Work done separating 2 masses -gravitation

Discussion in 'Physics & Math' started by Robittybob1, Nov 21, 2011.

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1. Robittybob1BannedBanned

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I'm not strong on maths.

3. TrippyALEA IACTA ESTStaff Member

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First piece of advice - ignore most of what's been said in this thread so far.

Second piece of advice. You're close, but there's two further points that you need to keep in mind.

Work done is equal to the change in energy of the system.

Second point you need to keep in mind, is that while you're right about g, you're wrong in a broader sense.

1. $f=\frac{Gm_1m_2}{r^2}$
Sub in:
2. $f=m_1a$
to get:
3.1 $m_1a=\frac{Gm_1m_2}{r^2}$
Solve for a to get:
3.2 $a=\frac{Gm_2}{r^2}$

The first thing to realize about this equation is that it demonstrates mathmatically why all objects fall with the same acceleration in a vacuum.

The second thing to realize is that the g in E = mgh is really just a 'special case' where m[sub]2[/sub] us the mass of the earth, and r is it's average radius.

Why is this important? Because it allows us to realize that eqn 3.2 allows us to calculate g at any height above the earth. And why is that important? Because it allows the :idea: moment that allows us to realize that 3.2 allows us to calculation the gravitational acceleration experience by any mass towards some other mass at any distance.

And why is that important? that allows us to calculate the change in GPE seperating m[sub]1[/sub] to m[sub]2[/sub] from any arbitrary distance, to any arbitrary distance for any combination of masses.

So, the general case is:
First we set r=d, and substitute into E=mgh to get:
3.3 $E_p=m_1gr$

Next, we realize that g=a, and substitute 3.2 into 3.3 to get:
3.4$E_p=m_1.(\frac{Gm_2}{r^2}).r$
Which simplifies to:
3.5$E_p=\frac{Gm_1m_2}{r}$

So, at some distance $r_1$ we have $E_p_1$ and at some distance $r_2$ we have $E_p_2$, and the minimum work required is $E_p_2-E_p_1$. which gives us:
4.1 $w= \frac{Gm_1m_2}{r_2} - \frac{Gm_1m_2}{r_1}$

Which after some 'jiggery-pokey' and 'hocus-pocus' rearranges to give us:
4.2 $w= Gm_1m_2\frac{r_1 - r_2}{r_1.r_2}$

Which I will admit, initially left me with some mild skepticism, until I realized that all it was saying was that lifting a 2kg mass requires more work than lifting a 1kg mass, and that lifting a 1kg mass on Jupiter requires more work than lifting a 1kg mass on earth. Both of which are correct.

Now, this is the minimum work. The total work should also include force used to accelerate and decelerate the mass, both of which are trivial to calculate (it's just w=f.d) where f is the force used to accelerate/deccelerate, and d is the distance over which it is done.

Last edited: Nov 21, 2011

5. OnlyMeValued Senior Member

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As far as we can tell, gravitational attraction is infinite. We can observe the interaction of galaxies and galactic clusters where the distances approach if not exceed the 100 million LYs you suggest.

To the best of my knowledge gravitation cannot be applied at subatomic scales. The most likely reason is that at those distances electromagnetic and nuclear forces exceed any gravitational influences. Gravitation seems at present to be insignificant at subatomic scales.

I'm pretty sure gravitation has been extended to the scale of an atom and remained consistent with macroscopic experience. But nothing below that scale seems either experimentally practical or functionally observable.

The hypothetical you propose involving a neutron, is inconsistent with everything that we current understand about subatomic particles. Theoretically you could separate a neutron into an electron and a proton (and an anti neutrino). At that time you would have electromagnetic or charge related forces far exceeding any gravitational interaction. Trying to divide a neutron in any other manner would result in short lived particles like pions, muons, neutrinos and perhaps photons. The exact composition and mixture would be better described by someone more familiar with QM than I.

7. ReikuBannedBanned

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Awwww... Trippy why?

I was wanting to see if Tach was even capable of doing this ''integral?''

8. TrippyALEA IACTA ESTStaff Member

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Sorry :3

I'm just a chemist, what do I know, maybe I should stick to water quality eh? Besides, I think the fact that he spent the best part of a page evading doing so is fairly indicative of the fact that even if he can, he won't.

9. ReikuBannedBanned

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Not at all. In the end, you were much more helpful I guess than Tach. Atleast you didn't come in and poke fun at Robbitybob for not being able to do it. :shrug:

10. TrippyALEA IACTA ESTStaff Member

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It sometimes amuses me, many physicists seem to view chemistry as being the wet and dirty part of physics. To the point where there is an individual on another forum I frequent that takes every opportunity to run them down.

But I recently came accross a quote that I like:
"Chemistry has been termed by the physicist as the messy part of physics, but that is no reason why the physicists should be permitted to make a mess of chemistry when they invade it."

-Frederick Soddy

11. ReikuBannedBanned

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I think chemistry is a fascinating subject... I think sometimes people forget how much physics is really in chemistry; the same goes for Biology.

In fact, the way I view Chemistry and Biology is akin to the ''mechanical nature'' of a much more primordial understanding of the nature of physics itself. Without either subject, physics would not have been able to develop as easily as it has.

12. ReikuBannedBanned

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It was only in hindsight of physics was I able to distinguish the more quantum mechanical aspect which would rear it's head in a Biology class. I remember my first year of college studying biology and I remember the teacher almost shuddering to explain what a photon was and how it effected the cells of a plant. No one in the class even knew what a photon was, apart from myself.

13. Robittybob1BannedBanned

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There ultimately will be the unification of the forces (I have seen some attempts at saying the strong force and gravity are the same on the quantum scale (I'm only saying what I have read). So as the strong force had a limited range so too might gravity. Someone suggested that 200 mly is the extent after which galaxies are free of each other. (Which made me feel game enough to bring this matter up for that could be one proton field connecting with another 100 + 100 = 200mly.
Look it is unlikely that I'll pursuit this investigation, but it was most interesting, for the other side of it was since I "understood" the structure of a proton I could "see" how the electron-proton interaction was accomplished. But that is another story, another day.

Now you talk of dividing a proton up into parts, but that is atomic accelerator stuff. What I imagine happening to a proton/neutron is just the way it is.

14. OnlyMeValued Senior Member

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It was you that reduced this to splitting a neutron into parts...

Trippy, gave you the mathematics required to solve the problem when dealing with "objects"..., having a mass and atomic and molecular structure.

I don't think that same approach can be applied to subatomic particles, let alone the component parts of subatomic particles.

15. Robittybob1BannedBanned

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I can see how you are right, I did write that, but half the problem with someone as green as I am is describing things accurately. When I say it "splits" it is something like a lizard poking it's tongue out. It goes out along long way but it is still attached. Where as the idea I got from what you wrote was like they dissected the lizard.
As to whether it actually happens in real life like I was trying to calculate in 1998 I have no idea. I was just passing on my observations. Someone can check it out sometime.
Now I will look at Trippy's work (thanks a lot Trippy) a little later for I've exhausted myself with all this science, and I'm slightly burnt out.

16. TachBannedBanned

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You should have been skeptical because it is wrong. Remember the initial conditions:

So, the correct answer is :

$W= \int_{R_1+R_2}^{L} \frac{Gm_1m_2}{r^2}dr=Gm_1m_2(\frac{1}{R_1+R_2}-\frac{1}{L})$

because the integral goes from $R_1+R_2$ (the spheres touching each other) to $L$ (L being the final distance between the spheres centers. When $L-> oo$ you have $W->\frac{Gm_1m_2}{R_1+R_2}$

Last edited: Nov 21, 2011
17. TrippyALEA IACTA ESTStaff Member

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While you're attenting those remedial calculus classes, you might want to get your reading comprehension checked.

18. OnlyMeValued Senior Member

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Where's the mathematical proof?

19. TrippyALEA IACTA ESTStaff Member

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Prove it.

Illustrate which of these steps:

Is reliant on m[sub]2[/sub] being the mass of the earth (answer - there isn't one)

Explain to me how this statement:
Indicates that m[sub]2[/sub] as considered in equation 3.2 is the mass of the earth,
Answer - it doesn't, in fact it implicitly contradicts your assertion by making the point that g=9.8 m/s/s is simply a special case of equation 3.2

Or to put it another way, it explains how to derive g from equation 3.2

At no time has any assumption I made required that m[sub]2[/sub]=m[sub]e[/sub] - except where I've explicitly used that assumption to explain the physical meaning of the equations.

Your point is moot, and I still urge you to seek a rebate for any remedial classes you may have paid for.

20. TachBannedBanned

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Why do you keep digging yourself deeper and deeper?

21. TachBannedBanned

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Here. You can't recognize a simple calculation for mechanical work via integration?

22. Robittybob1BannedBanned

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I'm hoping there eventuates an equation I can use. I get the feeling I must have used the wrong equation in the first place for it wasn't that difficult!

23. OnlyMeValued Senior Member

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You are looking at the wrong thing again! As usual....

The relevant exchange was,