Why is G.P.E negative? (1 Viewer)

[k02033]

 

[k02033]

And if we want to apply the change in potenial to the energy theorem , we have analyze an isolated system. (an isolated system is a collection of all objects being accelerated and all objects causing a force.) ie delta k + delta U = 0 is applicable to an isolated system ONLY.

For some object falling towards earth, isolated system is just object and earth.

delta k + delta U = 0

1/2mVf^2-1/2mVi^2+CHANGE IN POTENIAL DUE TO THE FORCE OF GRAVITY=0
1/2mVf^2-1/2mVi^2-GMm/Xf+GMm/Xi=0
1/2mVf^2-1/2mVi^2=GMm/Xf-GMm/Xi
Xf is smaller than Xi, so RHS is +ve, which means final KE is greater than initial KE, and this makes sense for objects falling towards earth.

 

[k02033]

oh notice that my chose of origin was arbitary, i set it to be centre of earth for convience, try setting it somewhere else, it will still work. This is the reason why only the chagne in potenial is of usefulness and not the potenial at a pt, since chosen axis and origin are arbitary.

 

[k02033]

If 6.8 x 109J of work is done on the probe lifting it to a higher orbit determine its gravitational potential energy in the new orbit.

since work done is given, we have to consider a non isolated system. for which the energy theorem is W= , where W is work done by external agent. (very important:0

using W=



but Vf and Vi are the orbital velocities.

so and

sub these 2 in and you should get



give me the initial potenial and i can get u xi, then sub it in to get xf

 

[youngminii]

Lol this is high school Physics. I don't think you need to go that in depth.
Appreciate the extra reading though.

 

[k02033]

the integrals are not hsc stuff, but the ideas of potenial functions and isolated system are applied in hsc. ie photoelectric effect . And calculations of the stopping voltage requires you to know how to applied the energy theorom to the isolated system consisting of E field and electron.

 

[k02033]

and you need to realise that all these confusion with signs involving GPE and what not is because you have this mentality of "its not in hsc, i dont need to put any thoughts to it"
not goin to be very successful with that kind of thinking..
and i think the hsc syllabus is really silly anyways, that thing is garbage, not real physics.
things like potenial integrals is what real physics is about.

 

[youngminii]

Lol I appreciate your explanations and everything but you have to realise, most high school Physics students aren't interested in anything that's too far ahead of the Syllabus (like the potential integral stuff, seems interesting to learn after HSC though).

By the way, the negative sign in GPE IS in the syllabus.

 

[Ostentatious]

So that's what integrals are used for...

 

[Ostentatious]

OK, so in essence, we're trying to find the GPE of moving a mass from a point in a gravitational field to somewhere far away (infinity). So, that Gravitational Field would be the earth's and GPE is negative because the gravitational field is exerting a force on the object downwards (or however one interprets the direction to the centre of the earth).

Maybe this is where I'm wrong, since I'm considering 0 to be at the earth's surface and trying to relate it to zero at infinity. And the gravitational field is the field of the point at infinity right, since gravitational fields don't really have a finite boundary. This is so frustrating. I can't believe I could understand relativity but not this!

 

[youngminii]

Don't consider 0 to be at Earth's surface.

Gravitational fields have a finite boundary in that it doesn't exist at a point of an infinite distance away.

 

[lolokay]

ostentatious:
the further away two masses are, the higher their GPE is. For convenience, we say that at the maximum distance, ie. as it approaches infinity, the absolute GPE is zero. So at any other point it will be less than this. The amount below zero it will be is equal to the amount of work done by it moving it from infinity to that point, according to conservation of energy. This work is GMm/r (as can be found using integrals, like k02033 showed). So the GPE is -GMm/r.

You don't consider mass at this infinite point, as we are only considering the 2 masses (their centres of mass)

[can someone comfirm that this is the correct explanation?]

EDIT: blah whats wrong with the codes above ^ ?

you need the [ tex ] -code- [ /tex ] thing

blahblahblah

lol you sure do like to argue

 

[k02033]

There are all these notions of GPE at a point in space, on earth, at infinity whatever..
ok someone comes up to me and tells me that the GPE of a certain object at a point is 3 joules, that actually gives me no useful information what so ever. let me give an analogy as to why.

Also terms like "GPE is -ve" make no sense, what really is -ve? GPE at a point or change in GPE? two very diff things.

 

[k02033]

"their GPE" makes no sense, keep in mind that change in GPE is something that tries to measure the exchange of energy within some system.

 

[k02033]

thanks for the coding help

 

[k02033]

Im actually a teacher in training, i am doing B sci (physics)/B edu
and i tutor at parramatta lib every week on sats...
if enough people are interested in coming, i would very much like to set up a lesson on the energy theorom at parra lib on sat for free (no catch, i would just like to get some experience teaching a group)

 

[Omium]

GPE can be calculated using two equivalent equations, right? Work=mgh and Ep=-Gm1m2/r^2? Doesnt this ccontradict itself, because one eqn yields positive result and the other a negative result?

I haven't read this thead.

But essentially U = mgh is technically "wrong".

Now, If you were to sketch U = mgh and Ep = -Gm1m2/r^2 at low radii (essentially everything we see in everyday lfie).
You would see that mgh is approximately equal to -Gm1m2/r^2

At large distances from earth, this approximation is not valid

 

[Ostentatious]

youngmini: I didn't realise that was the explanation for "infinite distance" thank you!

lolok: Thanks for explaning why GPE was negative, it helped a lot!

k02: No offence to the other helpers, but THIS REALLY HELPED ME THE MOST! Now I understand the whole infinity thing I think. Thank you for posting this! I'd be really interested in attending your little lessons but I'm not very familiar with the Parramatta area. Thanks again!

 

[k02033]

Glad to be of help