This is a very interesting thread, I could read this stuff all day.

I think some aspects kind of come back to "nothing is at rest in the Universe", wouldn't everything already have some momentum as well as inertia ?

We all on Earth are moving through space so we must have momentum.

As far as building vehicles that can carry people to other Galaxies and even to the outer reaches of our own, I'm not convinced it would be safe to power vehicles at high speed through space without some kind of protection from collision with objects ect. Great speed may overcome problems with time but introduce other problems.

We should remember there are different definitions for a "vacuum". I think the only true definition of the word vacuum on it's own is 2, a: in bold, unless a "partial vacuum" is specified or "the vacuum of space" is specified then - vacuum means "a space absolutely devoid of matter", when we are talking of deep space we are talking about a "partial vacuum" to some degree, surely. Aren't we ? The differences between 2,a: and 2,b: would be small but they would exist.

http://www.merriam-webster.com/dictionary/vacuum

Full Definition of VACUUM
1
:  emptiness of space

2
a :  a space absolutely devoid of matter
b :  a space partially exhausted (as to the highest degree possible) by artificial means (as an air pump)
c :  a degree of rarefaction below atmospheric pressure

3
a :  a state or condition resembling a vacuum :  void <the power vacuum in Indochina after the departure of the French — Norman Cousins>
b :  a state of isolation from outside influences <people who live in a vacuum…so that the world outside them is of no moment — W. S. Maugham>

4
:  a device creating or utilizing a partial vacuum; especially :  vacuum cleaner

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Is it even possible to create a true vacuum ? I would be surprised if it is. Is this statement accurate ? In interstellar space, vacuums can approach 1 molecule per liter, which for all practical intents and purposes is perfect vacuum.

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This is a very interesting thread, I could read this stuff all day.

I think some aspects kind of come back to "nothing is at rest in the Universe", wouldn't everything already have some momentum as well as inertia ?

We all on Earth are moving through space so we must have momentum.

As far as building vehicles that can carry people to other Galaxies and even to the outer reaches of our own, I'm not convinced it would be safe to power vehicles at high speed through space without some kind of protection from collision with objects ect. Great speed may overcome problems with time but introduce other problems.

We should remember there are different definitions for a "vacuum". I think the only true definition of the word vacuum on it's own is 2, a: in bold, unless a "partial vacuum" is specified or "the vacuum of space" is specified then - vacuum means "a space absolutely devoid of matter", when we are talking of deep space we are talking about a "partial vacuum" to some degree, surely. Aren't we ? The differences between 2,a: and 2,b: would be small but they would exist.

http://www.merriam-webster.com/dictionary/vacuum
 

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Is it even possible to create a true vacuum ? I would be surprised if it is. Is this statement accurate ? In interstellar space, vacuums can approach 1 molecule per liter, which for all practical intents and purposes is perfect vacuum.

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This is why Dyson is building vacuums in space.  They are perfect...or nearly so. (Some would say they really suck)

Just kidding.

But seriously, I do not think there is anything in nature as a perfect vacuum, sort of like using the word instantaneous.  Neither exist but, some things get pretty close.

No MarkE!  Here's a video demonstration since you apparently don't have the ability to properly visualize something in your head.  The first rock thrown determines the velocity's direction.  The second rock thrown in the opposite direction stops the motion of the boat, and the process repeats itself with a net motion.

IOW: The position is the integral of the velocity, just as I described.

 Remember, in our hypothetical, the mass of the boat remains unchanged after the ejection of the rocks and there is no friction between the boat and the water.  It has nothing to do with the reduced mass of the boat after the rocks are thrown, nor does it have anything to do with the ratios of the masses after the ejection.

There you are absolutely wrong.  Let us use two examples:

Mass of the boat plus occupants:  100kg
Mass of each rock: 2kg
Ejection speed of the first rock with respect to the frame of reference:  10m/s
Case 1: Ejection speeed of the second rock with respect to the frame of reference: -10m/s
Case 2: Ejection speed of the second rock with respect to the boat: -10m/s
Initial condition:  Boat is at rest.

Eject first rock to the right, velocity change to the boat is: 10m/s * 2kg + V_{change_boat} * (100kg + 2kg) = 0  V_{change_boat} = -10m/s*2kg/(102kg)
Boat position now integrates at -0.19608 m/s.
Some time later the second rock is ejected.

Case 1: Rock is ejected left with respect to the frame of reference at -10m/s.
Velocity change to the boat is:  (-10m/s - V_start )* 2kg + V_{change_boat} * 100kg = 0  V_{change_boat} = (10m/s - 0.196m/s)*2kg/(100kg)
Boat position now integrates at -0.19608m/s + 9.804m/s * 2kg/100kg = 0.0.  The boat is stopped.  The first rock is moving rightward with respect to the frame of reference at the same speed that the second rock is moving leftward, the boat is at rest in the frame of reference and the net momentum sums to the initial value of zero.

Case 2: Rock is ejected left with respect to the boat at -10m/s.
Velocity change to the boat is:  -10m/s* 2kg + V_{change_boat} * 100kg = 0  V_{change_boat} = 10m/s*2kg/(100kg)
Boat position now integrates at -0.19608m/s + 10m/s * 2kg/100kg = 0.00392m/s.  The boat is drifting rightward.  The total momentum is:

-10.19608m/s*2kg + 0.00392m/s*100kg +10.0m/s*2kg = 0.0m/s*kg.  CoM is again satisfied and the boat position will integrate from the point that the second rock was thrown rightward at 0.00392m/s indefinitely.

If you had more rocks to start with you could repeat this process electing how hard you throw each propellant rock, and stepping the velocity accordingly.  If you started with 900kg of rocks instead of 4kg the initial velocity change magnitudes would be about 1/10th of the cases above and, the distance travelled in any interval would similarly be about 1/10th of the cases above.  Integral calculus absolutely applies.

  In other words, integral calculus doesn't apply in our hypothetical.  It's as simple as moving the boat 100ft. by throwing a rock, then stopping the boat by throwing a rock in the opposite direction, and then repeating the process over and over again.  Don't let the narrowness of your scope over complicate things by introducing ideas that have no relevancy to a particular case, such as integral calculus, which will cause an improper visualization.

Gravock

Doctor heal thyself.

IOW: The position is the integral of the velocity, just as I described.There you are absolutely wrong.  Let us use two examples:

Mass of the boat plus occupants:  100kg
Mass of each rock: 2kg
Ejection speed of the first rock with respect to the frame of reference:  10m/s
Case 1: Ejection speeed of the second rock with respect to the frame of reference: -10m/s
Case 2: Ejection speed of the second rock with respect to the boat: -10m/s
Initial condition:  Boat is at rest.

Eject first rock to the right, velocity change to the boat is: 10m/s * 2kg + V_{change_boat} * (100kg + 2kg) = 0  V_{change_boat} = -10m/s*2kg/(102kg)
Boat position now integrates at -0.19608 m/s.
Some time later the second rock is ejected.

How does ejecting a 2kg rock from the boat increase the boat's mass from 100kg to 102kg?

Also, why are you using -10m/s before the 2 second rock is ejected to integrate the boat's position at -0.19608m/s?  Is this how you do math MarkE?  Come on, get it right........then we can see how it draws an improper visualization.

Gravock

If the boat had started out with a much bigger pile of propellant rocks, say 900kg, then when the first rock is ejected, the boat would move to the left at: 2kg*10m/s / 998kg = ~ 0.020m/s, roughly one tenth the speed as in the first case.  As the pile of propellant goes down, the change in velocity of the boat with each uniform toss gets larger and larger.

If you think I have made any mistakes, then you are free to show your own work.

According to you, the boat is oscillating back and forth away from the starting point after both rocks are ejected, and the distance the boat travels between the oscillations grows larger and larger with each additional cycle of rocks being thrown.  This is not in-line with the original hypothetical as shown by the video demonstration.  Your words have already proven it gives the wrong visualization.  Thank you!

Gravock

@all,

In the hypothetical, the process proceeded without the loss of the boat's mass after the ejection of the rocks.  MarkE however wants to ignore this part of the process in the hypothetical so he can give a false perception that the use of integral calculus absolutely applies.  However, by doing this, he'll form a wrong visualization.  Velocity, mass, and time has no relevancy in the hypothetical, since the second rock is thrown after the boat travels a distance of 100m.  The second rock thrown will then stop the boat.  So, there is no reason to use the position to integrate the velocity in regards to the original hypothetical.

Gravock

You have not shown any error in my analysis or math.  You cannot eject mass from something and still have the same mass as when you started.  Position is the integral of the velocity.  The velocity is in turn the integral of the acceleration.  The boat accelerates due to reaction force from accelerating each propellant rock as it is ejected.  Each rock that is ejected reduces the remaining mass on the boat.  As I have shown: the net change in velocity that results from each ejection gets bigger and bigger as the boat plus remaining propellant mass shrinks.  Where the mass of each rock is small with respect to the mass of the boat the magnitude of adjacent velocity steps is small but real.

MarkE,

In a hypothetical, you can eject mass from something and still have the same mass as when you started, thus the reason it's called a hypothetical (are you really this disconnected?).

LOL, you eject the mass and keep it do you?

  In the real world, a person in a second boat could be placing rocks in the boat as they're being ejected in order to maintain the same mass of the boat.

There are multiple problems with that proposal.  The first is: Some energy must be expended in order for them to track the movements of the first boat so that they are in place to transfer new rocks.  The second is that the act of transferring rocks from one boat to the other effects a momentum exchange.

  As I have shown in my previous post, the velocity, mass, and time are irrelevant in the hypothetical since the mass of the boat is unknown and remains unchanged during the ejection of the rocks.

You have done no such thing.  The mass of the boat includes the propellant that remains at any given moment.  If the boat starts out with a single 2kg rock or a million 2kg rocks, the momentum of the laden boat changes equally and oppositely to the momentum change of the ejected rock.  The smaller the mass is of the ejected rock to the remaining mass of the boat plus all of its contents, the smaller the proportional velocity change of the boat versus the ejected rock.

The only thing given in the hypothetical is the distance the boat travels before the second rock is thrown.  How are you going to apply integral calculus when the only values given in the hypothetical is the distance?

You are so completely lost.

Gravock

Miss Information is being joined by Miss Direction today.