Lerts, you will not create energy this way, only transfer potential energy into kinetic energy and back.
The formula for calculating the potential energy is at the end of this section (ignore the constant gravity discussion at the beginning): http://en.wikipedia.org/wiki/P ... tational_potential_energy

Quote from wien :

Lerts' lunacy aside, if you don't have the required brainpower to contribute in any meaningful way, that usually means it's time to shut up and move along to icanhascheezburger.com before you hurt yourself. Your input is neither wanted nor required.

I find that quite rude, coming from you.

I wasn't insulting Lert's, nor was I "complaining" about it.

I was just saying that maybe if he posts something that most of us can understand, maybe more people will bother to contribute. FYI I don't even browse "icanhascheezburger.com". So I guess I can say touché.

thanks i had already gone through that link but dont really understand it

it says u=-g*m1*m2/r

u whats u, potential energy? it cant be cause distance, r, is dividing, so the bigger the distance the smaller the potential energy?¿¿¿

g the gravitational constant, m1 and m2 the masses and r the distance but as i said how can potentional gravitational energy decrease with distance as the r is dividing?!!!1111

s14 the concept i ask about is extreamly simple: closer objects are pulled harder by gravity, therefore the closer weight will be pulled more with the unusual outcome of lifting the center of gravity of the masses

this is quite odd and who knows if anybody else thought of this problem before

Using the standard notation that GPE (gravitational potential energy) is zero at infinite distance, then increaslingly negative as you approach the center of mass of a solid sphere. GPE become increasingly negative at an increasing pace until you reach the surface of the earth. Afterwards, it continues to become more negative, but with a negative cosine(distance from center) type curve shape. For example, the excape velocity from the center of the earth is higher than the escape velocity at the surface of the earth.

So in this case, the connected weights move to minimize GPE (make it most negative) which is the normal situation.

thats what i dont understand how can potential energy be 0 at infinite distance?

shouldnt it be 0 at the center of the earth and a big amount at infinite distance?

i mean thats exactly the oposite to what i thought, the highest you are the more potential energy

then according that formula at the center of the earth there would be infinite potential energy

come on this science must come from the petrol companies

edit:

i had aplied that formula before without understanding it which i still dont understand yet and yes the cog will raise becoming potential energy more negative

but some questions

does it make any sense talking of a negative energy? what does it do negative work? seems to me intended to confuse

because actually the potential energy of an object at infinity is the work necesary to take that object from the center of the planet to infinity

so from this point of view which has more energy the cog of the bar in the center of the earth or the cog raised by gravity?

i dont know ill have to learn to integrate to find out but the fact that the cog raises by gravity gives me some clues

Potential energy from the centre of the earth (ignoring all other celestial bodies) is, iirc, Ep=mgh, whilst kinetic energy is Ek=0.5mv2.

Your equation of u=-(gMm)/(r2) is the force of attraction between two bodies (commonly called Fg), not Ep or Ek.

Edit: And in most cases, at least at your level of understanding, you can ignore the minus. Two bodies separated by a distance cause a positive force of attraction, rather than a force of repulsion.

g=6.67300 × 10-11 m3 kg-1 s-2

I can't be bothered to add the approprate ^ because they always look messy. At this level of simplicity I'm sure you can work out instantly where a ^ would actually be appropriate.

Quote from lerts :

thats what i dont understand how can potential energy be 0 at infinite distance?

because its the potential energy of the ssystem not of a single body in it

Quote :

come on this science must come from the petrol companies

and now youre just embarrassing yourself... again

Quote from S14 DRIFT :

No offense, Lerts, but why not find something scientifical that the majority of the forum can understand, rather than the minority?

Maybe then you'd get more people taking part in the discussion, rather than people mocking you etc, because, frankly, most of us don't have a bloody clue.

Lots of people know physics and the laws of Gravitational attraction (F = Gm1m2/r^2). G = 6.67E-11, m1 = mass of earth, m2 = mass of little ****ing ball, r = distance of both mass's centres. Figure it out

Quote from tristancliffe :

Potential energy from the centre of the earth (ignoring all other celestial bodies) is, iirc, Ep=mgh, whilst kinetic energy is Ek=0.5mv2.

Tristan, using Ep=mgh is only accurate to a specific distance. When working out conservation of energy (atleast basic), using -Gm1m2/r^2 is more better at larger distances (think satellites) (or so I've been taught).

Therefore, by the sounds of it, Lerts is just getting confused with different shit that are similar, as I did with this unit in physics this year.

Quote from dawesdust_12 :

Lots of people know physics and the laws of Gravitational attraction (F = Gm1m2/r^2). G = 6.67E-11, m1 = mass of earth, m2 = mass of little ****ing ball, r = distance of both mass's centres. Figure it out

:banghead:

Quote from S14 DRIFT :

:banghead:

Little british boy confused by basic math formula?

Yes, actually.

How? This is basic physics, like I did this stuff in my first year of learning it in high school.

Cool.

This is A-level physics, and unfortunatly (fortunatly for me) I didn't do it. We did nothing like this in high school here. Unless it was only one lesson long and I happened to not be there.

Quote from S14 DRIFT :

Cool.

This is A-level physics, and unfortunatly (fortunatly for me) I didn't do it. We did nothing like this in high school here. Unless it was only one lesson long and I happened to not be there.

A-Level physics?! Wow, britain is really ****ed up then. This is expected for us to know for our final exam in Physics 11 and Physics 12. Some of the more shocking topics we get into is electricity (v=ir, and then electric circuits), and Kinematics in 2 dimensions (vavg = d/t, etc), projectile motion, Work, Energy, Power, Momentum, Orbital motion, circular motion, Forces.

Not quite as ****ed up as America/Canada. America is so fat it's literally flipping on it's axis, and as we know by now, if something in the world goes wrong... BLAME CANADA.

Really... I'd debate how really ****ed up britain is, but maybe I'll just write an essay for some thread and watch all the pussies cry.

Get it ready for the copy/paste.

*runs away screaming war!111//!/11!!!!@@@*

I need to write a compare/contrast essay for school, what better than writing one that says how britain is shit and should float away.

Trans-atlantic relations at an all time low here on LFS Forums..

Really, it's trans-Pacific+trans-Asian relations from here bitch

Quote from dawesdust_12 :

Tristan, using Ep=mgh is only accurate to a specific distance. When working out conservation of energy (atleast basic), using -Gm1m2/r^2 is more better at larger distances (think satellites) (or so I've been taught).

True - as g varies with distance then so mgh will become an intergral, which is a tiny bit harder. But the -G one isn't for potential energy, but for force, so you'd need to factor in some other bits and bobs to deduce g (the acceleration) at any given 'altitude'.

S14 - this is physics you should have learnt when you were about 11 or 12, possibly before.

Ahh ok Tristan. That's where I've been going wrong all these years.

Quote from tristancliffe :

But the -G one isn't for potential energy, but for force

sure is
The gravitational potential energy of a system of masses m1 and m2 at a distance R is