# 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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Can anyone help out with the equation for the work done separating two masses gravitationally attracted through a distance?
It is a variation of "E = mgh" but g there is the g (9.8) of the earth and we are looking at the situation where it is no where near the earth.

3. ### TachBannedBanned

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you need to overcome the attraction force between the two bodies:

$f=\frac{Gm_1m_2}{r^2}$

The total work is the integral between limits $r_1$ and $r_2$ of the elementary work $f dr$. Do you know how to calculate integrals?

5. ### Robittybob1BannedBanned

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No I don't know about integrals. But if the two masses are in contact at the beginning and then separate to a distance of 100 million light years how does one integrate that?

Last edited: Nov 21, 2011

7. ### ReikuBannedBanned

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Are you even sure you know how to conduct an integral TACH? You messed up in your marvelous involving the integral in a discussion with Pryzk.

8. ### TachBannedBanned

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Actually, I did not, Green Destiny. It is going to be interesting to see how long you'll manage to evade your permaban, sockpuppet.

9. ### ReikuBannedBanned

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By the looks of things, TACH, you might be well on your way for one youself if you keep wasting their valuable time down there, continue to delete posts when you have spotted an error in your own work, or keep back-tracking on your pathetic attempts to explain a physical set-up. You don't even comprehend some of the more easier physical set-up's which the likes of RJBeery can even catch you out on.

Damn, I'd even place money on Pincho Paxton to give one of your threads a little look over to see if there are any contradictions... mind you, he might not suceed. You are infamous to deleting those posts

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11. ### TachBannedBanned

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That's too bad, you will need to learn first. Is this homework? What grade are you in?

12. ### ReikuBannedBanned

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Oh no, you don't get off that easy TACH. Let's see if you can put your money where your mouth is?

this is a physics forum where people come to get answers. You can't prance around saying ''oh, do you know how to do an integral?'' And then not even offer to give any valuable help.

You seem to have forgotten what a forum is for. So without further ado, please, show us how you would perform this integral. Don't just say you can. You do a lot of talking but never putting up a lot of the time. If not, don't bother going into someone elses thread and question them on what they can or can't do, if you can't even do it yourself.

13. ### ReikuBannedBanned

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Don't try and weasle your way out of pretending you know how to do something.

You are the one prancing about saying you know s**t, so put up or shut up! Admit to Robbitybob that you where just poking at him for fun of not knowing integrals and that you also had no intentions of helping?

You don't need to admit this to me. I saw the direction your post was going, hence why I stepped in.

14. ### ReikuBannedBanned

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No Tach, that's called passing the buck. Can you find any other arguement than stealing my own? I said you should put up or shut up. You are the one with a stick, who came in here saying ''Oh, can you do integrals?''

Then when he honestly replied ''no'' You basically said ''oh too bad, go get an education.''

I have nothing to prove except your own fallacies. I am confident in what I know. I want to see if you do! Your blatent misuse of what a forum is for makes me angry.

15. ### Robittybob1BannedBanned

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Can anyone help out with the equation for the work done separating two masses gravitationally attracted through a distance?
It is a variation of "E = mgh"

16. ### Robittybob1BannedBanned

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therefore,

U = -M * Integral G*m/r

Where r = 0 to 100 million light years.

17. ### Aqueous Idflat Earth skepticValued Senior Member

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Getting back to the question, I think he wanted to confirm mgh.

The answer is yes, force times distance equals work.

18. ### OnlyMeValued Senior Member

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If this is homework the following may be of no use and could even confuse the problem. If it is homework it might be helpful to provide a better description of the problem.

If what you are asking for is the total force or work done, when moving an object or mass from the surface of a significant gravitational object or the earth, to a distance some specific distance from the gravitational source or the earth.., and there are no other forces or gravitationally significant objects involved and assuming that even at that distance there remains some interaction between the two masses....., there are two answers...

1 determine the force require to accellerate the mass to be moved away from, in this case the earth, at the escape velocity required to overcome the Earth's gravitational interaction with the object. In this case the object will with a single accelleration coast to the desired distance of separation.., and then continue moving away from the earth.

2 taking the above determined force, subtract the effective force of the gravitational interaction between the earth and the object at the desired distance and you will have the force required to move the object, to the specified distance..., at which time it will come to rest and the mutual gravitational interaction between the two objects will draw them back together.

The only way to address this issue is in the abscence of all other forces except those between the two masses involved and there is no solution that does not involve a final stable orbit that does not result in the smaller object either continuing to move away from, or be attracted back toward the larger object.

19. ### Robittybob1BannedBanned

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This not a homework problem, but some concept I have discussed earlier. on another thread.

That is exactly the solution. The problem was thinking of related to a neutron that is able to be split into its components. Will the gravitational attraction of the parts pull them together again. But just think of these parts as mass. and one part (the smaller mass) is converted to energy how far will that energy go before it is pulled back down by its own gravitational attraction to the remaining part?

I have a feeling my original calculations were wrong (but I don't know where these are now) but I remember the result and the height I got was 100 million light years.
It is a bit like the Schwarzschild radius of a black hole, what is the Schwartzschild radius of a singular neutron?
I just see now there is an article on the web called "The Schwarzschild Proton". I haven't read it.

But I was thinking, a graviton may be an extension of mass from a proton or neutron that extends into space and then drawn back, just like a photon might till it is drawn back by the gravitational pull of a Black Hole.
But the physics of the work I had done in 1998 is now so rusty and I was struggling but I think you have shown me how it could be done.

Last edited: Nov 21, 2011
20. ### originTrump is the best argument against a democracy.Valued Senior Member

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your on the right track -

integrate $f=\frac{Gm_1m_2}{r^2}$
with respect to r.

The rest is constant so all you have to integrate is $r^-^2$

The limits of r will be the closest that the centers of the masses are from each other (NOT 0) and the 100 million light years (hint: the constants over 100 million light is effectively 0).

21. ### Aqueous Idflat Earth skepticValued Senior Member

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I obviously misunderstood the question.

You are not only not on earth, you are nearly at the span of a filament.

I'm not sure what you really were asking.

Yes work and energy are equivalent. No, not the integral of that.
G m1 m2 /r is the integrated result

calculate over the span from a to b:

U = G m1 m2 (1/b - 1/a)

I don't think you'e on the right track though

you can't start from a distance of zero, as this requires infinite energy to overcome
and the other term is way small.

A lot of assumptions seem to be at play, maybe those need to come out

maybe you could rephrase the question
what are you really after?

22. ### Aqueous Idflat Earth skepticValued Senior Member

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OK my lag time is making me irrelevant

Hey - how come you were doing physics in 1998 and can't remember integrals?

Most people remember that and forget the rest!

23. ### Robittybob1BannedBanned

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To you I will be as open as I can be, right. I know very little about subatomic particles but from the make up of a proton and neutron I envisioned that the basic structure is a 27 loop spiral of energy like a spring pulled around and joined back to itself. That part was made from 2 of the like components (quarks).
The other part ran through the centre of this torus like a particle in an accelerator through the coils. Now this centre piece was a possible source of gravity if it became unbound and shot out but returns as well.
So that is from where I started to think how far this energy could go before it was pulled back to itself.

Can you follow that? In 1998 no one had any idea what I was on about so I stopped looking into it.