How does the formula for gravitation potential energy work?

In summary: I would be really grateful.-- ShounakHi Shounak! :smile:So, the article is saying that if we chose our "zero" level of potential energy to be at the singularity (r=0), then all other values of potential energy would be infinitely large (either positive or negative), making calculations impossible. So, instead we choose our "zero" level to be at infinity, which avoids this problem and makes calculations much easier.H and h are just variables used in the derivation. H represents the initial height or distance from the center of the Earth at which we define our "zero" level of potential energy. h represents the change in height or distance from the initial point. Does that make sense
  • #1
shounakbhatta
288
1
Hello,

Can anybody please explain me:

While going through gravitation potential energy, I came across:

F=G.m1.m2/r^2

From there it follows:

g=-GM/r^2.r

How does it follow? Specially the -G case?

-- Shounak
 
Physics news on Phys.org
  • #2
They're defining the "gravitational field" to be:

[tex]\mathbf{g}=-\frac{GM}{r^3}\mathbf{r}[/tex].

Basically this means that if you take the gravitational field at a point, [itex]\mathbf{g}(\mathbf{r})[/itex], and you multiply it by the mass [itex]m[/itex] of an object at that point, you get the force on that object:

[tex]\mathbf{F}=m\mathbf{g}[/tex]

The reason there's a "-" sign in the definition of the field is because the vector [itex]\mathbf{r}[/itex] points out from the "gravitating" mass. Since the gravitational field effectively tells you the acceleration that small, free particles experience, the [itex]\mathbf{g}[/itex] needs to point in the opposite direction of [itex]\mathbf{r}[/itex] (because gravity is attractive).
 
  • #3
It explains:

"We know that the further you get from an object, the higher your GPE relative to it. (As something must have done more work against gravity to get you there). Thus when you are infinitely far away, you have as high a GPE relative to it as possible. We choose (arbitrarily) to make the value of GPE of all bodies at infinity zero. Then since this is the highest value of GPE, all real values of GPE (closer than infinity) must be negative. Therefore the minus sign in the equation is NOT optional; it must always be included and all values of potential energy in a gravitational field are negative. (This is not the case when we come on to electric fields, because they can be repulsive too)."

I want to understand "all real values of GPE (closer than infinity) must be negative."

What does that mean?

You have mentioned that the vector r points out from the gravitating mass. If I draw a diagram, a point m, the lines will be pointing out from the point...That means the vectors is not drawing inwards rather outwards, hence negative right?

Is there any difference between POTENTIAL ENERGY AND GRAVITATIONAL POTENTIAL ENERGY?

Thanks,

-- Shounak
 
  • #4
Hi Shounak! :smile:
shounakbhatta said:
Is there any difference between POTENTIAL ENERGY AND GRAVITATIONAL POTENTIAL ENERGY?

gravitational potential energy is potential energy

(and potential energy is minus the work done by a conservative force)

gravitational potential is potential energy per mass

(just as electric potential is potential energy per charge)
… We choose (arbitrarily) to make the value of GPE of all bodies at infinity zero. Then since this is the highest value of GPE, all real values of GPE (closer than infinity) must be negative. … (This is not the case when we come on to electric fields, because they can be repulsive too)."

I want to understand "all real values of GPE (closer than infinity) must be negative."

What does that mean?

potential energy is relative

we measure it relative to a test mass at infinity, whose PE we define to be zero

since nothing can be at infinity (or further away!), that means that all real values of gravitational potential energy must be less than zero

for electric potential energy, we use a (positive) test charge at infinity … so for any real positive charge, the electric potential energy will also be less than zero, but for any real negative charge, the electric potential energy will always be greater than zero :smile:
You have mentioned that the vector r points out from the gravitating mass. If I draw a diagram, a point m, the lines will be pointing out from the point...That means the vectors is not drawing inwards rather outwards, hence negative right?

not following you :confused:
 
  • #5
Ok, I got it now. It is just as we want to avoid infinite, it is better even to get -ve values. If we consider all the non zero values it would be extremely difficult for us to compute. Am I right?

What I am trying is just to draw a vector with points pointing outwards and hence -G.
 
  • #6
Hi Shounak! :smile:
shounakbhatta said:
It is just as we want to avoid infinite, it is better even to get -ve values. If we consider all the non zero values it would be extremely difficult for us to compute. Am I right?

no

we're not trying to avoid infinity

it's just that we want a convenient formula

if we chose radius R as our "zero" level of potential energy, then the magnitude of the potential energy at a general distance r would be GM(1/R - 1/r)

that's a rather cumbersome formula*, so we prefer to put R = ∞, which makes it GM(1/∞ - 1/r), = GM(0 - 1/r), = -GM/r :wink:

* of course, if we're happy with an approximation … which we usually are … then we can use R, eg as the radius of the Earth

here's an extract from the PF Library, to show how we get the usual mgh out of GM(1/R - 1/r) :wink: …​

Derivation of mgh:

[tex]\Delta (PE)\ =\ \Delta(-mMG/r)[/tex]

[tex]=\ \frac{-mMG}{r_{earth}\,+\,H\,+\,h}\ -\ \frac{-mMG}{r_{earth}\,+\,H}[/tex]

which is approximately:

[tex]\frac{-mMG(H\,-H\,-\,h)}{r_{earth}^2}\ =\ \frac{mMGh}{r_{earth}^2}\ =\ mgh[/tex]

What I am trying is just to draw a vector with points pointing outwards and hence -G.

nope, still not following you :redface:
 
  • #7
Thank you very much. It cleared me all up. Just 2 more questions:

Wikipedia writes:"The singularity at r=0 in the formula for gravitational potential energy means that the only other apparently reasonable alternative choice of convention, with U=0 for r=0, would result in potential energy being positive, but infinitely large for all nonzero values of r, and would make calculations involving sums or differences of potential energies beyond what is possible with the real number system. Since physicists abhor infinities in their calculations, and r is always non-zero in practice, the choice of U=0 at infinity is by far the more preferable choice, even if the idea of negative energy in a gravity well appears to be peculiar at first."

Hence I was asking about avoiding infinity.

Secondly, if you can please explain H and h.

Thanks.
 
  • #8
shounakbhatta said:
Wikipedia writes:

"The singularity at r=0 in the formula for gravitational potential energy means that the only other apparently reasonable alternative choice of convention, with U=0 for r=0, would result in potential energy being positive, but infinitely large for all nonzero values of r …

wikipedia is simply saying that if you choose R = 0 as your "zero" of potential energy, then all other potential energies would be infinite! :biggrin:
Secondly, if you can please explain H and h.

H is the height above Earth at which you choose the "zero" of potential energy for any particular experiment.

h is the extra height above that. :wink:

(so if you move a height h above H, the PE is +mgh)
 
  • #9
One more thing:

U=-m(G.M1/r1+G.M2/r2)

Are you putting GM(1/R - 1/r) in here or somewhere else?
 
  • #10
It also says:

"the choice of U=0 at infinity is by far the more preferable choice, even if the idea of negative energy in a gravity well appears to be peculiar at first"
 
  • #11
shounakbhatta said:
U=-m(G.M1/r1+G.M2/r2)

where does this come from?

what is it supposed to be? :confused:
shounakbhatta said:
It also says:

"the choice of U=0 at infinity is by far the more preferable choice, even if the idea of negative energy in a gravity well appears to be peculiar at first"

it's saying that you can choose U = 0 at any value of R, but that R = ∞ is (usually) the most convenient (for the reasons i gave above)

it's adding that some people find negative energy peculiar

well, that's obviously correct, because you do! :biggrin:
 
  • #12
What I am trying to say is:

GM(1/R - 1/r) how to put in which equation?
 
  • #13
shounakbhatta said:
GM(1/R - 1/r) how to put in which equation?

r is the (variable) distance at which you're finding the potential energy

R is the (fixed) distance which you arbitrarily choose as your "zero" level for potential energy :smile:
 
  • #14
To sum up:If r is non-zero, then there cannot be infinity, hence r is always non zero.

If U i.e. GPE =0 at infinity then anything less than zero is always negative, hence it is negative.

Right?
 
  • #15
shounakbhatta said:
If U i.e. GPE =0 at infinity then anything less than zero is always negative, hence it is negative.

Right?

right :smile:
If r is non-zero, then there cannot be infinity, hence r is always non zero.

no, r is non-zero because every massive object has a non-zero size, so you can't be zero from the centre of it
 
  • #16
Ok, thank you very much for clearing all the doubts and answering the questions.

I am on it.

Thanks once again.
 

What is gravitation potential energy?

Gravitation potential energy is the energy an object possesses due to its position in a gravitational field. It is the energy that can be released or harnessed when an object moves from a higher position to a lower position in the gravitational field.

How is gravitation potential energy calculated?

The formula for calculating gravitation potential energy is PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height or distance from the object to the ground. This formula assumes that the gravitational field is constant.

In what units is gravitation potential energy measured?

Gravitation potential energy is typically measured in Joules (J) in the SI system of units. However, it can also be measured in other units such as kilojoules (kJ) or calories (cal).

What factors affect the amount of gravitation potential energy an object has?

The amount of gravitation potential energy an object has is affected by its mass, the strength of the gravitational field it is in, and its height or distance from the ground. The greater the mass and height of the object, and the stronger the gravitational field, the more gravitation potential energy it will have.

How is gravitation potential energy related to kinetic energy?

Gravitation potential energy and kinetic energy are forms of mechanical energy and are related through the law of conservation of energy. When an object falls from a higher position to a lower position, its gravitation potential energy decreases while its kinetic energy increases. At the bottom of its fall, all of the object's gravitation potential energy will have been converted into kinetic energy.

Similar threads

Replies
10
Views
1K
Replies
12
Views
2K
Replies
27
Views
11K
Replies
2
Views
762
Replies
10
Views
901
Replies
11
Views
2K
Replies
2
Views
1K
Replies
6
Views
2K
Replies
9
Views
2K
Back
Top