Momentum Not Conserved, in the use of a medieval battle flail

Not at infinite (or acceptably asymptotically zero gravity distances).

Again, not unless it's caught by gravity.

Right, creating an energy that wasn't in the system to begin with.

Again, energy that wasn't in the system to begin with. You like to calculate. Calculate the initial energy in the system.

Ideally, but not very efficiently in non-relativistic collisions.

No, it's not. You've as much as stated above that the force of gravity induces kinetic energy from the system. Energy that wasn't there either negatively or positively to begin with!

Calculate it.

I didn't say that. That's your analogy of my system.

Which is a problem because the attractive force comes from a source outside the system. It's not isolated! The marbles aren't doing the work!

If the top is zero, then the bottom is what? It'd have to be more than zero! Neither marble begins outside of the attractive force.

Okay.

It's not analogous to my model at all. As stated above, neither ball starts from a point outside of the attractive force. Gravity is as constant at the zero distance. as it is in the depression.

I think you should be smart enough to see the weaknesses of your analogy.

This post has been edited by ubavontuba on Sep 8 2007, 07:26 AM

aim: To demonstrate that Energy is conserved in a closed system, irrespective of the height of the mass..

definitions:

E_p = Gravitational Potential; enegry of a body.
E_k = Kinetic energy of a body.
M_a = Mass of body being accelreated
M_e = Mass of the Earth
g = Acceleration due to Gravity
h_2 = Height above the Earth.
r_∞ = Gravitational Inifinity the point at which a body can be considered to be infinitely distant from the Earth.

Starting point/assumptions:

1) f = ma
2) f = G.m_1.m_2/r^2
3) E_p = mgh
4) E_k = ½mv^2
5) v = d/t
6) a = (v_f - v_i)/t

Proof:
Derive E_k, E_p, and g as functions of height.

7) a=g; m=m_2=M_a; m_1=M_e; r = h_2

Substitute 7 (where appropriate) into 1-6 (as appropriate)

Substitute 1 into 2 and solve for g.

9.1) M_a . g = (G . M_a . M_e)/h_2^2
9.2) g = G.M_e/h_2^2

Substitute 9.2 into 3.

10.1) E_p = M_a.g.h_2
10.2) E_p = M_a.((G . M_e)/h_2^2).h_2
10.3) E_p = G.M_a.M_e/h_2

Rearrange 6, substitute 5 into 6, and solve for d to obtain:

11) v = √2gh

Where h = the distance fallen (r_∞ - h_2)

Substitute 11, and 9.2 into 4.

12.1) E_k = ½mv^2
12.2) E_k = ½M_a.(√2g(r_∞ - h_2))^2
12.3) E_k = M_a.g.(r_∞ - h_2)
12.4) E_k = M_a . (G.M_e/h_2^2) . (r_∞ - h_2)
12.5) E_k = (G.M_a.M_e.r_∞/h_2^2) - (G.M_a.M_e/h_2)

Pay close attention to 12.5, compare it to something. What does it remind you of? Let me show you.

Substitute 9.2 and 10.3 into 12.5

13.1) E_k = (G.M_a.M_e.r_∞/h_2^2) - (G.M_a.M_e/h_2)
13.2) E_k = (G.M_a.M_e.r_∞/h_2^2) - E_p
13.3) E_k = g.M_a.r_∞ - E_p
13.4) E_k + E_p = g.M_a.r_∞

Now, if g was a constant, which is an acceptable approximation for most common applications, then we would have just proved that Potential Energy + Kinetic Energy = Constant, but, we all know that g varies with height, so let's put it on the left hand side with the other things that vary with height, and see what we can do.

14.1) (E_k + E_p)/g = M_a.r_∞
14.2) (E_k/g) + (E_p/g) = M_a.r_∞

We know from 10.1 that:

15) E_p/g = M_a.h_2

Rearranging gives us:

15.1) g = Ep/M_a.H_2

Substituting 9.2 into 13.1 tells us that:

16.1) E_k = g.(M_a.r_∞ - m_a.h_2)
16.2) E_k = g.M_a.(r_∞ - h_2)
16.3) g = E_k/M_a.(r_∞ - h_2)

So, now we know that

17) E_k/M_a.(r_∞ - h_2) = g = Ep/M_a.H_2

If we substitute 15.1 and 16.3 into 14.2 we discover that:

18.1) E_p/(E_k/M_a.(r_∞ - h_2)) + E_k/(Ep/M_a.H_2) = M_a.r_∞
18.2) (E_p.M_a.(r_∞ - h_2))/E_k + (E_k.M_a.H_2)/E_p = M_a.r_∞
18.3) (E_p^2.M_a.(r_∞ - h_2) + E_k^2.M_a.h_2)/E_p.E_k = M_a.r_∞
18.4) (E_p^2.M_a.r_∞ - E_p^2.M_a.h_2 + E_k^2.M_a.h_2)/E_p.E_k = M_a.r_∞

Examining 13.2 we find that

19.1) E_k = (G.M_a.M_e.r_∞/h_2^2) - E_p
19.2) E_k + E_p = G.M_a.M_e.r_∞/h_2^2
19.3) h_2^2 . (E_k + E_P) = G.M_a.M_e.r_∞
19.4) r_∞ = (h_2^2 . (E_k + E_P))/G.M_a.M_e

Substituting 19.4 into 18.4 we get

20.1) (((E_p^2.M_a).(h_2^2.(E_p+E_k)/GM_a.M_e)) - E_p^2.M_a.h_2 + E_k^2.M_a.h_2)/E_p.E_k = M_a.r_∞
20.2) (E_p^2.h_2^2.(E_p + E_k)/E_p.E_k.G.M_e) - (E_p.M_a.h_2/E_k) + (E_kM_a.h_2/E_p) = M_a.r_∞

Rearranging 10.3 and solving for h_2 we get

21) h_2 = G.M_e.M_a/E_p

Substituting 21 into 20.2 puts us on the home stretch, and gives us

22.1) (E_p^2(G.M_e.M_a/E_p)^2.(E_p + E_k)/E_p.E_k.G.M_e) - ((m_a.E_p.(G.M_e.M_a/E_p))/E_k) + ((E_k.M_a.(G.M_e.M_a/E_p))E_p) = M_a.r_∞
22.2) ((G.M_e.M_a^2.(E_p + E_k))/E_p.E_k) - (G.M_e.M_a^2/E_k) + ((G.M_e.M_a^2).E_k/E_p^2) = M_a.r_∞
22.3) G.M_e.M_a^2.(((E_p + E_k)/E_p.E_k) - (1/E_k) + (E_k/E_p^2) = M_a.r_∞
22.4) (((E_p + E_k)/E_p.E_k) - (1/E_k) + (E_k/E_p^2) = M_a.r_∞/G.M_e.M_a^2
22.5) (E_p^2 + E_p.E_k - E_p^2 + E_k^2)/E_p^2.E_k = r_∞/G.M_e.M_a
22.6) (E_p.E_k + E_k^2)/E_p^2.E_k = r_∞/G.M_e.M_a
22.7) E_k.(E_p + E_k)/E_p^2.E_k = r_∞/G.M_e.M_a
22.8) (E_p + E_k)/E_p^2 = r_∞/G.M_e.M_a

22.8 is the money formula.
22.8 proves that I was right, as was Dallas
22.8 proves that at every EVERY height there exists some constant that represents the total of the Potential Energy and the Kinetic energy, and (usefully) relates this number to the other constants of the system.

Holding the asteroid 'beyond infinity' then strapping a booster pack on to it, and igniting it does not invalidate this formula if you realize that at any time (or height) the total kinetic energy of the mass being accelerated by Gravity and the Kinetic energy that was provided to it by the rocket pack, so in the equations we replace the measured E_k with E_k - E_ko where E_ko is the kinetic energy provided by the rocket pack.

For the record? This is not copied from a website. I have no idea if this appears anywhere on the web (and really, I'm too lazy to go looking - Ironic huh?) And it's derived entirely from first principles, assuming that equations 1-6 are correct, and that there exists some r_∞.

I'm willing to discuss relevant questions, but, there's your proof.
And no, it hasn't taken me 4 days to dream this up. It took me 3 days to find the motivation through curiosity to attempt the proof in the first place, and it took me about 6 hours of work to get through the derivations, INCLUDING a 75% complete previous attempt to post the proof here, and a whole bunch of wrong terms, dead ends (including a couple that ended up proving that 1=1) and explorations out of curiosty.

For the Record:

(Kg.m^2.s^-2)/(kg^2.m^4.s^-4) = m/m^3.kg^-1.kg.kg.s^-2
1/(kg.m^2.s^-2) = 1/(kg.m^2.s^-2)

So, because I'm still to lazy to go and try and find any proofs on the web, I have performed a dimensional analysis (illustrated above) which should prove (to anybody) that the formula I have derived, is in fact correct.

Very nicely done!
Unfortunately it will blow right over Eric's "the tuba" bald spot. Watch his answer

I can hear circkets...

And I know Ubavontuba is still around, because he's posting on other threads...

And yet no response from him here...

Give me a break. I allowed you more than a full four days for your reply without comment. I'd decided to give you as long as you wanted. If you didn't respond, I was merely going to drop this thread as I haven't a lot of time lately to keep up with it. Notice that I also left the CERN thread alone too.

Okay then. A crack of the knuckles and away we go...

Let's use your 22.8.

Would you mind plugging in some values for me and telling me what you get?

In consideration of my thought experiment, let's start with two equal masses (you can choose any reasonable value you like).

Let's start with zero relative momentum.

Let's start with asymptotically zero gravity.

Please calculate the total system energy.

Actually...

Yes, I would.

I simply can not be bothered doing your work for you.

If you question specific steps? State the step and why.

Otherwise, be graceful, and admit defeat.

If you want to plug a series of values into the equation, and test them to see if they're right? Feel free to do so, but bare in mind that if you claim I have made an error, I'm going to want to see what values you fed into the equation, and your justification for using those specific values.

The other thing to bare in mind, is that as a final step I should have inverted the equation, because at the moment it will give you an answer of 1/Joules.

Or have you forgotten how to use a calculator/spread sheet?

Do I detect a copout?

You're really that stupid that you think taking the time to type out a three page proof of why you're wrong is a cop out?

Are you really that scared of the math that you can't do a little bit of finger work for yourself?

I'm sorry, but it sounds to me like you're the one that's coping out.

I'd like to demonstrate that you're wrong with your own math, performed by you, checked by you, and rechecked by you. That way, when you see the startling conclusion you will have no recourse but to accept it.

Let's see it , Eric. Should be fun, more of your trying to pass your comical blunders as "physics".

I smell a cop out.

If there's a problem with the logic. Name the step.
If there's a problem with one of my equations. Name the step.

Otherwise, quit the bluster, and admit you were wrong.

right....

Posting 56,000 points of data broke the system

I have as a spread sheet, the data for something falling from 8 million kilometers towards the earth.

I can upload it for you if you really want to see it.

Buggered if I know why I'm doing your work for you, but it does prove that you're completely and utterly wrong.

So here's some selected points.

8 million km was pretty arbitrary, but I chose it because it gave g in the mm/s/s range, and f in the mN range. This is for a 1kg mass falling towards the earth.

So, you ready to admit you're wrong yet? Or do I need to start posting graphs?

This post has been edited by Trippy on Sep 21 2007, 10:24 AM

And, because I have no doubt that you'll cry and moan and complain and try and argue that because the last data set deals with a 1kg mass falling towards the earth, that it's some how irrelevant and meaningless and doesn't validate my claims because it doesn't represent your 'thought experiment'.

Here's two 1kg masses falling towards each other from a distance of 10km, tabulate every 10m.

notice, again, how the last two colums are constant, and the same all the time?

Now, are you ready to admit that you were wrong yet?

Edit: Ugh, the formating's all screwed up, but, I don't really care.

This post has been edited by Trippy on Sep 21 2007, 11:29 AM