Revisiting a commonly presented image...

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rvlvr

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Re: Revisiting a commonly presented image...
« Reply #300 on: February 14, 2018, 04:12:27 AM »
How many here are not bots? Totallackey? Are there others?

And don't you think it is funny your FE masters banned you? I mean does that not mean they consider you unfit to represent them?

The recoil topic is less interesting than the railgun issue. Still, how do recoil compensators fit in this "air rushing back in the barrel" scenario? They open up more airways. Shouldn't that mean, by the model you follow, the recoil effect is stronger as the air gets in faster? Or am I completely wrong here?

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Papa Legba

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Re: Revisiting a commonly presented image...
« Reply #301 on: February 14, 2018, 04:19:27 AM »
The jackblack AI algorithm has not been programmed to understand Newton's third law..

Force pairings are unnecessary to it, and force vectors can be reversed as needed to fit whatever mad bullshit it is spamming.

Total physics illiterate.
You are the one wanting to discard Newton's third law and pretend the gas needs to push of something,

The need for something extrinsic to push off is the totality of Newton's third law.

Looks like the Confused Voodoo Priest still doesn't understand the totality of Newton's third law.

It could be said that once the burnt fuel reaches the bell of the rocket it is "extrinsic" to the rest of the rocket!

Though where did Newton mention "extrinsic" anyway? An object can split in two and one half can push off the other.

So the Deceptive Voodoo Priest fails again!  ;D ;D So sorry about that. ;D ;D

The exhaust of a rocket is a force exerted by the rocket itself.

Therefore it will require a body extrinsic to the rocket for it to create a force pairing and thus produce motion.

Well, according to Newton's third law it will, anyway...

Physics illiterates like the jackblack and rabbibot AI algorithms are not programmed to understand this.
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rabinoz

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Re: Revisiting a commonly presented image...
« Reply #302 on: February 14, 2018, 04:59:44 AM »
The exhaust of a rocket is a force exerted by the rocket itself.
Once it has left rocket it is not,  just as the bullet is not part of the gun.

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Papa Legba

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Re: Revisiting a commonly presented image...
« Reply #303 on: February 14, 2018, 05:16:04 AM »
The exhaust of a rocket is a force exerted by the rocket itself.
Once it has left rocket it is not,  just as the bullet is not part of the gun.

So rocket exhausts do not exert a force then?

And the exhaust propels solid metal bullets out the back to get to space instead?

And the rocket does all this without exerting a force, somehow?

Yeah, you don't understand anything, let alone Newton's third law.
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Re: Revisiting a commonly presented image...
« Reply #304 on: February 14, 2018, 12:55:12 PM »
The exhaust of a rocket is a force exerted by the rocket itself.
Therefore it will require a body extrinsic to the rocket for it to create a force pairing and thus produce motion.
Maybe according to Papa's imaginary laws of physics, not according to Newton.

According to Newton (and reality), there exists a force pairing between the rocket and the exhaust.
The rocket pushes the exhaust backwards while the exhaust pushes the rocket forwards.
No external entity required.

If you like, you can consider the exhaust the "extrinsic body".

So rocket exhausts do not exert a force then?
No, they do exert a force, on the rocket.
What happens later is fairly irrelevant for a rocket.
Just like when the bullet hits it's target and is slowed down, the gun doesn't magically recoil as if there is some magic connection between the bullet and gun; the same is true for the rocket. The rocket's exhaust pushing against something wont magically make the rocket go more, at least not once it has left the rocket.


Yeah, you don't understand anything, let alone Newton's third law.
There you go projecting again.


Here are a few nice simple questions for you:
You are out in space, you have the rocket and the "exhaust" trapped together.
The exhaust then starts moving out the back of the rocket.
What force causes the exhaust to accelerate?
What is the equal and opposite force required by Newton's third law?
What is accelerated by this latter force?

Or even simpler, just one question:
What causes the curve in the photo in the OP?
« Last Edit: February 14, 2018, 12:56:50 PM by JackBlack »

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Papa Legba

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Re: Revisiting a commonly presented image...
« Reply #305 on: February 14, 2018, 01:40:32 PM »
If you like, you can consider the exhaust the "extrinsic body".

If you're mental you can

A rocket and it's exhaust are demonstrably not two separate objects.

The rocket and the exhaust are part of the same system, exerting a force vector rearwards.

This rearwards force vector creates a force pairing with the mass of the atmosphere, thus propelling the rocket in the opposite direction, i.e. forwards.

This accords with Newton's third law i.e. f1=-f2.

Proof that the rocket and exhaust are part of the same system can be attained through simple observation, where it is seen that they both move together at all times:



It will also be noted that the rocket exhaust forces up huge clouds of dirt at launch, proving that the exhaust is creating a powerful force pairing with its external environment.

If you cannot provide similar visual evidence for your mad model, using a rocket, do not reply.
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sokarul

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Re: Revisiting a commonly presented image...
« Reply #306 on: February 14, 2018, 02:04:55 PM »
And yet you have no mechanism to transfer a force to the rocket.

Also you will notice no change in acceleration as the rocket leaves the launchpad. The ground should accelerate the rocket way more than air in you fantasy model.
« Last Edit: February 14, 2018, 02:26:52 PM by sokarul »
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rabinoz

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Re: Revisiting a commonly presented image...
« Reply #307 on: February 14, 2018, 02:20:49 PM »
Yeah, you don't understand anything, let alone Newton's third law.
Who doesn't understand anything again, Mr Delusional Voodoo Priest Controlled Puppet?

Here go and read about someone whose theories predated NASA by over 50 years, The Tsiolkovsky formula.

But Konstantin Tsiolkovsky's theories have sums and thingos that would completely mystify a Delusional Voodoo Priest Controlled Puppet.
Out of his work comes the all important "Rocket Equation", ΔV=ve⋅ln(Minitial/Mfinal).

This time Mr Deluded Voodoo Priest, go and do your homework!

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rabinoz

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Re: Revisiting a commonly presented image...
« Reply #308 on: February 14, 2018, 02:33:32 PM »
Proof that the rocket and exhaust are part of the same system can be attained through simple observation, where it is seen that they both move together at all times:



It will also be noted that the rocket exhaust forces up huge clouds of dirt at launch, proving that the exhaust is creating a powerful force pairing with its external environment.

If you cannot provide similar visual evidence for your mad model, using a rocket, do not reply.
I don't take orders from a Delusional Voodoo Priest.

Incorrect! All it proves is "that the rocket exhaust forces up huge clouds of dirt at launch" - big deal!

The exhaust velocity far exceeds the velocity of sound, so no influence can be transmitted back up the exhaust column.  So there is no way for the ground forces to be transmitted back up to the rocket.

Ever actually looked at where the exhaust gases go during a shuttle launch?

IMAX // Shuttle launch (Hubble 2010 - STS 125), FalkoJ89
See from 0:55 to 1:05 where the exhaust is deflected to the side.

Go and read a bit of elementary stuff on it: Compressible flow..
Then a bit on: High Speed Aerodynamics.

Until you understand that, you will never properly understand rocket engines and De Laval nozzles so run away and play with the only thing you understand:
Voodoo Fetishes.

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Papa Legba

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Re: Revisiting a commonly presented image...
« Reply #309 on: February 14, 2018, 02:47:07 PM »
Proof that the rocket and exhaust are part of the same system can be attained through simple observation, where it is seen that they both move together at all times:



It will also be noted that the rocket exhaust forces up huge clouds of dirt at launch, proving that the exhaust is creating a powerful force pairing with its external environment.

If you cannot provide similar visual evidence for your mad model, using a rocket, do not reply.
I don't take orders from a Delusional Voodoo Priest.

So you could not provide similar visual evidence as I did that the exhaust is a force exerted by the rocket and chose to launch into a mad tirade of abuse instead?

That is so unlike you, rabbibot!

Anyhoo, neither forces exerted by the body nor forces internal to the body can be included in any consideration of how said body creates force pairings, rabbibot...

So your de laval nozzle is irrelevant, as any forces therein would be internal.

So, please make a free body diagram of a rocket that includes neither the exhaust nor the combustion chamber/nozzle, so we can finally see where the force pairings are created.

You know, do something an engineer might?

Kthxbai!
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sokarul

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Re: Revisiting a commonly presented image...
« Reply #310 on: February 14, 2018, 02:52:23 PM »

I notice you cut it from my reply, as if you did not want anyone to watch it...



Chatbot fail.  Don't cut his post where he is addressing you, Especially when you call out others doing it.

lol

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Papa Legba

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Re: Revisiting a commonly presented image...
« Reply #311 on: February 14, 2018, 02:55:38 PM »
I notice you cut it from my reply, as if you did not want anyone to watch it...
Chatbot fail.  Don't cut his post where he... (rest of mad bot shitpost cut out)

Sorry what was that again, botty boy?
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sokarul

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Re: Revisiting a commonly presented image...
« Reply #312 on: February 14, 2018, 02:59:30 PM »
When you have nothing, post in the style of papa legba to continue to drag the thread on.
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Re: Revisiting a commonly presented image...
« Reply #313 on: February 14, 2018, 03:50:17 PM »
If you like, you can consider the exhaust the "extrinsic body".
If you're mental you can
No. If you are sane you can.

A rocket and it's exhaust are demonstrably not two separate objects.
Considering they separate, they are quite demonstrably two (or more) separate objects.

The rocket and the exhaust are part of the same system, exerting a force vector rearwards.
Nope.
The exhaust leaves the rocket. This means it needs a force vector pushing it rearwards.
This needs to be balanced by a force vector pushing forward, and the only thing to push forward on is the rocket.
This is quite simple physics.

This accords with Newton's third law i.e. f1=-f2.
My analysis does. Your analysis does not.
Your analysis has the exhaust rocket magically fly backwards with no force at all.

Explain what is the origin of the exhausts backwards movement.
Where does the force come from which pushes the exhaust backwards?
What is it paired to?

Proof that the rocket and exhaust are part of the same system can be attained through simple observation, where it is seen that they both move together at all times:
Nope.
The exhaust is thrown backwards and leaves the rocket. New exhaust comes out from it.
It doesn't carry its exhaust with it.

Try to find a rocket that is no longer powered with its exhaust still moving with it.

If you cannot provide similar visual evidence for your mad model, using a rocket, do not reply.
i.e. you will completely ignore the arguments presented.

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Papa Legba

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Re: Revisiting a commonly presented image...
« Reply #314 on: February 14, 2018, 04:15:07 PM »
No not Nope not no Nope doesn't no ignore

Cut out all your mad AI algorithm Pseudoscience and just left the negativity...

Cos that's what you're all about really isn't it?

Just a massive downer...

I'll call you Negatron from now on.
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rabinoz

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Re: Revisiting a commonly presented image...
« Reply #315 on: February 14, 2018, 05:04:17 PM »
Proof that the rocket and exhaust are part of the same system can be attained through simple observation, where it is seen that they both move together at all times:



It will also be noted that the rocket exhaust forces up huge clouds of dirt at launch, proving that the exhaust is creating a powerful force pairing with its external environment.

If you cannot provide similar visual evidence for your mad model, using a rocket, do not reply.
I don't take orders from a Delusional Voodoo Priest.

So you could not provide similar visual evidence as I did that the exhaust is a force exerted by the rocket and chose to launch into a mad tirade of abuse instead?

That is so unlike you, rabbibot!

Anyhoo, neither forces exerted by the body nor forces internal to the body can be included in any consideration of how said body creates force pairings, rabbibot...

So your de laval nozzle is irrelevant, as any forces therein would be internal.
Incorrect! In the de Laval nozzle soninc flow in the throat and supersonic flow in the bell prevents any outside forces, such as from the exhaust hitting the ground affecting the thrust.

Quote from: Papa Legba
So, please make a free body diagram of a rocket that includes neither the exhaust nor the combustion chamber/nozzle, so we can finally see where the force pairings are created.
No free body diagram is necessary! if you want one, go scribble one out yourself.

Quote from: Papa Legba
You know, do something an engineer might?
Read about supersonic flow again and again till you understand it!

The exhaust velocity far exceeds the velocity of sound, so no influence can be transmitted back up the exhaust column. 
So there is no way for the ground forces to be transmitted back up to the rocket.

Ever actually looked at where the exhaust gases go during a shuttle launch?

IMAX // Shuttle launch (Hubble 2010 - STS 125), FalkoJ89
See from 0:55 to 1:05 where the exhaust is deflected to the side so could not affect the thrust anyway..

Go and read a bit of elementary stuff on it: Compressible flow..
Then a bit on: High Speed Aerodynamics.

Until you understand that, you will never properly understand rocket enginess and De Laval nozzles.

And Mr P. Legba, you are no more than a Deluded Voodoo Priest if you refuse to understand concepts like supersonic flow.

Here's a little present for your delectation and enjoyment:
Quote from: National Aeronautics and Space Administration
On this slide, we show a schematic of a rocket engine. In a rocket engine, stored fuel and stored oxidizer are ignited in a combustion chamber. The combustion produces great amounts of exhaust gas at high temperature and pressure. The hot exhaust is passed through a nozzle which accelerates the flow. Thrust is produced according to Newton's third law of motion.

The amount of thrust produced by the rocket depends on the mass flow rate through the engine, the exit velocity of the exhaust, and the pressure at the nozzle exit. All of these variables depend on the design of the nozzle. The smallest cross-sectional area of the nozzle is called the throat of the nozzle. The hot exhaust flow is choked at the throat, which means that the Mach number is equal to 1.0 in the throat and the mass flow rate m dot is determined by the throat area. The area ratio from the throat to the exit Ae sets the exit velocity Ve
and the exit pressure pe. You can explore the design and operation  of a rocket nozzle with our interactive thrust simulator program which runs on your browser.

The exit pressure is only equal to free stream pressure at some design condition. We must, therefore, use the longer version of the generalized thrust equation to describethe thrust of the system. If the free stream pressure is given by p0, the thrust F equation becomes:

F = m dot * Ve + (pe - p0) * Ae

Notice that there is no free stream mass times free stream velocity term in the thrust equation because no external air is brought on board. Since the oxidizer is carried on board the rocket, rockets can generate thrust in a vacuum where there is no other source of oxygen. That's why a rocket will work in space, where there is no surrounding air, and a gas turbine or propeller will not work. Turbine engines and propellers rely on the atmosphere to provide air as the working fluid for propulsion and oxygen in the air as oxidizer for combustion.

The thrust equation shown above works for both liquid rocket and solid rocket. engines. There is also an efficiency parameter called the specific impulse which works for both  types of rockets and greatly simplifies the performance analysis for rockets.

From: NASA Rocket Thrust Equation.

Now, if you disagree, please provide your own thrust equation which shows what thrust a rocket would generate at various altitudes up to 100 km.

Re: Revisiting a commonly presented image...
« Reply #316 on: February 14, 2018, 05:19:27 PM »
Cut out all your mad AI algorithm Pseudoscience and just left the negativity...
You mean cut out all the rational arguments showing you to be wrong.

Now why don't you try answering one of 2 simple questions:
What causes the exhaust to move backwards?
What causes the curve in the image in the OP?

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sandokhan

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Re: Revisiting a commonly presented image...
« Reply #317 on: February 14, 2018, 10:46:09 PM »


Nasa has never sent a single rocket into outer space: the experimental data simply does not exist to support the free stream pressure equation.

It is very easy to prove that each and every NASA space mission was faked.


A design as revolutionary as the phase conjugate mirror for the Sagnac effect, in the field of rocket design science, has been been published: it shows, beyond a shadow of a doubt, that rockets cannot work in full vacuum:

https://www.theflatearthsociety.org/forum/index.php?topic=30499.msg1740524#msg1740524



The ring shaped exhaust will use only ETHER WAVES to propel the rocket: without the ether, no thrust is possible.

In full vacuum, no thrust is possible.

That is, the ring shaped exhaust will use the potential in Whittaker's proven equations, and not the vector fields to provide the thrust in the form of double torsion ether waves.

If Nasa did have a secret space program, then that program must have been based totally on the Biefeld-Brown effect which uses ether waves for propulsion.

The Nasa equation has never been tested in real life experiments in the assumed outer space vacuum: each and every space mission was faked. If there was a secret space program, then electrogravity must have been used; electrogravity = ether waves.



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sandokhan

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Re: Revisiting a commonly presented image...
« Reply #318 on: February 14, 2018, 11:00:14 PM »
Let's now take a closer look at the equation provided by Nasa:

https://www.grc.nasa.gov/WWW/k-12/airplane/rockth.html

https://www.grc.nasa.gov/WWW/k-12/airplane/Images/rockth.gif

The first term says Force = Mass x Velocity whereas NASA web sites say that
Force = Mass x Acceleration (Newton’s 2nd law of motion):

https://spaceflightsystems.grc.nasa.gov/education/rocket/BottleRocket/journey_newtona2.htm

"NASA sites also say that Mass x Velocity = Momentum which is not a force. Momentum is potential energy. If you throw a rock it has momentum. If you throw it harder it has more momentum. No force is generated until the rock hits something. Gas shot out of the back of a rocket very fast does not create a force until it interacts with something, which it never does in the vacuum of space. It remains high momentum gas streaking endlessly through space looking to do work but never getting the chance.

The second term (Pressure Difference between inside the rocket and the vacuum of space) x Nozzle Area violates the “free expansion” effect, part of the first law of thermodynamics by which pressurized gas moves into a vacuum without any work being done. It does not matter how highly pressured the gas is inside the rocket nor how fast it comes out. Because it is going into a vacuum the gas makes the trip “for free” and does not do any work, does not expend any energy and does not create any force or thrust.

The NASA space rocket equation has two terms the first of which is incorrect and so is the second."

https://www.grc.nasa.gov/WWW/k-12/airplane/momntm.html

https://er.jsc.nasa.gov/seh/f.html

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sokarul

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Re: Revisiting a commonly presented image...
« Reply #319 on: February 14, 2018, 11:03:39 PM »
You show up late, ignoring everything previously posted, and ignored all the other threads on the topic. Cool. If you scroll up you will find the real thrust equations.

So now there is no ether in a vacuum? Better tell Biefeld and Brown.
« Last Edit: February 14, 2018, 11:09:50 PM by sokarul »
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Papa Legba

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Re: Revisiting a commonly presented image...
« Reply #320 on: February 15, 2018, 12:28:14 AM »
Cut out all your mad AI algorithm Pseudoscience and just left the negativity...
You mean cut out all the rational arguments showing you to be wrong.

I meant cutting out all the mad AI algorithm Pseudoscience that violates both Newton's laws of motion and the laws of Thermodynamics, Negatron:

https://www.theflatearthsociety.org/forum/index.php?topic=67626.0
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rabinoz

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Re: Revisiting a commonly presented image...
« Reply #321 on: February 15, 2018, 04:10:30 AM »
Let's now take a closer look at the equation provided by Nasa:

https://www.grc.nasa.gov/WWW/k-12/airplane/rockth.html

https://www.grc.nasa.gov/WWW/k-12/airplane/Images/rockth.gif

The first term says Force = Mass x Velocity
No it does not! The first term is m dot and it the mass flow rate of exhaust gases.
Please read the fuller explanation in: NASARocket Thrust Equation.

Quote from: sandokhan

whereas NASA web sites say that
Force = Mass x Acceleration (Newton’s 2nd law of motion):

https://spaceflightsystems.grc.nasa.gov/education/rocket/BottleRocket/journey_newtona2.htm

"NASA sites also say that Mass x Velocity = Momentum which is not a force. Momentum is potential energy. If you throw a rock it has momentum. If you throw it harder it has more momentum. No force is generated until the rock hits something.
Incorrect!
Your claim would mean that a fired artillery round would not cause any recoil until the round stiles some object,  which is clearly ridiculous.

Quote from: sandokhan
Gas shot out of the back of a rocket very fast does not create a force until it interacts with something, which it never does in the vacuum of space.
That is again is totally baseless. There is absolutely nothing in either Newton's 3rd law or Conservation of Momentum to suggest that.
CoM states that the total momentum of an isolated system is conserved. Here the "isolated system" can be taken as the rocket and any expelled exhaust gasses.

Your requirement of hitting something violates this isolated system premise.

Quote from: sandokhan
It remains high momentum gas streaking endlessly through space looking to do work but never getting the chance.
And the high momentum of this gas has to be matched by the same magnitude momentum change in the rocket.

Quote from: sandokhan
The second term (Pressure Difference between inside the rocket and the vacuum of space) x Nozzle Area violates the “free expansion” effect, part of the first law of thermodynamics by which pressurized gas moves into a vacuum without any work being done. It does not matter how highly pressured the gas is inside the rocket nor how fast it comes out. Because it is going into a vacuum the gas makes the trip “for free” and does not do any work, does not expend any energy and does not create any force or thrust.
No, you need to ask just what the system that has no work done on it.

Quote
Joule-Thomson Expansion - Free Expansion of a Gas
Imagine a gas confined within an insulated container as shown in fig 1. The gas is initially confined to a volume V1 at pressure P1 and temperature T1. The gas then is allowed to expand into another insulated chamber with volume V2 that is initially evacuated. What happens? Let’s apply the first law.

We know from the first law for a closed system that the change in internal energy of the gas will be equal to the heat transferred plus the amount of work the gas does, or ∆U = Q + W. Since the gas expands freely (the volume change of the system is zero), we know that no work will be done, so W=0. Since both chambers are insulated, we also know that Q=0. Thus, the internal energy of the gas does not change during this process.
     
Fig 1 Expansion into box
from reference below

From: Joule Thomson



Now, how does this relate to a rocket in space?
For an ideal gas, free expansion does no work, but what does this mean?  It is simply that the temperature of the gas is unchanged during the expansion.  But, the upper diagram does not represent our rocket in free space.  The right half of this should be replaced by "the infinite vacuum of space", more as in fig 2.

Joule-Thomson expansion simply says that the temperature of an (ideal) gas does not change, but this in no way affects Newton's Laws of motion.  The whole system is the rocket plus the "near-infinite vacuum of space".
There is nothing in the Joule-Thomson free expansion to "countermand" the momentum of the gas heading right (in the lower diagram) imparting like momentum to the rocket heading left.

As obviously expected there is no conflict between Newton's Laws and the Joule-Thomson free expansion.
     
Fig 2 Expansion into space
modofied from reference

Quote from: sandokhan
The NASA space rocket equation has two terms the first of which is incorrect and so is the second."

https://www.grc.nasa.gov/WWW/k-12/airplane/momntm.html

https://er.jsc.nasa.gov/seh/f.html
No, neither term is incorrect. Not only that, but it is hardly "The NASA space rocket equation".

The simple version (without the pressure term) formed the basis of Konstantin Tsiolkovsky's "rocket equation", .

Maybe you should read up on both Konstantin Tsiolkovsky in say Konstantin Eduardovich Tsiolkovsky (1857-1935)
and Robert Goddard in Robert Goddard and His Rockets
  or in Rocket Fundamentals.

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rabinoz

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Re: Revisiting a commonly presented image...
« Reply #322 on: February 15, 2018, 04:22:03 AM »

Nasa has never sent a single rocket into outer space: the experimental data simply does not exist to support the free stream pressure equation.
Totally incorrect!

Quote from: sandokhan
It is very easy to prove that each and every NASA space mission was faked.

Really? NASA was not the first to launch space mission and there are now many other countries that have launched them.

Well, you prove that none of the
Soviet Union, United States of America (NASA, Space X or Blue Origin), France, Europe (ESA, Ariane), Japan, China, Britain, India, Russia, Ukraine, Israel, Iran and North Korea
have ever launched any space missions.

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sandokhan

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Re: Revisiting a commonly presented image...
« Reply #323 on: February 15, 2018, 04:34:33 AM »
Nasa has put this equation in front of its readers:

https://www.grc.nasa.gov/WWW/k-12/airplane/rockth.html

https://www.grc.nasa.gov/WWW/k-12/airplane/Images/rockth.gif

The first term says Force = Mass x Velocity

It is being described as the mass flow rate, but the terms in the m dot expression are: mass and velocity.

whereas NASA web sites say that

Force = Mass x Acceleration (Newton’s 2nd law of motion):

https://spaceflightsystems.grc.nasa.gov/education/rocket/BottleRocket/journey_newtona2.htm

A clear contradiction.

As for Nasa having faked all of its space missions, it is very simply to prove this: the Allais effect, the DePalma spinning ball experiment, the Biefeld-Brown effect show that there is no such thing as the law of universal gravitation, supposedly used by Nasa to compute the trajectories.

Dark flow totally disproves Newtonian gravitation.

Please explain how the Nasa space missions took place without using the law of universal gravitation.

« Last Edit: February 15, 2018, 04:36:17 AM by sandokhan »

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Papa Legba

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Re: Revisiting a commonly presented image...
« Reply #324 on: February 15, 2018, 05:10:40 AM »
Lol the rabbibot thinks that an object moving in one direction solely due to momentum is somehow imparting a force in the opposite direction at the same time...

It is mental.
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Re: Revisiting a commonly presented image...
« Reply #325 on: February 15, 2018, 05:16:36 AM »
"NASA sites also say that Mass x Velocity = Momentum which is not a force. Momentum is potential energy. If you throw a rock it has momentum. If you throw it harder it has more momentum. No force is generated until the rock hits something. Gas shot out of the back of a rocket very fast does not create a force until it interacts with something, which it never does in the vacuum of space. It remains high momentum gas streaking endlessly through space looking to do work but never getting the chance.
A rock has momentum after you throw it, which is not a force.  Why are you conveniently forgetting the force being applied to the rock when you throw it?  The act of throwing is analogous to the rocket engine working.  Both are applying force to their respective projectile to move it forward.

Looking at the rock's momentum after the throwing action is complete would be comparable to looking at the rocket's momentum after engine shutdown, and yes, at that point there is no longer any force acting on the rocket. 

Are you intentionally confusing an unpowered object (the rock) with a powered object (the rocket with it's engine working) or do you actually not understand the difference?

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sandokhan

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Re: Revisiting a commonly presented image...
« Reply #326 on: February 15, 2018, 05:56:59 AM »
If gas has an effect on objects in a vacuum we would expect to find an example in nature.

"Saturn's moon Enceladus, for example, shoots a jet of water ice 500 KM into space. The diameter of the moon itself is only 500 KM. Does this jet have any effect? No. The jet as tall as the moon is wide goes harmlessly off into space."

https://www.space.com/22181-saturn-moon-enceladus-water-geysers.html

https://saturn.jpl.nasa.gov/science/enceladus/





« Last Edit: February 15, 2018, 05:59:45 AM by sandokhan »

Re: Revisiting a commonly presented image...
« Reply #327 on: February 15, 2018, 06:04:25 AM »
If gas has an effect on objects in a vacuum we would expect to find an example in nature.

"Saturn's moon Enceladus, for example, shoots a jet of water ice 500 KM into space. The diameter of the moon itself is only 500 KM. Does this jet have any effect? No. The jet as tall as the moon is wide goes harmlessly off into space."

https://www.space.com/22181-saturn-moon-enceladus-water-geysers.html

https://saturn.jpl.nasa.gov/science/enceladus/

Okay?  Does Force still equal Mass x Acceleration?  If so, who cares how tall the plume is or what the diameter of the moon is?  I don't see height or diameter in the only relevant equation.  What is the mass of the water being expelled, what is the acceleration, and what is the mass of the moon?  Once you have those numbers, plug them in and calculate just how much force you think should be applied to the moon and what the resulting movement should be.  If after all that, you still think something is wrong, let us know and we can discuss that. 

« Last Edit: February 15, 2018, 06:10:24 AM by ItsRoundIPromise »

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Re: Revisiting a commonly presented image...
« Reply #328 on: February 15, 2018, 06:21:45 AM »
"NASA sites also say that Mass x Velocity = Momentum which is not a force. Momentum is potential energy. If you throw a rock it has momentum. If you throw it harder it has more momentum. No force is generated until the rock hits something. Gas shot out of the back of a rocket very fast does not create a force until it interacts with something, which it never does in the vacuum of space. It remains high momentum gas streaking endlessly through space looking to do work but never getting the chance.
The act of throwing is analogous to the rocket engine working.  Both are applying force to their respective projectile to move it forward.

The force vector of the exhaust is opposite to the force vector of a hand throwing an object.

So your analogy is the opposite of the truth.

kek
 
The rocket and exhaust are ONE object, creating ONE force vector, rearwards.

When this one force vector strikes the second object of the atmosphere, a second force vector is created in the opposite direction, thus propelling the rocket in that direction, i.e. forwards.

Newton's third law - f1=-f2, two objects, equal and opposite,  action/reaction.

Simple enough...

You're not programmed to understand that though, so on we slog, forever...
I got Trolled & Shilled at the CIA Troll/Shill Society and now I feel EPIC!!!

Re: Revisiting a commonly presented image...
« Reply #329 on: February 15, 2018, 08:00:59 AM »
"NASA sites also say that Mass x Velocity = Momentum which is not a force. Momentum is potential energy. If you throw a rock it has momentum. If you throw it harder it has more momentum. No force is generated until the rock hits something. Gas shot out of the back of a rocket very fast does not create a force until it interacts with something, which it never does in the vacuum of space. It remains high momentum gas streaking endlessly through space looking to do work but never getting the chance.
The act of throwing is analogous to the rocket engine working.  Both are applying force to their respective projectile to move it forward.

The force vector of the exhaust is opposite to the force vector of a hand throwing an object.

So your analogy is the opposite of the truth.

kek
 
The rocket and exhaust are ONE object, creating ONE force vector, rearwards.

When this one force vector strikes the second object of the atmosphere, a second force vector is created in the opposite direction, thus propelling the rocket in that direction, i.e. forwards.

Newton's third law - f1=-f2, two objects, equal and opposite,  action/reaction.

Simple enough...

You're not programmed to understand that though, so on we slog, forever...
The force vector of the exhaust is away from the rocket, so the force vector of the rocket is away from the exhaust.
F1=-F2. 

Simple enough...