# Does Potential Energy = 0 at Earth Centre?

## Main Question or Discussion Point

Hello,

I am new here so I hope this is posted correctly.
If you assume a perfectly round earth (with uniform density etc.), is the potential energy of an object = 0 there?
I assumed the object was just a small sphere of concentrated mass with no dimensions or whatever makes it simple.

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Filip Larsen
Gold Member
Yes. The gravitational potential with the assumptions you mention will decrease [STRIKE]linearly[/STRIKE] from some value at the surface to zero at the centre.

Edit: Sorry, I was thinking about the force when I said linearly. If the potential is set to U(r) = -GM/r then U(r)/U(R) = (r/R)2 with R being the surface radius. That is, the potential varies with the square of the radius from the surface and down.

Last edited:
niranjan_learner
russ_watters
Mentor
Err, well, potential energy is relative to any pre-defined reference point. For a great many applications, that reference point is the surface of the earth.

I suppose you ccould say the lowest reference point for an earth-centered potential energy problem is the center of the earth.

sophiecentaur
Gold Member
Err, well, potential energy is relative to any pre-defined reference point. For a great many applications, that reference point is the surface of the earth.

I suppose you ccould say the lowest reference point for an earth-centered potential energy problem is the center of the earth.
To move the object away from that point would involve doing work on it. That means it's a potential minimum.
But referring to the centre of a planet is not a lot of use for any practical application. For launching rockets and building Space Elevators, the Earth's surface is a good reference level with the potential increasing in value as you go up but, for discussing orbits and gravitational problems in general, it's usual to define an 'absolute' potential, with respect to infinity. That's because the work done in moving from one distance (R2) from one 'point mass' to another distance (R1) is
W = 1/R2 - 1/R1
which goes to infinity if you refer to the start point being at the centre (R2 = 0). It's easier to use R2 as infinite so the potential becomes -1/R1. The negative sign comes in because the journey from infinity inwards involves negative work being done.
This has the advantage of not needing to know the radius of a celestial body - just its mass and you can also ignore the fact that, once you go below the surface, the potential doesn't follow a simple 1/R law.

Dale
Mentor
Potential energy is 0 wherever it is convenient to set it to 0. Only differences in potential energy are meaningful, so you can set the 0 anywhere you wish as long as the differences are unchanged.

Inder kumar
Thanks for all the replies.
I now realize the significance of "work" and "frame of reference" in this discussion.

stewartcs
Potential energy is 0 wherever it is convenient to set it to 0. Only differences in potential energy are meaningful, so you can set the 0 anywhere you wish as long as the differences are unchanged.
Exactly.

This is the basis of Thermodynamics...the change in energy not the absolute values.

CS

Filip Larsen
Gold Member
I feel it is a bit like waving off the original question to say that since you can set the reference level of your potentials any way you like then, sure why not set it at zero at the centre of the earth and be done with it. While technically correct, the question seems to have a more interesting answer.

The gravity potential at the centre of earth is interesting not because it can arbitrarily be set to zero (or not) but because it has a minima at that position. If the OP instead had asked "is the potential for gravitational work zero at the centre of earth" I bet there wouldn't have been all this referring to potential reference levels.

If you assume a perfectly round earth (with uniform density etc.), is the potential energy of an object = 0 there?

I'm assuming the OP means "gravitational potential" energy....

In introductory physics texts, isn't the usual zero reference point taken as an infinite distance from the center of attraction? and the gravitational potential is consequently negative at any finite distance.

Isn't gravitational potential at the centre negative infinity? I mean that is what the standard formula for potential suggests. I am also a little confused about the potential at the canter of the earth.

Isn't gravitational potential at the centre negative infinity? I mean that is what the standard formula for potential suggests. I am also a little confused about the potential at the canter of the earth.
As per previous responses, it all comes down to your reference point.

If it's the centre of the earth then that will be 0 GPE. But if it's the surface of the earth then the centre will have a different value.

As per DaleSpam, it's the difference between those two points that matters in calculations, not the values themselves.

Potential energy is 0 wherever it is convenient to set it to 0. Only differences in potential energy are meaningful, so you can set the 0 anywhere you wish as long as the differences are unchanged.
I clearly see the importance of the frame of reference here but from what you are saying it seems there can be no absolute value for P.E which implies no absolute for K.E as well which a scary thought for a mathematician don't you think???

sophiecentaur
Gold Member
Of course KE must also be 'relative'. Two asteroids belting through the Solar system on almost parallel courses and with small velocity difference won't do much damage if they hit each other. The only KE that is relevant to the collision / energy exchange is due to relative velocity.

sophiecentaur
Gold Member
Hold on there. The Earth is not a point mass, which would exhibit the inverse law for potential. The Shell Theorem tells you that, once below the surface of a sphere, the bits 'outside' provide no potential at all. Once you get to the centre, there is zero potential: not -∞, which some of you seem to be suggesting.

As I said earlier, take the potential as the work done bringing a kg from infinity to your point of interest. Everything then falls out nicely for you and you've no need to worry.

The maximum (negative) potential and, hence, the greatest (attraction) force is, therefore, on the surface.

An equation is worth a thousand words.

To get from the centre of the earth will take energy so you must be at a negative potential whilst at the centre.

Think of it like being in a hole - no matter what you do it takes energy to get you out. You are at a negative potential in relation to the surface.

You can certainly put yourself at zero, but the surface will always be at a higher potential - and you need to do work to get there.

sophiecentaur
Gold Member
Yep. Right.
It's just that the force law is linear and not 1/r2.
The extra PE is
-∫Gρr dr from r=0 to r = R
which is
-GρR2/2

The potential wrt to the shells outside you may be zero but, as soon as you start to climb, there are shells below you, which gives you the result above. But I am still having a bit of a problem with the definition of potential. Inside, it's definitely zero inside a shell but suddenly it becomes non-zero. That's the problem with intuition - you can't trust it.

D H
Staff Emeritus
Inside, it's definitely zero inside a shell but suddenly it becomes non-zero.
There is always an arbitrary constant with respect potential energy. You can define the potential energy for a hollow spherical shell of mass such that the potential inside the shell is zero.

That is not the typical definition of gravitational potential. The canonical form has potential energy equal to zero at an infinite distance from the mass in question. With that definition, the gravitational potential energy for any mass distribution is given by

$$\phi(\vec r) = \int \frac{-G\rho(\vec x)}{||\vec x - \vec r||} d\vec x$$

In the case of a thin (infinitesimally thin) spherical shell of mass m and radius a, the gravitational potential is given by

$$\phi(\vec r) = \begin{cases}\frac{-G M}{r}\quad&r>a \\ \frac{-G M}{a}&r<a\end{cases}$$

Note that inside the shell the potential is constant but it is not zero.

K^2
Potential inside the shell is constant, not zero. What that constant is depends on your reference.

Edit: Never mind. Didn't see the second page.

sophiecentaur
Gold Member
Gottit
Thanks. I had failed to spot that you don't actually 'lose' potential when you go inside a shell, you just keep the same potential and the Force is zero because there is no potential gradient.

Dale
Mentor
I clearly see the importance of the frame of reference here but from what you are saying it seems there can be no absolute value for P.E which implies no absolute for K.E as well which a scary thought for a mathematician don't you think???
Well, you are correct that there is no absolute KE either, but I don't really see how that is implied by the fact that there is no absolute PE.

sophiecentaur
Gold Member
Isn't it all to do with the fact that work is done / energy is transferred because of differences in position or velocity? If two objects share both position and velocity then there can be no energy transfer.

Well, you are correct that there is no absolute KE either, but I don't really see how that is implied by the fact that there is no absolute PE.
because K.E = T.E - P.E
am i missing something here

D H
Staff Emeritus
because K.E = T.E - P.E
am i missing something here
Yes, you are missing something here. You are missing the fact that C-C=0.

Dale
Mentor
because K.E = T.E - P.E
am i missing something here
If you add the same amount to both TE and PE then KE is unchanged. So, in principle, it would be possible for KE to be "absolute" as long as TE and PE vary by the same amount.

If you add the same amount to both TE and PE then KE is unchanged. So, in principle, it would be possible for KE to be "absolute" as long as TE and PE vary by the same amount.
But according to the law of conservation of energy the total energy of the system remains same so the total energy cannot change or are you saying that changing the reference point will also effect the total energy of the system?