Why is G.P.E negative? (2 Viewers)

[sinophile]

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?

 

[cutemouse]

No,

Recall the definition of GPE. It is the work done to move an object from infinity to a place in the gravitational field.

That means as an object moves towards a mass, say Earth, then its GPE should decrease. Therefore it has to be negative, since at an infinite distance away from Earth its GPE is zero (as it is not attracted to any masses).

The definition that GPE=mgh is used only for Junior Science. I don't know why they teach that, probably to help facilitate for the concept of GPE.

EDIT: W=mgh is correct, assuming that 'g' is constant, however it is not used in HSC Physics as 'g' is not considered to be constant (as g=GM/r² -- if 'r' increases then obviously 'g' becomes less). But note that GPE=/=mgh.

 

[Affinity]

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?

It depends on where you set your zero.

the first one sets as zero GPE at the surface, while the second one "at infinity" these are chosen to make the equations look nice.. without any funny terms. One can add a constant to GPEs and the theory will not be any different


as mentioned in the previous post... the 2 are not the same

the first, mgh is the gravitational potential energy in a constant field, while the second gives the GPE in a gravity field which decreases according to the inverse square law. Neither are less valid in principle, it depends on the gravitational field under consideration.

the first one is used for calculations say close to the surface of the earth, where gravity doesn't change by any appreciable amount and thus can be considered constant. but for larger distances one can no longer ignore the change in gravitational field strength
and by the way, it's -Gm1m2/r not r^2

 

[cutemouse]

lol.. Affinity you seem to recall all that HSC watered down Physics from 2003...

I take it that you've done some sort of Physics/Maths related course at uni? If so, I'd have imagined that you used calculus, but anyhow

 

[92LUCAS]

GPE is always negative because as the distance between the centre of the gravitational field and the mass in the field decreases (i.e. it gets closer to the centre of the larger mass), GPE is lost and decreases
» So, a mass that is an infinite distance away from Earth has GPE of 0. When it moves closer to earth, its kinetic energy came from GPE. Thus, it has lost GPE, and therefore, GPE now is negative. So, it will ALWAYS be negative.

 

[viviansidiss]

Because the formula is divided by distance "r", the larger "r" gets the closer GPE approaches 0 until such a point at infinite meters GPE will be 0.

 

[lolokay]

when we use GPE in calculations we would be dealing with changes in the GPE, so it doesn't matter whether your GPE is positive or negative. it's just set it to which is most convenient

 

[cutemouse]

Erm, no... I've seen 1 or 2 mark questions in trials asking you the calculate the GPE of an object. So the negative sign does matter!

Why would you give such misleading advice?

 

[sinophile]

Wait, according to the equation GPE=-m1m2/r, we set "zero" GPE at an infinite distances. Then what is the GPE at the earth's surface? Infinite?

 

[cutemouse]

Depends on how far you are from centre of the Earth.

 

[lolokay]

Erm, no... I've seen 1 or 2 mark questions in trials asking you the calculate the GPE of an object. So the negative sign does matter!

Why would you give such misleading advice?

if they ask you to calculate GPE of an object then they would specify that it's absolute GPE, in which case it's defined to be negative anyway.

GPE doesn't necessarily mean absolute GPE and the OP was asking why it can be used as both a positive and negative term

As affinity said, it just depends on where you set your zero, and you can add a constant to the GPE at all points and the theory won't really be any different

also, i would have assumed that PE=mgh is used beyond HSC, as it's a good approximation assuming g is fairly constant

 

[jtyler]

In time you'll come to realise that mathematics goes only so far as a model for reality, and nuances like this really have no appreciable consequences. Consider that someone had to decide how to define GPE, in the context of a purely invented quantity that can complement other concepts of energy (which is more of an abstraction than most people know). Can you think of a better way to define GPE? In the end, the answer is no, and thats why GPE is negative.

Edit: GPE = mgh is basically an invention designed to let concepts of energy be taught earlier in schooling. In reality, it's only an approximation, and things like where the zero is set determined solely by the simplicity that results.

Energy is a very tricky concept, the rigorous definition and nature of which often eludes very accomplished students of physics.

 

[cutemouse]

if they ask you to calculate GPE of an object then they would specify that it's absolute GPE, in which case it's defined to be negative anyway.

If you were dealing with CHANGES in GPE, you would get the correct MAGNITUDE if you omitted the negative signs, but this is very inconvenient, as some questions ask to calculate the GPE of an object, given its GPE at its current position (value of GPE, which is negative), and if it moves up (or down) by a certain distance, in which case the negative sign DOES matter.

Notice what you said "when we use GPE in calculations we would be dealing with changes in the GPE, so it doesn't matter whether your GPE is positive or negative. it's just set it to which is most convenient"

You basically said that in Physics when you do calculations with GPE, it's all to do with changes, which is NOT the case. Thus, you gave misleading advice!

 

[Ostentatious]

From what I interpret,

GPE = -G(m1m2)/r

as r -> infinity, GPE -> 0

where r is the distance between the centre of the Earth and the centre of infinity.

So even if this "infinite point" had a mass, GPE = 0 anyway. Sorry, I kinda hijacked the thread there because I had trouble understanding it also.

 

[Ostentatious]

That's strange... What if the mass at the point of infinity was just as infinitely large,

GPE = -G(m) ???

Am I interpreting this correctly?

 

[youngminii]

If you were dealing with CHANGES in GPE, you would get the correct MAGNITUDE if you omitted the negative signs, but this is very inconvenient, as some questions ask to calculate the GPE of an object, given its GPE at its current position (value of GPE, which is negative), and if it moves up (or down) by a certain distance, in which case the negative sign DOES matter.

Notice what you said "when we use GPE in calculations we would be dealing with changes in the GPE, so it doesn't matter whether your GPE is positive or negative. it's just set it to which is most convenient"

You basically said that in Physics when you do calculations with GPE, it's all to do with changes, which is NOT the case. Thus, you gave misleading advice!

In the HSC, all questions based on GPE is only concerned with the change, not the value itself.

That's strange... What if the mass at the point of infinity was just as infinitely large,

GPE = -G(m) ???

Am I interpreting this correctly?

Lol you're thinking too hard.
Also, it's not between the centre of the Earth and centre of infinity. Just the centre of whatever mass we're dealing with and the centre of infinity.
And yes, it doesn't matter how massive an object is, at infinity the GPE will be 0 ('cause there must be a 'point' in space where there is no Gravitational Potential Energy).

 

[cutemouse]

In the HSC, all questions based on GPE is only concerned with the change, not the value itself.

You either are very ignorant, arrogant, or stupid and obviously haven't seen any trials, which are typically harder than the actual HSC examination.

Did you read what I said:

some questions ask to calculate the GPE of an object, given its GPE at its current position (value of GPE, which is negative), and if it moves up (or down) by a certain distance, in which case the negative sign DOES matter.

May I note that the syllabus requires students to define Ep as a negative IIRC so it'd be EXTREMELY stupid just to take your luck and just consider the magnitude just because the HSC examinations up to date have only asked for changes in Ep. You'd be basically throwing away marks if you actually got a question that required you to take into consideration the signs.

Here's a question from a trial:

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.

So, are you still going to say that the negative sign doesn't matter? And, are you still going to support the misleading advice that is clearly is not beneficial?

 

[youngminii]

Lol calm down son.
I never said the negative sign in GPE doesn't matter, I simply stated that in the HSC, there won't be a question with the actual value of GPE. I never said anything about trials.
Also, that specific question you gave is out of context. It's probably part of a bigger question which already gives you the GPE of the probe in the 'current' orbit. Hence, you're just using the value of work done given to calculate the change in GPE, which you then use (along with the original GPE) to calculate the 'new' GPE.

Edit: You gave me a negative rep for 'A novice giving misleading advice'? Get off your high fucking horse you arrogant bastard. Who made you king of Physics? Read my posts more carefully next time, pretentious cunt.

 

[cutemouse]

I never said anything about trials.[/quote]

Also, that specific question you gave is out of context. It's probably part of a bigger question which already gives you the GPE of the probe in the 'current' orbit.

Yes that is a part of a question. The previous question asked to calculate the Ep of an object. So, the negative sign in the final answer DOES matter. I'll post that question later.

Edit: You gave me a negative rep for 'A novice giving misleading advice'? Get off your high fucking horse you arrogant bastard. Who made you king of Physics? Read my posts more carefully next time, pretentious cunt.

Me arrogant? You're pretending to be a fortune teller for next year's HSC paper for Physics:

in the HSC, there won't be a question with the actual value of GPE.

Because, you are yet again giving misleading advice. How do you know for sure that they wont ask questions asking for the value of GPE? So, who made YOU the king of Physics?

All I'm trying to do is help others by not throw away easy marks like this in their examinations by trying to say that you could get a question asking you to calculate the Ep of an object.

According to you, if HSC Physics only deals with changes in Ep, as you say, then why don't they give you the formula just for the change in Ep? They give you the formula to calculate Ep for a reason. Plus, the syllabus doesn't limit the scope of Ep just to changes.

So yes, I was trying to help others not believe your misleading information by saying that the negative sign in Ep DOES MATTER. So what did I do wrong?

So you tell me, who's really on their high horse?

Lol calm down son.

I think there's a pot calling the kettle black.

 

[k02033]

Ok um lets start at the very beginning of this potenial energy stuff...
Lets assume a 1 dimensional case:
So if a particle has a net force F imposed on it, by newton's 2nd law there will be an accerleration, ie a change in the velocity of the particle. Now if the object's speed were to also change in the process, then we can say Vf^2-Vi^2=2a(delta S), where Vf is final speed Vi is initial speed and delta S is the CHANGE in position of the particle in relation to the origin of axis you have chosen ie the particle's displacement. It is not simply the distance the particle has travelled. mutiplying both sides of the equation Vf^2-Vi^2=2a(delta S) by 1/2m we get (delta K)=F(delta S) where (delta K is change in kinetic energy). So this mysterious combination of F(delta S) actually keeps track of the particle's change in KE. Since its useful we can define -F(delta S)=(delta U) where delta U is change in potenial energy. Its purpose is to keep track of the particle's energy changes. now notice that F should be a constant in order for the equation -F(delta S)=(delta U) to work, but if the force varies... says high altitudes above earth, then we have to take limits and integrate over the particle's journey to get the potenial function. ie integral of -Fds with end pts Xf and Xi (the final position of particle and initial position) equals to the change in potenial between Xf and Xi. Now lets set the origin to the centre of earth and the integral for F=-mg (where g is 9.8, and the minus sign to signify direction of force is in -ve x direction) is int(-mgdx) over Xf and Xi. this yields(delta U)= mg(Xf)-mg(Xi), if the object is falling towards earth, then Xf<Xi, this mean (delta U) is -ve. someone said mgh is +ve, that is wrong. and if we used F=-GMm/x^2 then the integral becomes int(GMm/x^2dx) over Xf and Xi then we get delta U = -GMm/Xf+GMm/Xi, if we integrated without end pts we get the familar hsc equ U(x) = -GMm/x