of it as the gas leaving and pushing the rocket forward. Or you can thin of the rocket throwing the gas out the back, and thereby moving forward... it's the same thing.
You can also either think of the gas as a uniform continuous fluid with mass and net velocity and heat capacity, or as a collection of particles (ideal gas or more complex models). Again, you will still have a reasonable description of how/why a rocket moves, or how a balloon moves when the end is opened.
"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)."
It is pressure gradient force that is the external force that causes the exhaust to overcome its inertia. The gas that exits has momentum because of pressure gradient force. Hence no force imposed on the rocket by the exiting gas. Like when you drop a ball from your hand, force of gravity causes the falling ball to gain momentum as it falls, thus no opposite force on your hand.
How do explain this? Are you saying that the exhaust movement is not caused by pressure gradient force?
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Hence no force imposed on the rocket by the exiting gas.
The gas doesn't expand only to the rear of the rocket: it expands in all directions. The portion of the gas moving towards the inside of the rocket nozzle has momentum too. The expanding gas that is pushing against the rocket nozzle is what causes the rocket to experience a force that propels it.
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Hence no force imposed on the rocket by the exiting gas.
The gas doesn't expand only to the rear of the rocket: it expands in all directions. The portion of the gas moving towards the inside of the rocket nozzle has momentum too. The expanding gas that is pushing against the rocket nozzle is what causes the rocket to experience a force that propels it.
Yes but the imbalanced force is created by the gas pushing on the front of the rocket. It is because the exiting gas is not pushing on the rocket is what causes the imbalanced force. The opening at the rear causes less surface area on rear side than the front side in which the gas pushes on.
The momentum causing the gas to move into the nozzle is caused by pressure gradient force.
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Yes but the imbalanced force is created by the gas pushing on the front of the rocket. It is because the exiting gas is not pushing on the rocket is what causes the imbalanced force. The opening at the rear causes less surface area on rear side than the front side in which the gas pushes on.
That is quite simply wrong.
There is no doubt that the pressure differential between the reaction chamber and the outside is the driving force of the rocket exhaust, but it is the reaction to the exhaust momentum that causes the rocket to move.
The principle is over 300 years old. Do you believe that with all the rockets designed and flown over the last 100 years we still do not know the most basic aspect of rocket propulsion? Think about how absurd that is.
I don't know where you got your bogus information, but I recommend going to reputable sites to learn about science!
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Yes but the imbalanced force is created by the gas pushing on the front of the rocket. It is because the exiting gas is not pushing on the rocket is what causes the imbalanced force. The opening at the rear causes less surface area on rear side than the front side in which the gas pushes on.
That is quite simply wrong.
There is no doubt that the pressure differential between the reaction chamber and the outside is the driving force of the rocket exhaust, but it is the reaction to the exhaust momentum that causes the rocket to move.
The principle is over 300 years old. Do you believe that with all the rockets designed and flown over the last 100 years we still do not know the most basic aspect of rocket propulsion? Think about how absurd that is.
I don't know where you got your bogus information, but I recommend going to reputable sites to learn about science!
So you are saying that the imbalanced forces explanation is incorrect? This was the explanation that I was told many years ago and there are plenty of reputable source that explain rocket propulsion in this manner.
Even NASA’s balloon vector diagram states this:
“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.”
The exiting exhaust is not pushing on the balloon, it simply causes imbalanced internal pressure.
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Yes but the imbalanced force is created by the gas pushing on the front of the rocket. It is because the exiting gas is not pushing on the rocket is what causes the imbalanced force. The opening at the rear causes less surface area on rear side than the front side in which the gas pushes on.
That is quite simply wrong.
There is no doubt that the pressure differential between the reaction chamber and the outside is the driving force of the rocket exhaust, but it is the reaction to the exhaust momentum that causes the rocket to move.
The principle is over 300 years old. Do you believe that with all the rockets designed and flown over the last 100 years we still do not know the most basic aspect of rocket propulsion? Think about how absurd that is.
I don't know where you got your bogus information, but I recommend going to reputable sites to learn about science!
So you are saying that the imbalanced forces explanation is incorrect? This was the explanation that I was told many years ago and there are plenty of reputable source that explain rocket propulsion in this manner.
Even NASA’s balloon vector diagram states this:
“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.”
The exiting exhaust is not pushing on the balloon, it simply causes imbalanced internal pressure.
The imbalanced internal pressure is, itself, a result of conservation of momentum. The hot gasses in the combustion chamber are made of molecules with individual momenta. It is the change in momentum of a molecule when it strikes the wall of of the chamber that produces the force on the wall.
For a rocket, you might imagine a person standing in a long skateboard, with a wall erected on one end. He throws medicine balls in both directions, two at a time. Since the pair of ball are moving in opposite directions at opposite speeds, throwing them produces no net change in his momentum. However, one of the Ball hits a wall, bounces off and ends up moving in the same direction as the other. The result is a change of momentum of the ball of 2mv (with m being the mass of the ball and assuming an perfectly elastic collision). The wall (and the skateboard the person is standing on) moves in the opposite direction to conserve momentum. The end result ends up being the same as if he had just thrown both balls towards the end with no wall.
Now coat the wall with a thick rubber coating. The compression of the rubber "softens" the impact of the ball hitting the wall, making the bounce of the ball less sudden, in turn reducing the force of the impact. But as long as the bounce is still perfectly elastic, the ball ends up still changing it momentum by 2mv, and transferring the same amount of momentum to the skateboard.
If we assume a non-elastic collision, for example the rubber absorbs some of the impact energy and converts it to heat, you can still make the impact force be equal to that for the perfectly elastic rubber, but now the ball will not change its momentum by the full 2mv, and neither will the skateboard be propelled by as much.
In the end, it is the total end momentum of the balls that determines how fast the skateboard ends up moving.
The expanding gas in the chamber is the mechanism for the exchange of momentum, but in the end, it is the exchange and conservation of momentum that results in the rocket's motion.
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So you are saying that the imbalanced forces explanation is incorrect?
Yes, that is exactly what i am saying.
This was the explanation that I was told many years ago and there are plenty of reputable source that explain rocket propulsion in this manner.
Whoever told you that was wrong. Reputable sources do not say that.
Even NASA’s balloon vector diagram states this:
“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.”
That is odd, because when I look at the NASA website this is what it says:
Rocket Principles
A rocket in its simplest form is a chamber enclosing a gas under pressure. A small opening at one end of the chamber allows the gas to escape, and in doing so provides a thrust that propels the rocket in the opposite direction. A good example of this is a balloon. Air inside a balloon is compressed by the balloon's rubber walls. The air pushes back so that the inward and outward pressing forces are balanced. When the nozzle is released, air escapes through it and the balloon is propelled in the opposite direction.
The site that the above quote is from is: https://www.grc.nasa.gov/www/k-12/rocket/TRCRocket/rocket_principles.html
Could you supply the website that you got your quote from because it was clearly a different NASA site.
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Yes but the imbalanced force is created by the gas pushing on the front of the rocket. It is because the exiting gas is not pushing on the rocket is what causes the imbalanced force. The opening at the rear causes less surface area on rear side than the front side in which the gas pushes on.
That is quite simply wrong.
There is no doubt that the pressure differential between the reaction chamber and the outside is the driving force of the rocket exhaust, but it is the reaction to the exhaust momentum that causes the rocket to move.
The principle is over 300 years old. Do you believe that with all the rockets designed and flown over the last 100 years we still do not know the most basic aspect of rocket propulsion? Think about how absurd that is.
I don't know where you got your bogus information, but I recommend going to reputable sites to learn about science!
So you are saying that the imbalanced forces explanation is incorrect? This was the explanation that I was told many years ago and there are plenty of reputable source that explain rocket propulsion in this manner.
Even NASA’s balloon vector diagram states this:
“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.”
The exiting exhaust is not pushing on the balloon, it simply causes imbalanced internal pressure.
The imbalanced internal pressure is, itself, a result of conservation of momentum. The hot gasses in the combustion chamber are made of molecules with individual momenta. It is the change in momentum of a molecule when it strikes the wall of of the chamber that produces the force on the wall.
For a rocket, you might imagine a person standing in a long skateboard, with a wall erected on one end. He throws medicine balls in both directions, two at a time. Since the pair of ball are moving in opposite directions at opposite speeds, throwing them produces no net change in his momentum. However, one of the Ball hits a wall, bounces off and ends up moving in the same direction as the other. The result is a change of momentum of the ball of 2mv (with m being the mass of the ball and assuming an perfectly elastic collision). The wall (and the skateboard the person is standing on) moves in the opposite direction to conserve momentum. The end result ends up being the same as if he had just thrown both balls towards the end with no wall.
Now coat the wall with a thick rubber coating. The compression of the rubber "softens" the impact of the ball hitting the wall, making the bounce of the ball less sudden, in turn reducing the force of the impact. But as long as the bounce is still perfectly elastic, the ball ends up still changing it momentum by 2mv, and transferring the same amount of momentum to the skateboard.
If we assume a non-elastic collision, for example the rubber absorbs some of the impact energy and converts it to heat, you can still make the impact force be equal to that for the perfectly elastic rubber, but now the ball will not change its momentum by the full 2mv, and neither will the skateboard be propelled by as much.
In the end, it is the total end momentum of the balls that determines how fast the skateboard ends up moving.
The expanding gas in the chamber is the mechanism for the exchange of momentum, but in the end, it is the exchange and conservation of momentum that results in the rocket's motion.
In your example you have made the exhaust movement the balls, the person on the skateboard is the pressure gradient force and the wall is the rocket. The conservation of momentum explanation is that the rocket is the person on the skateboard. Because the imbalanced force is applied to the wall. The skateboard would not move unless the ball bounces off the wall and hits the person on the skateboard or the skateboard itself.
Your analogy needs more clarification or correction.
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Gas leaves the rocket. Gas has mass. Gas has velocity. Momentum is mass x velocity. There can be no change in total momentum so (mass x velocity)gas = (mass x velocity)rest of the rocket if you integrate over the entire flight. Instantaneous calculation is a bit more difficult because the system is accelerating and the masses aree changing whilst the fuel is burning, but von Ohain's equations describe all the phases of the flight of an ideal rocket if you have an interest in such matters.
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Gas leaves the rocket. Gas has mass. Gas has velocity. Momentum is mass x velocity. There can be no change in total momentum so (mass x velocity)gas = (mass x velocity)rest of the rocket if you integrate over the entire flight. Instantaneous calculation is a bit more difficult because the system is accelerating and the masses aree changing whilst the fuel is burning, but von Ohain's equations describe all the phases of the flight of an ideal rocket if you have an interest in such matters.
When you drop a ball from the cusp of you had from a height, then the ball has momentum (mass * velocity). But the hand does not have an equal momentum of (mass * velocity). It is because gravity provided the ball’s momentum. Not the hand
Similarly pressure gradient force provide the exhaust’s momentum. Not the rocket.
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Oh yes it does. Ball goes down quickly because it has a small mass. Hand plus body plus planet goes up slowly because it has large mass. Same with gun recoil, astronauts in spacewalk, everything. Σ(mv) = 0, always.
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Oh yes it does. Ball goes down quickly because it has a small mass. Hand plus body plus planet goes up slowly because it has large mass. Same with gun recoil, astronauts in spacewalk, everything. Σ(mv) = 0, always.
So you are saying that a 175lbs person who picks up a 200lbs weight and drops it from a height will get lifted up from the ground?
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So since you did not reply where this quote came, is it safe to say it is not a quote from the NASA site and that you dishonestly made it up?
Even NASA’s balloon vector diagram states this:
“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.”
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So you are saying that a 175lbs person who picks up a 200lbs weight and drops it from a height will get lifted up from the ground?
No. He's saying that the ground itself will be lifted up by its gravitational attraction to the 200 pound weight, and therefore lift the 175 pound person up in the process. All due to conservation of momentum.
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So you are saying that a 175lbs person who picks up a 200lbs weight and drops it from a height will get lifted up from the ground?
No. He's saying that the ground itself will be lifted up by its gravitational attraction to the 200 pound weight, and therefore lift the 175 pound person up in the process. All due to conservation of momentum.
But the momentum is given to the earth, not the person. What if the person was on an airplane in the air?
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But the momentum is given to the earth, not the person.
Some of the Earth's momentum is imparted to the person because they were standing on the ground.
What if the person was on an airplane in the air?
I would expect the upward movement of the Earth to cause a slight increase in air pressure, thus pushing the plane upward somewhat.
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Even NASA’s balloon vector diagram states this:
“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.”
So let's review where we are in this thread:
1. You do not understand a basic law of physics that has been known for over 300 years.
2. You will lie about sources to support your incorrect notions.
3. When you are caught lying you don't have the fortitude to admit you lied.
OK, I guess if that's what we have to work with it is going to make this discussion sort of useless...
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What if the person was on an airplane in the air?
This is actually a nice example. Suppose we have an airplane with mass of 100,000 kg, and it is in level flight (equilibrium). I am standing in the middle of that airplane holding a 100 kg rock (I switched to metric since a pound is not a unit of mass). The mass is dropped from a meter above the floor. The aircraft will immediately be reft of the weight of this mass and will move upwards. The rock hits the floor and stops this movement, all according to conservation of momentum. Afterwards, the airplane will be 1mm higher because this was done. Similarly if I lift the rock from the floor and put it 2 meters up in the overhead bins (2 m? Who put those way up there??), the airplane will move downward 2 mm in the process. All this assumes that the controls of the airplane are never modified from the state in which it was in equilibrium.
The exact same movement of the aircraft (1mm up, then 2mm down) would occur if the experiment was done in outer space, except lacking gravity, some other force like a pulled thread would need to move the rock from here to there.
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What if the person was on an airplane in the air?
This is actually a nice example. Suppose we have an airplane with mass of 100,000 kg, and it is in level flight (equilibrium). I am standing in the middle of that airplane holding a 100 kg rock (I switched to metric since a pound is not a unit of mass). The mass is dropped from a meter above the floor. The aircraft will immediately be reft of the weight of this mass and will move upwards. The rock hits the floor and stops this movement, all according to conservation of momentum. Afterwards, the airplane will be 1mm higher because this was done. Similarly if I lift the rock from the floor and put it 2 meters up in the overhead bins (2 m? Who put those way up there??), the airplane will move downward 2 mm in the process. All this assumes that the controls of the airplane are never modified from the state in which it was in equilibrium.
The exact same movement of the aircraft (1mm up, then 2mm down) would occur if the experiment was done in outer space, except lacking gravity, some other force like a pulled thread would need to move the rock from here to there.
What about if you drop it out of the airplane?
According to your explanation the plane would move significantly upwards.
Why doesn’t that happen when people jump off to parachute?
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Why doesn’t that happen when people jump off to parachute?
Whoever said that it didn't?
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What about if you drop it out of the airplane?
According to your explanation the plane would move significantly upwards.
It does, until whatever is controlling the aircraft notices the change in flight path from the sudden decrease in payload, at which point it will presumably take corrective measures to bring it back to level flight.
Why doesn’t that happen when people jump off to parachute?
It does. Same with bombers, where a small plane with a heavy bomb (torpedo say) results in a marked upward acceleration as the bomb releases. It happens to any aircraft on a gradual basis as it loses mass from fuel consumption. The flight computer/pilot must make fairly frequent corrections to keep the aircraft on course as its mass changes, just as I make continuous corrections to the steering wheel of my car to keep it in its lane as the road and wind conditions change.
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What about if you drop it out of the airplane?
According to your explanation the plane would move significantly upwards.
It does, until whatever is controlling the aircraft notices the change in flight path from the sudden decrease in payload, at which point it will presumably take corrective measures to bring it back to level flight.
Why doesn’t that happen when people jump off to parachute?
It does. Same with bombers, where a small plane with a heavy bomb (torpedo say) results in a marked upward acceleration as the bomb releases. It happens to any aircraft on a gradual basis as it loses mass from fuel consumption. The flight computer/pilot must make fairly frequent corrections to keep the aircraft on course as its mass changes, just as I make continuous corrections to the steering wheel of my car to keep it in its lane as the road and wind conditions change.
Then why wasn't the pilot paying attention to the parachute people jumping off? Yes I have been. Adjustments would have to be timed to prevent sudden movement upwards. The force is applied as soon as you drop. Many people jumped off before I did and I didn't feel the plane move up.
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Then why wasn't the pilot paying attention to the parachute people jumping off? Yes I have been. Adjustments would have to be timed to prevent sudden movement upwards. The force is applied as soon as you drop. Many people jumped off before I did and I didn't feel the plane move up.
The weight of the people would have been small compared to the weight of the airplane. Any net change in altitude would therefore have been small.
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Then why wasn't the pilot paying attention to the parachute people jumping off? Yes I have been. Adjustments would have to be timed to prevent sudden movement upwards. The force is applied as soon as you drop. Many people jumped off before I did and I didn't feel the plane move up.
The weight of the people would have been small compared to the weight of the airplane. Any net change in altitude would therefore have been small.
Let's say 10 people with average weight of 70kg jump out? That's a force of 6860 N. The plane weighs 50000kg, so planes acceleration will be 0.137 m/s^2. How long is the acceleration? Should it be as the same time the people are accelerating?
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Let's say 10 people with average weight of 70kg jump out? That's a force of 686 N. The plane weighs 50000kg, so planes acceleration will be 0.0137 m/s^2.
Unless all 10 people jumped out at the same time, the acceleration should be lower than that. This is because, presumably, the pilot is continually making adjustments to keep a constant altitude. Even without that assumption, an acceleration of 1.37 centimeters per second is indeed small. You'd only experience a weight increase of about 0.14%.
How long is the acceleration?
However long it takes for the pilot to compensate for it.
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Let's say 10 people with average weight of 70kg jump out? That's a force of 686 N. The plane weighs 50000kg, so planes acceleration will be 0.0137 m/s^2.
Unless all 10 people jumped out at the same time, the acceleration should be lower than that. This is because, presumably, the pilot is continually making adjustments to keep a constant altitude. Even without that assumption, an acceleration of 1.37 centimeters per second is indeed small. You'd only experience a weight increase of about 0.14%.
How long is the acceleration?
However long it takes for the pilot to compensate for it.
How long does the acceleration last? How long does the pilot have to keep adjusting? Until the person reaches the ground? Because the conservation of momentum has to continue so long as the persons are still falling due to gravity?
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Yes but the imbalanced force is created by the gas pushing on the front of the rocket. It is because the exiting gas is not pushing on the rocket is what causes the imbalanced force.
Of course there are no imbalanced forces in physics. That's Newton's Third Law. You may have heard of it. The force that pushes on the rocket (from the gas pressure times the cross-sectional area it acts over) causes an equal and opposite force on the exhaust which accelerates the exhaust in the opposite direction; and so the rate of change of the momentum of the rocket is always equal to the rate of change in momentum of the exhaust.
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How long does the acceleration last?
Like I said, however long it takes for the pilot to readjust. Probably a second or two.
How long does the pilot have to keep adjusting? Until the person reaches the ground?
No.
Because the conservation of momentum has to continue so long as the persons are still falling due to gravity?
That's not how that works. Conservation of momentum here simply means that the net force (lift) pushing the plane up has to be equal and opposite to the weight of the skydivers at the moment that they jump out. Canceling out that change in momentum is possible by using the control surfaces, which transfers that gain in momentum to the air around the plane.
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How long does the acceleration last?
Like I said, however long it takes for the pilot to readjust. Probably a second or two.
How long does the pilot have to keep adjusting? Until the person reaches the ground?
No.
Because the conservation of momentum has to continue so long as the persons are still falling due to gravity?
That's not how that works. Conservation of momentum here simply means that the net force (lift) pushing the plane up has to be equal and opposite to the weight of the skydivers at the moment that they jump out. Canceling out that change in momentum is possible by using the control surfaces, which transfers that gain in momentum to the air around the plane.
What defines a moment. 1 second 0.1 seconds?
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Even NASA’s balloon vector diagram states this:
“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.”
Either supply the source for this quote of withdraw the claim. I will have no choice but to report you if you continue to dodge my legitimate inquiry.
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What defines a moment. 1 second 0.1 seconds?
It would be instantaneous. The plane would begin to accelerate upward as soon as it was no longer supporting the weight of the skydiver.
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What defines a moment. 1 second 0.1 seconds?
It would be instantaneous. The plane would begin to accelerate upward as soon as it was no longer supporting the weight of the skydiver.
Going back to the skateboard example, the force applied to the ball will stop as soon as the ball is released from the hand and thus acceleration will stop . The force applied by the ball onto the person on the skateboard will also cease at the same time the ball is released and acceleration will stop. However, the force of gravity is always there and causing acceleration until the object hits the ground. Hence the acceleration of the plane should keep occurring until the object falling stops accelerating
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What defines a moment. 1 second 0.1 seconds?
It would be instantaneous. The plane would begin to accelerate upward as soon as it was no longer supporting the weight of the skydiver.
Going back to the skateboard example, the force applied to the ball will stop as soon as the ball is released from the hand and thus acceleration will stop . The force applied by the ball onto the person on the skateboard will also cease at the same time the ball is released and acceleration will stop. However, the force of gravity is always there and causing acceleration until the object hits the ground. Hence the acceleration of the plane should keep occurring until the object falling stops accelerating
No. The Earth continues to accelerate until the object stops falling.
The plane or skateboard, or whatever, experiences a change in its environment when (as*) the object departs, but once gone, it doesn't matter what happens to the object, the plane's or skateboard's trajectory and the object's trajectory are independent as soon as contact is lost (unless we also consider gravitational interactions of the two, but let's assume neither the object nor the plane is more massive than an aircraft carrier, so that is negligible)
*If we really want to consider how long the acceleration on the plane lasts, we have to consider how the object is leaving. For example: If it just rolls out of the plane on the horizontal, then there is some small window of time where the contact between the plane and the object goes from full contact to no contact. That window is determined by many factors, such as the shape and rigidity of both the object and the edge of the craft. As the object falls off, is it rigid enough that it all accelerates simultaneously? Or does it distort as it falls (imagine pushing a sealed but half-filled sack of potatoes off the plane--the plane will accelerate for each potato that goes over, and it could take several seconds for the whole things to tumble off).
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What defines a moment. 1 second 0.1 seconds?
It would be instantaneous. The plane would begin to accelerate upward as soon as it was no longer supporting the weight of the skydiver.
Going back to the skateboard example, the force applied to the ball will stop as soon as the ball is released from the hand and thus acceleration will stop . The force applied by the ball onto the person on the skateboard will also cease at the same time the ball is released and acceleration will stop. However, the force of gravity is always there and causing acceleration until the object hits the ground. Hence the acceleration of the plane should keep occurring until the object falling stops accelerating
No. The Earth continues to accelerate until the object stops falling.
The plane or skateboard, or whatever, experiences a change in its environment when (as*) the object departs, but once gone, it doesn't matter what happens to the object, the plane's or skateboard's trajectory and the object's trajectory are independent as soon as contact is lost (unless we also consider gravitational interactions of the two, but let's assume neither the object nor the plane is more massive than an aircraft carrier, so that is negligible)
*If we really want to consider how long the acceleration on the plane lasts, we have to consider how the object is leaving. For example: If it just rolls out of the plane on the horizontal, then there is some small window of time where the contact between the plane and the object goes from full contact to no contact. That window is determined by many factors, such as the shape and rigidity of both the object and the edge of the craft. As the object falls off, is it rigid enough that it all accelerates simultaneously? Or does it distort as it falls (imagine pushing a sealed but half-filled sack of potatoes off the plane--the plane will accelerate for each potato that goes over, and it could take several seconds for the whole things to tumble off).
If momentum is conserved by earth moving towards the falling object, what conserved the plane’s momentum as it moves up? Mass*velocity of falling object = mass*velocity of the earth.
Where does the plane fit into the picture?
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The earth, the object and the plane (and the atmosphere) are all in the same system.
The earth interacts with everything (and everything interacts with the earth).
The object and plane only interact when they are touching.
Energy is conserved.
Mass is conserved.
Momentum is conserved.
Angular momentum is conserved.
Charge is conserved.
As long as the different models agree on these assumptions, it doesn't matter how we rationalize the interactions (rockets don't work because of the conservation of momentum any more than because of electromagnetism.)
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The earth, the object and the plane (and the atmosphere) are all in the same system.
The earth interacts with everything (and everything interacts with the earth).
The object and plane only interact when they are touching.
Energy is conserved.
Mass is conserved.
Momentum is conserved.
Angular momentum is conserved.
Charge is conserved.
As long as the different models agree on these assumptions, it doesn't matter how we rationalize the interactions (rockets don't work because of the conservation of momentum any more than because of electromagnetism.)
So a person throwing a ball is equivalent of gas leaving a rocket due to pressure gradient?
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So a person throwing a ball is equivalent of gas leaving a rocket due to pressure gradient?
Insofar as momentum is conserved by splitting kinetic energy between masses that are accelerating away from each other in each case? yes!
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So a person throwing a ball is equivalent of gas leaving a rocket due to pressure gradient?
Insofar as momentum is conserved by splitting kinetic energy between masses that are accelerating away from each other in each case? yes!
So if a hot air balloon starts to lift off from the ground, does it push off the ground first?
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So if a hot air balloon starts to lift off from the ground, does it push off the ground first?
No.
It’s a good question. Given what’s been said, think about it and see if you can work it out.
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Deleted, question already answered.
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Deleted, question already answered.
Didn’t see that, but it did appear there was sufficient info to work it out if not.