How can we choose any level for gravitational potential energy to be zero?

Question

In my book, I read that we can choose any level as Zero Gravitational P.E. and measure height of objects above it and call its energy 'mgh'. But by saying that all the points on that level is of zero potential, it can be inferred that it is an Equipotential surface but we know that all the points have different potentials (I know , a concentric sphere around earth is an equipotential surface but we often choose zero level that is not earth's surface )

Please explain as it is really confusing me :)

What do you mean when you say, "...all the points have different potentials?" All what points? and how do you know that they all must have different potentials? It almost sounds a as if you are arguing that no two points in Earth's gravitational field can have the same potential. — Solomon Slow, Sep 01 '20 at 14:08
See here I know they can have same potential. By points i mean all those points that lie on that level. How can you say that I have drawn an arbitrary line and all the points on it have same potential that is zero. They are not lying on an equipotential surface — Aryaman, Sep 01 '20 at 16:38

Gert · Accepted Answer · 2020-09-01T15:33:41.553

1

In my book, I read that we can choose any level as Zero Gravitational P.E. and measure height of objects above it and call its energy 'mgh'.

This works because we're almost always only interested changes in potential energy, rather than absolute values.

Let me show below that the reference point is of no importance in that case.

Let $h_r$ be a reference point where $U=U_r$. You can assume $U_r$ to be unknown. Now we look at an object of mass $m$ that is moved from $h_1$ to $h_2$ and want to know its change in potential energy.

We know that: $$U_1=U_r+(h_1-h_r)mg$$ $$U_2=U_r+(h_2-h_r)mg$$ The change $\Delta U$ is: $$\Delta U=U_2-U_1=U_r+(h_2-h_r)mg-[U_r+(h_1-h_r)mg]$$ $$\Delta U=mg(h_2-h_1)$$ So choosing an arbitrary, non-zero reference point gives the same, correct change in potential energy.

Another, more 'absolute' way of looking at it is by looking at the potential energy in a gravitational field$^\dagger$, far away from Earth's surface:

$$U=-\frac{GMm}{r}$$

Here obviously for $r\to +\infty$ then $U=0$. So here $r= +\infty$ is a good zero reference point.

$^\dagger$ http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html

edited Sep 01 '20 at 15:33

answered Sep 01 '20 at 15:18

Gert

35,289

A MathJax issue in your last line. – badjohn Sep 01 '20 at 15:22
thx Gert , but the problem is that How can I justify that the line I have drawn is an equipotential surface having zero potential. – Aryaman Sep 01 '20 at 16:41
You can't really and it isn't necessary. In the case of a central gravitational field, $U(r)$ only depends on the distance of the point to the centre of the field. Equipotential surfaces are thus concentric spheres around that centre. $U=0$ only for infinite distance of centre. the surface of the earth is approximated by a flat surface. – Gert Sep 01 '20 at 16:49
Gert , Can we say that the line we have drawn for a simple problem (as per our convenience) is so small that we should not compare it with any concentric sphere type of thing ( as for our earth any concentric sphere will be so large). So it may be considered as a part of a very big sphere and thus we call it a straight line. Hope you got my point :) – Aryaman Sep 02 '20 at 14:01
Yes, that is correct. In addition any height changes $\Delta h$ are to be small compared to the radius of the Earth, so that we can use the $\Delta U \approx mg \Delta h$ approximation, instead of the more accurate $U=-\frac{GMm}{r}$. – Gert Sep 02 '20 at 14:27

How can we choose any level for gravitational potential energy to be zero?

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