# Work done separating 2 masses -gravitation

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

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1. ### Aqueous Idflat Earth skepticValued Senior Member

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Robb

I got lost when an extra radius popped up.

I thought you were going from zero to 100 MLY

the integral is evaluated between two points 100 MLY apart

I thought that's what you asked

and beyond that, the problem is unfathomable, because we can't be at zero

so what's the other position, the one that can't be zero?

because it's the one that's the killer as far as work is concerned.

that inverse square law is a beast at close range

for two bodies of mass m1 and m1 kg, starting from a separation of a meters and moving to a separation of b meters will require

-G m1 m2 (1/b-1/a) Joules of work to get from a to b. (a>0)

Last edited: Nov 22, 2011

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3. ### OnlyMeValued Senior Member

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It was a joke Tach. You so often demand others use math to prove their point that I just asked for the math...

At least.., I think you finally got it!

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5. ### ReikuBannedBanned

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I can sit in the space of an hour and type about 8 times as much as he did. It was such an elementary integral, the only question I have now is why it took him so long.

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7. ### ReikuBannedBanned

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Oh we covered this Tach. You were the one who came in here poking fun at the OP for knowing how to conduct an integral. Now if I can post equations describing how to swap timelike and spacelike events, or I know the mathematics behind neutrino's, dirac particles, photon behaviour, general relativity then I doubt anyone would believe I don't know how to conduct an integral on something.

You on the other hand, can mess up integrals and have done in the past with Pryzk. I don't need to troll. If I have a question, I am honest enough to ask. You on the other hand prance around as if you know it all. For all we know, you submitted a question to Ask an Astrophysicist in the many hours of debating this without putting up anything.

Last edited: Nov 22, 2011
8. ### Robittybob1BannedBanned

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If "Zero" is a point in the centre of the neutron, well it doesn't come from there but still emanates from below the surface of the neutron. It would be like taking something from the upper core - mantle boundary of the Earth and taking it into space. It will require some work to dig it out and bring it to the surface. I hadn't taken that into consideration before either. I think there is an estimate for the size of a proton. R1 maybe about 0.6 of the radius of a proton.
R2 would depend on the energy of the graviton.
This is all very hypothetical but challenging.

I think it surprised you a little when you thought of us being partly here on Earth and also a thin filament in space extending to max of 100 million light years that may or may not connect. But wasn't that the same with Newtonian gravity and with the String Theory in those cases everything was being connected. But at least in my concept it was kept to a single galaxy and not the entire universe.

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9. ### PeteIt's not rocket surgeryRegistered Senior Member

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The two expressions are actually equivalent, except for a change of sign. The key is that r1 and r2 in Trippy's post are not the same as R1 and R2 in Tach's post.

In [post=2860816]Trippy's post[/post], which describes the general case:
r1 = initial distance between centres
r2 = final distance between centres

In [post=2860920]Tach's post[/post], which describes the special case in the OP:
R1 = Earth radius
R2 = Radius of the other mass.
R1+R2 = initial distance between centres
L = final distance between centres

Substituting:
r1 = R1 + R2
r2 = L
We find that:
$Gm_1m_2\frac{r_1 - r_2}{r_1.r_2} = -Gm_1m_2(\frac{1}{R_1+R_2}-\frac{1}{L})$

10. ### ReikuBannedBanned

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Oh yes!

Well spotted Pete.

11. ### Robittybob1BannedBanned

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Thanks for the connection between the two versions. They may never stop being sparring bulls.

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I'll try putting Trippy's version into an excel workbook and see if there are any sensible results from it. I'll be able to vary M1 and M2 on the condition M1 + M2 = Mn (Neutron mass) Ri is 0.6 of the radius of a neutron, G = Gravitational constant. so the output will be the R2 factor. Here goes!

Last edited: Nov 22, 2011
12. ### Robittybob1BannedBanned

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The results do not confirm my hypothesis - Failed. Thanks

13. ### TrippyALEA IACTA ESTStaff Member

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The only digging is yours. I did not associate m[sub]2[/sub] with the mass of the earth, explicitly, or implicitly, except for a side note where I pointed out how the equation relates to reality.

You're wrong, and you're the only person that can't see it.

Deal with it.

14. ### TrippyALEA IACTA ESTStaff Member

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Use mine, in spite of what Tach says it's correct - ignore Tach, his objections are based on emo-rage. At all steps in my derivation m[sub]1[/sub] and m[sub]2[/sub] were refering to the masses of your two test masses, and in all cases r[sub]1[/sub] is the initial distance between them, and r[sub]2[/sub] is the final distance between them.

I really don't know what's crawled up Tach's ass.

Maybe it's because I'm just a chemist.

15. ### TrippyALEA IACTA ESTStaff Member

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Thankyou Pete.

16. ### TrippyALEA IACTA ESTStaff Member

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And awaiting the public apology from Tach on the grounds that the only thing I have had to say about his commentary was that he was wrong about mine - which has been demonstrated by Pete.

17. ### Robittybob1BannedBanned

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I used your equation but the amount of energy in the gravitational potential would not have been enough to stop the "graviton" going forever. If anything it is the Strong Force factor that would need to be deducted first. Newtonian physics did not account for any restricted gravitational range.

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18. ### TachBannedBanned

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R1 is not the radius of the Earth, it is simply the radius of the first body. The OP stipulated that the experiment is conducted far away from the Earth. As an aside, this stipulation excludes any derivation using stuff like $E=mgh$, a simple one line integration of the variable force $f=\frac{Gm_1m_2}{r^2}$ is sufficient, as already explained in post 2. As such, there was no justification for Trippy to instruct the OP to discard all the pre-existing solutions as incorrect.

..meaning that the mechanical work in Trippy's proof is negative, when in reality, is positive, $W=Gm_1m_2(\frac{1}{R_1+R_2}-\frac{1}{L})$.
This is obvious, since one needs to expend energy in order to separate the two bodies.

Last edited: Nov 22, 2011
19. ### TrippyALEA IACTA ESTStaff Member

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This is because I got r[sub]1[/sub] and r[sub]2[/sub] the wrong way around in the final equation.

oops

Meaning nothing more than this step:
Should have been:
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 $E_p_1-E_p_2$. which gives us:
4.1 $w= \frac{Gm_1m_2}{r_1} - \frac{Gm_1m_2}{r_2}$

:shrugs:

And the final equation becomes:
4.2 $w= Gm_1m_2\frac{r_2 - r_1}{r_1.r_2}$

Which, in reretospect I should probably have spotted sooner.

Oh well, mea culpa.

You still have some serious apologizing to do though Tach.

Last edited: Nov 22, 2011
20. ### Robittybob1BannedBanned

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Modified quote above. Are the substitutions that Pete said he made are they valid? Did anyone check these substitutions?

I couldn't use the second equation for there was no such thing as the radius of the second body. What is the radius of a graviton?

21. ### TachBannedBanned

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So, you bungled your solution, were rude to me:

in the context of my solution being correct all along and you still demand that I apologize to you?

Last edited: Nov 22, 2011
22. ### PeteIt's not rocket surgeryRegistered Senior Member

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Tach's got a point, Trippy.
In this thread, the flaming between you and Tach began with you.

Yes, Tach misinterpreted your post and said it was wrong for the wrong reasons.
But that was a simple mistake which didn't warrant a rude response.

Yes, Tach is never backward in escalating a flame war, and we know there's recent history involved... but I think we all have a responsibility to at least try to maintain politeness as best we can (though god knows it can be bloody hard!)

Anyway... my advice is for both of you to just let it go, or apologize to each other.

(I know, I know... I'm no saint either.)

23. ### TrippyALEA IACTA ESTStaff Member

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I bungled nothing. I made a minor that changed the sign, but not the magnitude of the work done.

You FAILED to correctly indicate my error.
You FAILED to understand my derivation.
You FAILED to see how your answer related to mine.
In short you FAILED to correctly understand the prose content of my post, and you FAILED to understand the quantitative content of my posted.

You asserted on two seperate occasions that I was wrong, but neither of those assertions were correct, and my rudeness was in response to those false assertions.

Not one of those assertions has been born out, so yes, I think I am entitled to an apology, and when you apologize, publicly, I will consider apologizing for my rudeness.