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	Astronomy Start‑class Mid‑importance
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This article was nominated for deletion on 22 March 2009 (UTC). The result of the discussion was no consensus.

Gravitational potential and gravitational potential energy

Gravitational potential is NOT the same as gravitational potential energy. The writer has mixed these two things. Needs fixoring ASAP. —The preceding unsigned comment was added by 130.234.6.46 (talk • contribs) on 06:59, 9 January 2006.

What, specifically, do you believe is incorrect about the article as it presently stands? --Christopher Thomas 07:46, 9 January 2006 (UTC)

The article is just fine, it's just named incorrectly. Gravitational potential energy (U) is equal to the mass of the body multiplied by the gravitational potential (Φ) => U = m * Φ -- the same — Preceding unsigned comment added by 130.234.202.80 (talk • contribs) on 15:39, 31 January 2006 (UTC)

Merge with potential energy

Most of the material on this page is covered at potential energy in the "gravitational potential energy" subsection. I've added "merge" tags to get discussion on what, if any, advantage there is to keeping this material on its own page. --Christopher Thomas 07:59, 9 January 2006 (UTC)

sign (+ or -) of the gravitational potential P

L.S.,

According to me (and others: see other pages of Misplaced Pages; also Vector Methods, D.E. Rutherford) the formula for P schould be: P=+GM/r and not P=-GM/r.

Here follows the reason. For P=+GM/r the accelleration is indeed given by the gradient of P. The components of the gradient of P are:(in ax the x is subscript, so the x-component) ax=-GMx/r3 ay=-GMy/r3 az=-GMz/r3. Suppose y=0 and z=0. Then ax=-GMx/r3. So for positive x, ax is negative (towards the left, so towards the pointmass) For negative x, ax is positive (towards the right, so also towards the pointmass. As it should be, of course: gravitation is an attractive force.

Now, if you should have have: P=-GM/r,then ax= + GMx/r3. For positive x that value is positive, so to the right, away from the point mass. For negative x that value is negative, so to the left,also away from the point mass. That would make gravity a repulsive force.

So, P=+GM/r gives the correct value for the acceleration by a=gradient P. The + or - sign is not trivial and should be correctly chosen in Misplaced Pages. There must me no ambiguity at this point(unless there should be ambiguity in the scientific litterature; in that case the ambiguity should be mentioned in the page).

Besides i remark, as is done by previous contributors, that the potential is not the same as the potential energy of a mass of 1kg (in that case you would not need the concept). The value of the concept is that it gives the acceleration by means of its gradient.

Fransepans (talk) 22:32, 21 September 2009 (UTC)

This is a case of differing sign convention. Per potential, the usual convention is to define a potential ${\displaystyle U({\vec {x}})}$ such that ${\displaystyle F=-\nabla \cdot U({\vec {x}})\cdot Q}$, where Q is the charge upon which the force is acting (in this case, mass). This convention gives the change in potential energy ${\displaystyle \Delta E_{p}}$ over some path ${\displaystyle P}$ as ${\displaystyle \Delta E_{p}(P)=\int _{P}\nabla \cdot U({\vec {x}})\cdot Q\cdot d{\vec {x}}}$. Because a force moves a particle in the direction of decreasing potential energy, a minus sign results in the expression for force. --Christopher Thomas (talk) 23:59, 21 September 2009 (UTC)

L.S., I can agree with the convention (although the literature is not unanimous at this point). I conclude that, according to that convention, in the article a minus sign must be added to the word "gradient": The gravitational field equals MINUS the gradient of the potential. We agree on that. I will edit the page accordingly.

Dimensions are very important

The dimensions such as force, mass, time, and distance are very important from an engineering point of view. Although the pure mathematician may think dimensions are irrelevant, there are engineering oriented people who read the Misplaced Pages. Therefore it is important that we give adequate attention to dimensions in Misplaced Pages articles such as this. RHB100 (talk) 02:23, 27 February 2010 (UTC)

This is a case where too much detail is as bad as too little. The article currently emphasizes the dimensions of the gravitational constant, which is not relevant to understanding the potential, and is written in a potentially confusing manner. Sławomir Biały (talk) 11:11, 27 February 2010 (UTC)

It certainly is not too much emphasis to merely state the dimensions of the gravitational constant. If anything it is too little emphasis since only the dimensions, not the units, are stated and the value is not stated. RHB100 (talk) 21:52, 27 February 2010 (UTC)

It is important that at least the dimensions be stated since it aids the reader in verifying the dimensional compatibility of the equation. One of the first and most important things that one should do upon encountering a new equation is verify the dimensional compatibility. Every well educated engineer with degrees from one or more of the better American universities understands this. As a licensed professional engineer I know the importance of dimensional compatibility. RHB100 (talk) 21:52, 27 February 2010 (UTC)

I find it distracting from the much more meaningful point that the potential has units of energy per unit mass. Also, when saying "which has dimensions", the referent is unclear: the reader is expecting dimensions of the potential (the subject of this article), but is instead given the dimensions of the gravitational constant. This sort of information belongs in a footnote, if in the article at all. Surely professionalism also demands the ability to follow footnotes (if not wikilinks to the gravitational constant article). Sławomir Biały (talk) 23:53, 27 February 2010 (UTC)

Update. I have started a dedicated section on Units and dimension. Please populate this with information that would be useful to "professionals". Sławomir Biały (talk) 00:17, 28 February 2010 (UTC)

It needs to be in the main section not relegated to a footnote. The reader needs to see the dimensions or units of all the factors on the right side of the equation and verify that they are compatible with the left side of the equation.

Sławomir Biały, don't you know, dx^3 is not a differential element of mass. Any competent mathematician should know this. Where did you study math?

Here ρ(x) is the distribution function, and d^3x is the volume element. Sławomir Biały (talk) 02:20, 28 February 2010 (UTC)

I believe this notation, using ρ for the mass density, is a fairly standard one. I will wait a few days for others to comment, and then restore the original version of the formula. Also, back to the original topic of the thread, I don't think the value of the gravitational constant G needs to be right next to the formula for V. It seems to be that the better place for that is in the new section that I have created for a fuller discussion of the units and dimensions. Sławomir Biały (talk) 03:15, 28 February 2010 (UTC)

I'd suggest that giving the exact value of G right beside the formula for V is distracting (the wikilink is sufficient). I also don't think the dimensions of G are necessary either. I do think it's a good idea to give the dimensions of the potential right after giving the formula. Regarding the notation switch from ρ(x)dx to dm, I'm ambivalent: in some sense dm is more conceptual, on the other hand, the notation ρ(x)dx is more friendly to the less experienced reader. So who is the target audience?

I'd also suggest that in the newly added expansion of the denominator the article should not assume that the reader is aware of the convention that if v is a vector quantity then v is its magnitude. Either something should be said about this, or the notation should be changed.

I also think the first thing someone should think about when seeing an equation for a physical quantity is what it says about the dependence of the quantity on others. I would place verifying the dimensional consistency of a >200 year old equation low on my list.

As a last comment, I find it generally better in a content dispute to revert the article back to the previously held "consensus" version of the article and to use the discussion page to build a new consensus (this would appear to be this). RobHar (talk) 03:57, 28 February 2010 (UTC)

Comments:

• I don't see any problem with giving the numerical value of G, including units, in the 'Mathematical form' section. It's true that the article is supposed to be about potential and not about the gravitational constant, but I don't think that giving the value is crossing the bounds of WP:TOPIC. Don't don't agree entirely with RHB100's reasons for including it, but since this is essentially a physics article it seems appropriate to include the values of constants used. However, the placement on the same line as the formula is awkward. Perhaps it can be incorporated into the text by rewording the paragraph.

I took a stab at a rewording that may work as a compromise. Not claiming it's perfect but hopefully it's a step towards addressing everyone's concerns. It's a small point but the previous wording had M being used both as a position in space and as a mass value; there's not much harm in confusing the two for a point mass but it shows the previous wording wasn't perfect either.--RDBury (talk) 05:09, 28 February 2010 (UTC)

• I've always seen the volume element written as dV, but this would be problematic in this article because V is also used for the potential. The dm form is correct but in practice it would just be factored as ρdV anyway. If there is a way to work around the coincidence of V being used to mean two different things then I think that would be the best way of writing it, otherwise I don't really have a preference.

It appears that Φ is also commonly used for the potential, so using it instead of V would allow using V for volume as is customary. So I vote to replace all the Vs with Φs and use dV as the volume element.--RDBury (talk) 05:39, 28 February 2010 (UTC)

• The 'Units and dimension' and section that was added, seems inappropriate per WP:TOPIC whether or not the value of G is included in the other section. This is an exercise in dimensional analysis anyway so I don't think it adds value.

--RDBury (talk) 04:45, 28 February 2010 (UTC)

The use of the differential element of mass, dm, is more common than what some people realize. Dynamics professor, Dr. Peter W. Likins, taught at UCLA before going on to become Dean of Engineering at Columbia, and President of Lehigh and Arizona. In his dynamics text, "Elements of Engineering Mechanics", Likins used dm wherever appropriate such as in defining angular momentum. In the dynamics text, "Methods of Analytic Dynamics" by Leonard Meirovitch, dm is used wherever appropriate. I think that in dynamics texts used in engineering schools, the use of dm is fairly standard. RHB100 (talk) 23:32, 28 February 2010 (UTC)

I, for one, am not contending that dm is uncommon. In fact, it is very common in sufficiently advanced physics texts as well (by which I mean textbooks that would be used in sophomore physics and beyond). However, ρ(x)dx is much more likely to be understood by anyone who has taken some calculus, without necessarily moving on to higher level physics courses. Another matter is that dm is really mostly a shorthand for ρ(x)dx. Were one to actually compute an integral, one would, in most cases, immediately rewrite dm as ρ(x)dx. RobHar (talk) 22:41, 28 February 2010 (UTC)

I have several problems with "dm". First, the integral is still over physical space, but the notation does not emphasize this. It should at the very least be ${\displaystyle \int _{\mathbb {R} ^{3}}dm(\mathbf {x} )}$. Secondly, I do not like the way in which the integral with respect to "dm" is treated as though it were something well-defined in its own right. If this is an ordinary Riemann integral, then it is the integral of a distribution function (and so we should write the simpler ρ(x)dx). If it is a Lebesgue integral of a mass measure, then that should be indicated instead. But it is not explained at all what the "integral over the extent of the differential mass elements, dm" means. Also, the fact that this is a convolution is significant, and should be mentioned, although that fact was removed in the recent round of edits. Sławomir Biały (talk) 23:08, 28 February 2010 (UTC)

The problems you have with dm are due to your own lack of understanding and education. You should go back and repeat undergraduate dynamics for rigid bodies. RHB100 (talk) 21:11, 1 March 2010 (UTC)

The expression, dx, is not a meaningful expression for a differential volume. dx dy dz is a proper expression for a differential element of volume, a differential cube. I think that a student who had studied calculus would be confused by dx. The elements of dx dx dx are not orthogonal, they are all in the same direction and it is thus not a meaningful expression of a differential element of volume and it is confusing to say the least. RHB100 (talk) 00:07, 1 March 2010 (UTC)

The notation dx is completely standard in physics texts, by the way, and it is not meant to mean dx dx dx=(dx), but rather the 3 symbolizes the fact that the integral is over three dimensions. It is true that from the point of view of making this article at least accessible to people who have done calculus, it would be better to use dxdydz. RDBury above also suggested dV instead which I have certainly seen a lot, but is not necessarily standard, so I might be reticent to use that. RobHar (talk) 00:28, 1 March 2010 (UTC)

I have not read all physics books but I did take a survey of 4 physics books that I own. All 4 used the differential element, dm, in connection with angular momentum and moments of inertia. I did not observe the notation, dx in these particular books but again I confess I have not read all physics books. The books I surveyed are "Elements of Physics" by Shortley and Williams, "University Physics" by Sears and Zemansky, "Fundamentals of Physics" by Haliday and Resnick, and "Physics for Science and Engineering" by McKelvey and Grotch. RHB100 (talk) 01:52, 1 March 2010 (UTC)

I have added an explanation of the unexplained notation, since from the earlier version, is was not even clear that the integration was over ordinary physical space. I also felt the need to explain what "dm" is, and link to the appropriate notion of integral. Sławomir Biały (talk) 12:01, 1 March 2010 (UTC)

Physics and engineering books use integrals with dm as the differential element and manage to make it clear to the intelligent student without a lot of extra explanation. We should use mathematics to explain not to confuse. RHB100 (talk) 21:11, 1 March 2010 (UTC)

There isn't "a lot of extra explanation", we simply state correctly (and now with a reference) what "dm" refers to in the notation. Saying that "dm" is a "mass measure" is mathematically much more satisfactory than referring to fictional "differential elements of mass", and is not likely to lead to any greater confusion than the latter. (However, it still seems simpler to me to work with a distribution function instead.) The latest version now also includes a brief discussion of the role of Poisson's equation, which was strangely lacking from the earlier revision. Also, I have corrected the recently-introduced errors that put R=r−r_i as incorrect and incoherent. Sławomir Biały (talk) 22:06, 1 March 2010 (UTC)

Look at the definition of R. Try to understand it. One of the things we now need to do is add a figure showing R, x, and r. To avoid ambiguity we need to talk in terms of point a and point b rather than just talk about the point, the point, ... .

—Preceding unsigned comment added by RHB100 (talk • contribs)

Please stop adding this, it is both incoherent and incorrect. Also, you continue to remove sourced content. The potential is a function of a point r in space. There is no need to talk about the center of mass of the system when defining it. Of course, if one picks a coordinate system that places that origin at the center of mass, then one recovers what you have written, but that does depend on choices that are irrelevant to the definition of the potential. Sławomir Biały (talk) 01:32, 2 March 2010 (UTC)

The current revision is wrong. Also, sourced content was removed, I don't know if you are aware of that. (I keep pointing this out, but you keep ignoring it, and just reverting every edit that I make, which is getting to be really aggravating.) Sławomir Biały (talk) 02:01, 2 March 2010 (UTC)

Comment. I tried to make various compromises in the article, including using RHB's preferred "dm" notation, but I also felt it important to include the case of a continuous mass distribution ρ, and a discussion of the Poisson equation. My efforts to achieve compromise have been systematically reverted, often without any edit summary or talk page explanation, by RHB. On top of that, RHB has introduced conceptual errors into the article, as well as unnecessary and unexplained formulas (like the formula that pretends to be an integral in spherical coordinates, but fails to say what any of the variables are), and removed sourced content. I ask that we should either go back to the earlier consensus version of the section, or that a new consensus should emerge. The current version of the article is totally unacceptable. Sławomir Biały (talk) 02:08, 2 March 2010 (UTC)

Let me add that this formula

${\displaystyle V(\mathbf {r} )=-\sum _{i}{\frac {GM_{i}}{|\mathbf {R} -\mathbf {r} _{i}|}}\,{\rm {where\ \mathbf {r} =|\mathbf {R} -\mathbf {r} _{i}|\ }}}$ and where R is a vector from the center of mass to point b.

is extremely unclear, since the vector r appears on the left-hand side as the argument of the function and on the right as a scalar (which is one reason I changed these rs to x: they were persistently being misunderstood by an editor). The equation r = |R − r_i| is also questionable, because on the left is something that does not involve i and the right clearly depends on i. Happily, I see that my preferred (correct) revision has been restored that has none of these problematic issues. Sławomir Biały (talk) 11:49, 2 March 2010 (UTC)

I have to add that I don't see how any formula involving the center of mass can be anything non-trivial. I thought about this last night, and I have an example that I'd like others here to think about: Suppose I'm standing at the origin. There are two point masses, each 1 kg. One point mass is 1 m directly to my left, and the other point mass is 1 m directly to my right. I'm at the center of mass of the system of the two point masses. But the potential is not singular there; instead it is singular at the locations of the two point masses. So the center of mass really doesn't appear naturally in the potential; you can always change your coordinate system, as I think RHB100 wants to do, but there's no reason why the center of mass is a better place than any other (from a mathematical perspective). Ozob (talk) 12:44, 2 March 2010 (UTC)

I think terms such as my left and my right are ambiguous. You need to talk about point a and point b. Another thing I rely on vectors. Vectors are independent of any coordinate system in which they happened to be expressed. I like to rely on vectors since it often eliminates the need for consideration of a coordinate system. It is important to give consideration to the center of mass when expanding the potential in terms of Legendre Polynomials as I have done below. RHB100 (talk) 03:14, 3 March 2010 (UTC)

I had a similar problem as well, and initially tried to post it, but my explanation failed utterly to convey the sense of it. For me the bottom line is that the potential is currently defined in a totally coordinate-independent manner, which is how it should be. Sławomir Biały (talk) 13:00, 2 March 2010 (UTC)

Which revision?

We need to develop a consensus. I propose my most recent revision as a suitable starting point. I have attempted to take into consideration the comments of the preceding section, including RHB's preference of the "dm" notation, as well as RDBury's suggestion of using "dV" and what seems to be a consensus that a distribution function should be presented. (Not to rehash old arguments, but most of the physics books that I checked which used "dm" in an intuitive setting almost immediately explained this with a distribution function.) In addition, I had made a number of edits to address some of RHB's misunderstandings of the text as it had stood. The other revision, which I feel to be substantially inferior, is this one (here is a diff between the two revisions). I would ask that some consensus should develop on which revision should be the "stable" one. Sławomir Biały (talk) 02:21, 2 March 2010 (UTC)

Hmm. While you wrote this, I was working on a new revision myself. It's pretty similar to yours; as of now it's on the page itself, but in case the article changes my version can be seen at . Ozob (talk) 04:47, 2 March 2010 (UTC)

Looks fine to me, clear progress over the version I had suggested. Sławomir Biały (talk) 11:22, 2 March 2010 (UTC)

Sławomir Biały, you are not qualified to accuse me of misunderstanding anything. I am a Licensed Professional Engineer in the field of control systems engineering. I hold advanced engineering degrees from two of the better quality engineering schools in America. I have had a career designing autopilots for American missiles. I don't believe you could get an undergraduate degree from one of the better quality American Universities. You are not qualified to accuse me of misunderstanding anything. The errors I have pointed out are due to sloppy writing and errors on your part not any misunderstanding on my part. RHB100 (talk) 23:43, 2 March 2010 (UTC)

Wow, that's some bold text.

I don't understand why you're even interested in qualifications. Isn't it more important to be correct? I for one plan to never reveal my identity or my qualifications. If I'm wrong, I want to be told I'm wrong, and I don't care whether the person doing it is a certified genius or a bored teenager.

RHB100, your version of the article is wrong. The formula V = -Σ GM_i / |R - r_i| is incorrect. I gave an example above: Suppose that you are standing at the origin and that there is a 1 kg mass 1 m to your left and a 1 kg mass 1 m to your right. What is the potential? Ozob (talk) 01:37, 3 March 2010 (UTC)

In the example above you had V(r), now you have V without an argument. V without an argument is correct. The notation with V(r) is incorect since r is a dummy variable of summation. I have added a diagram below for the continuous case which should also make it clear for the discrete case if you keep in mind that x in the continuous case replace the discrete, r_i. RHB100 (talk) 02:46, 3 March 2010 (UTC)

Could you explain for me how my example would look in your fomulas? Ozob (talk) 02:59, 3 March 2010 (UTC)

V(r) means V at point r. By the way, an additonal notation b is not needed for this point.11:50, 3 March 2010 (UTC)~

Mathematical form is poorly written

The mathematical form section is poorly written at present. Expressions such as "where r is a vector of length r pointing from the origin" indicate the general sloppiness with which the article is written since the origin is not even defined. The article below indicates some of the ways the article could be improved. A diagram has been added to this new version to clearly show the relationship of R, r, and x to supplement the verbal definition. Also the integral for the continuous case has been expanded into a series of Legendre Polynomials in the new version. The comments on mass measure serve only to add confusion and should be removed.

—Preceding unsigned comment added by RHB100 (talk • contribs)

Mathematical form, New

The potential V at point b located a distance r from a point mass at point a is given by:

${\displaystyle V=-{\frac {GM}{r}},}$

where M is the mass at point a and G is the gravitational constant (6.67428 × 10 N (m/kg)).

The potential, V, has dimensions of energy per unit mass, and it approaches an upper limit of zero as r approaches infinity.

More generally, the potential associated with any mass distribution is obtained by the superposition of potential associated with the individual masses. In the case of discrete point masses, this is the algebraic sum of the gravitational potentials. The potentialat point b is given by

${\displaystyle V=-\sum _{i}{\frac {GM_{i}}{|\mathbf {R} -\mathbf {r} _{i}|}}\,{\rm {where\ \mathbf {r} =|\mathbf {R} -\mathbf {r} _{i}|\ }}}$ and where R is a vector from the center of mass to point b.

For the continuous case the potential, V at point b, is computed as an integral over the distributed mass,

${\displaystyle V=-\int {\frac {G}{|\mathbf {R} -\mathbf {x} |}}\ dm}$ where dm equals density times differential volume, where x is a vector from the center of mass to the differential element of mass, and where r is a vector from the differential element of mass to point b, i.e. ${\displaystyle \mathbf {r} =|\mathbf {R} -\mathbf {r} |.}$

Expanding the denominator by taking the square root of the square, carrying out the dot product, and factoring R out of the denominator we get the more useful expression,

${\displaystyle V=-\int {\frac {G}{\sqrt {R^{2}-2\mathbf {R} \cdot \mathbf {x} +x^{2}}}}\,dm=}$

${\displaystyle -{\frac {1}{R}}\int G\,/\,{\sqrt {1-2cos(\beta ){\frac {x}{R}}+\left({\frac {x}{R}}\right)^{2}}}\,dm.}$ If we expand in Legendre Polynomials using Theorem 2 on page 454 of section 10.8 of then we get the result,

${\displaystyle V=-{\frac {G}{R}}\int \sum _{n=0}^{\infty }\left({\frac {x}{R}}\right)^{n}P_{n}(cos(\beta ))\,dm\ {\rm {}}}$ where the ${\displaystyle P_{n}(cos(\beta ))}$ are Legendre Polynomials of degree n and argument ${\displaystyle cos(\beta )\,.}$ This is a form which is quite useful for applications as shown in section 11.7 Gravitational Potential ... of .

The gravitational field, and thus the acceleration of a (small) body in the space around the massive object, is given by the negative gradient of the gravitational potential. The gravitational field associated to a point mass follows an inverse square law: the magnitude of the observed acceleration is

${\displaystyle \left|\mathbf {a} \right|={\frac {GM}{r^{2}}}.}$

The direction of the acceleration is towards the point mass. Thus the acceleration vector is

${\displaystyle \mathbf {a} =-{\frac {GM}{r^{3}}}\mathbf {r} =-{\frac {GM}{r^{2}}}{\hat {\mathbf {r} }}.}$

—Preceding unsigned comment added by RHB100 (talk • contribs)

Perhaps it is better to put the power series in a separate section, and to work out the first few terms.--Patrick (talk) 11:30, 3 March 2010 (UTC)

1. C. R. Wylie, Jr. 1960, Advanced Engineering Mathematics (McGraw-Hill Book Company)
2. Leonard Meirovitch 1970, Methods of Analytical Dynamics (McGraw-Hill Book Company)

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