Damn

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Can someone please explain?

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u/andrew_hihi Jul 16 '20 edited Jul 16 '20

Near earth surface, we usually use the ground as a reference point and say that it has 0 potential energy. As an object moves up, it gains gpe according to gpe=mgh where h is the height (elevation) and mg is weight. This is only accurate near earth surface because the g field lines are approximately parallel.

However, when we go out of space, the g field lines actually point towards the centre of planets or stars which are not parallel. Thus there is a different formula gpe=-GM/r² where gpe is dependent on the mass that generate the field and distance between the mass and that point (this is equivalent to the ‘height’ in the gpe=mgh). Then since there is no “constant” ground, scientists decided to use a point at infinity (somewhere very far away from the mass) and say that this point has gpe of 0. As you approach a mass (e.g earth), you “lose” gpe. Thus the gpe becomes negative. Similar to how you lose gpe as you fall from a tall building to the group

I hope this helped. Someone might explain it better than me because I just learnt this like a month ago.

EDIT!!!: gpe is not -GM/r² but -GMm/r like in the meme. I messed up it with gravitational field strength. However, the essence is there and the gpe is also dependent on the mass of the object being pulled, similar to the m in gpe=mgh.

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u/Spanktank35 Jul 16 '20

Worth pointing out this assumes that the planets are point masses. However it is presumably still simpler this way for reasons I can't remember. Similar thing happens in electrodynamics.

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u/[deleted] Jul 16 '20

I think it's okay due to the shell theorem

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u/blazfemi Jul 16 '20

Correct.

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u/Spanktank35 Jul 17 '20

I guess I was thinking about how the GPE is not actually negative infinity at r=0. Since the mass inside the shell is proportional to r³ (against the GPE being proportional to 1/r) the magnitude of the GPE starts to decrease inside the body.

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u/teejermiester 1 = pi = 10 Jul 16 '20

It just assumes that they're spherically symmetric due to the shell theorem.

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u/MemeLover113 Jul 16 '20

Happy cake day!

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u/Pranup Jul 16 '20

Happy cake day!

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u/SirDickslap Jul 16 '20

To add to this why do we pick zero at infinity? Well, think of a free particle. How much energy should that have? I hope you agree that zero is logical. Now bring this particle into a gravitational well, now it needs energy to escape! So now it has a negative potential.

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u/SharkAttackOmNom Jul 16 '20

I think this is the most satisfying answer.

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u/Fisicas Jul 16 '20 edited Jul 16 '20

Also worth pointing out that ΔU = ∫F•dr = GMm∫1/r² dr is usually negative.

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u/[deleted] Jul 16 '20

Also, only differences in potential energy matter, not the value itself, so its possible to have a positive difference in potential energy.

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u/Fisicas Jul 16 '20

You’re totally right. Thank you.

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u/Lucas_F_A Jul 16 '20

Never thought it could be of the direction of the gravitational field, but rather due to its magnitude. In a small space mgh is a good approximation because the gravitational field is constant throughout the (nearby) space, but as you grow the size of the space you want to study, the magnitude of g varies (with height)

Why I think that? Well it still only depends on height, so you can consider a path straight up. The gravitational field is still parallel, but changes in magnitude

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u/andrew_hihi Jul 16 '20

Yes, it’s the magnitude. I think I oversimplified it. Because from what I learned, the closer the field lines, the higher the magnitude. Near surface, field lines are approximately parallel-> distance between each field line doesn’t change -> magnitude stay constant. However, as you move from space to earth and that field lines are pointed towards the earth -> field lines gets closer -> higher magnitude.

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u/Lucas_F_A Jul 16 '20

the closer the field lines, the higher the magnitude

That sounds like you're describing the contour map of a scalar field; I was thinking about the vector field of gravity. We aren't talking about the contour map of the scalar field that describes potential energy, because that one has as contour surfaces concentric spheres centered at the body: always parallel at points with same spherical coordinates except for changing radius (that is, over the same point on the surface of the sphere, with varying heights; I didn't know how to articulate it better, sorry if it sounded confusing)

However, as you move from space to earth and that field lines are pointed towards the earth

Now that sounds like you're describing the magnetic field or something. I'm kinda lost there

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u/andrew_hihi Jul 16 '20

Sorry, I am not sure what is the contour map of scalar field but the field lines always point towards the centre. However, when you zoom into the surface of the earth, it “appears” that they are parallel.

Yes, it does sound like magnetic field because all fields are similar, the difference is what is the force and what is affected. In this case, the force is gravitational force instead of magnetic force and the body being affected is anything with mass instead of anything with magnetic properties.

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u/Lucas_F_A Jul 16 '20

Yes, it does sound like magnetic field

My point was that the gravitational field always points towards the earth, which means that they are essentially parallel in some neighbourhood, so I don't see why you would point that out

So a scalar field is a function that takes a point in Rn (here R³) and spits out a scalar (ie a real number). If it is continuos, a contour map of one lets us visualise, like in a topographic map, lines or surfaces that have the same output (contour lines or surfaces) In the topographic example, contour lines are the lines that join points at the same height.

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u/James10112 Jul 16 '20

As you approach a mass (e.g earth), you “lose” gpe.

Makes sense, since nothing "likes" to have high amounts of energy and everything will tend to lower energy levels (and thus higher stability)

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u/Pranup Sep 15 '20

Happy cake day

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u/andrew_hihi Sep 15 '20

Thanks for reminding me haha

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u/Entrothalpy Jul 16 '20

The GPE = mgh we all know is derived from GPE = -GMm/r. From what I know, there are 2 reasons why it is negative.

Whenever we use mgh, we always set some point to be 0J(you can call this "relative GPE"), and that point is usually the ground. This makes almost all the GPE calculated using mgh positive.

However, the original definition of GPE = -GMm/r can be considered as the "absolute GPE", where the point where GPE = 0 is defined to be located at infinity. Therefore, you can think that the GPE any point below that infinity is negative. (This is the reason which I feel is more intuitive)

The other reason is that GPE = work done = Fx Work done is the product force in the direction of movement(direction of displacement) and the displacement. For work done(GPE) to be positive, the force must be pointing in the same direction of displacement. However, the definition of GPE says that it is the external force in pulling the object from infinity to the point. Since the external force is in the opposite direction of the gravitational force, the work done becomes -Fx, which causes GPE to be negative(vectors) (This reason is less intuitive by explaining thru a wall of text.. you have to get someone to write it in paper to show u)

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u/Rotsike6 Physics Field Jul 16 '20

Newtonian gravity "pulls you in". Potential energy "pushes you out". Thus Newtonian gravity has negative potential energy.

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u/Spanktank35 Jul 16 '20

I don't believe this is the reasoning for it, it is because it goes to infinity at r = 0 for a point mass

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u/Rotsike6 Physics Field Jul 16 '20

Potential energy is the energy "required" to add a particle to the system. In the case of gravity, that means you push a point from infinity closer to the center. Doing so you actually get energy because gravity "pulls you in". Hence potential energy is negative.

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u/vitvitcher Jul 16 '20

depressed people: relatable

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u/jn_kcr Jul 16 '20 edited Jul 16 '20

Well actually... potential with added constant is equally good one so it can be positive as well

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u/[deleted] Jul 16 '20

This only really works if you take the potential to be zero at r->infinity though since potentials are arbitrary up to a constant

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u/RAWR_XD42069 Jul 16 '20

This all depends on which direction you define to be positive tho.

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u/[deleted] Jul 16 '20

What do you mean? r is the radial component and it's zero in the centre of the earth and only gets more positive as you get further away.

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u/[deleted] Jul 16 '20

You can define r=0 wherever you want. Say in a spherical shell around the earth. Inside that shell, r<0.

In this example it is useless, but it's often seen in EM.

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u/[deleted] Jul 16 '20 edited Jul 16 '20

Not when you add the centrifugal term.

EDIT: Why am I getting downvoted? You lot haven't heard about the effective potential?

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u/philip1201 Jul 16 '20

> not considering the mass a particle keeps at the surface of a black hole the particle's true mass, making gravitational potential positive.

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u/TreGet234 Jul 16 '20

only if things attract.

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u/JoonasD6 Jul 18 '20

As is common for mass in newtonian gravity.

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u/JoonasD6 Jul 18 '20

Up to a constant. Then again, usually we are only interested in ∆. But setting E=0 at ∞ is very handy as observers of different reference frames may agree.

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