Talk:Newton's law of universal gravitation

Gravitational force of exoplanets and other planets/moons

What is the formula to determine the gravitational force of a planet or moon either in terms of units or in relation to Earth? If you have the answer, could you put a message on the discuss page of Chiss Boy? Chiss Boy 12:15, 25 April 2007 (UTC)

meaning of the formula

At the time of writing, the introduction is in an inconsistent state, with the formula being

$F=G{\frac {m_{1}m_{2}}{r^{2}}}$

and the commentary saying that it is considered a repulsive force for consistency with Coulomb's Law. Clearly you can both give the "F" value a negative sign and say you consider the force repulsive, or you can give it a positive sign and say you consider it attractive.

This really matters only when you consider the vector form of the law. In this case, once you have defined which force and which vector distance you is mean by F₁₂ and r₂₁, the sign is not conventional: it must agree with the attractive nature of the force.

At the time of writing the article defines:

F₁₂ is the force on object 2 due to object 1

$\mathbf {\hat {r}} _{21}\equiv {\frac {\mathbf {r} _{2}-\mathbf {r} _{1}}{\vert \mathbf {r} _{2}-\mathbf {r} _{1}\vert }}$ is the unit vector from object 1 to 2.

Then the correct formula would be:

$\mathbf {F} _{12}=-G{m_{1}m_{2} \over r_{21}^{2}}\,\mathbf {\hat {r}} _{21}$

That is, the force due to object 1 on object 2 acts in the opposite direction to the position of object 2 with respect to object 1.

Stefano 21:52, 11 May 2006 (UTC)

Note - this has been changed, the mention of "repulsive force" is gone and the law is stated in its usual way (all quantities positive, and forces attractive). Much less confusing this way.

67.36.184.126 02:49, 3 February 2007 (UTC)

Just for the record - I made the correction mentioned above. I was not logged in at the time.

Cgwaldman 17:59, 5 February 2007 (UTC) The formula is actually:

$F=G{\frac {m_{1}+m_{2}}{r^{2}}}$

Donteras 07:08, 15 August 2007 (UTC)

No, it is correct as given in the article, and as in any high school physics text. Multiply the masses. —Duae Quartunciae (talk · cont) 07:49, 15 August 2007 (UTC)

2 overlapping forcefields add together, not multiply by each other. What universe do you live in? Donteras 06:53, 16 August 2007 (UTC)

Duae_Quartunciae is correct. Cardamon 07:26, 16 August 2007 (UTC)

I have removed some off topic additions. This is an extreme step, but multiple advice has been given on the point. Removal of irrelevant comments from talk pages is permitted by the Wikipedia:Talk page guidelines. Stick to discussing the article. —Duae Quartunciae (talk · cont) 08:04, 18 August 2007 (UTC) For the record: here is the edit in which I deleted off topic comments. [1] Even before this step I had tried to help at User talk:Donteras#Newton's gravity. —Duae Quartunciae (talk · cont) 08:08, 18 August 2007 (UTC)

Law or theory?

I've usually heard this described as a law, and indeed that is the title of the article. However the first line of the article says that it is a theory. This should be either changed, or explained in the article. Harley peters 01:35, 27 October 2006 (UTC)

What ranges of r, m amd M has this Law/ Theory been tested experimentally? 81.152.87.111 (talk) 01:52, 13 September 2008 (UTC)

Guys, "Newton's law of universal gravitation" is invalid for almost a century!

Wikipedia shouldn't propose a "law" that is outdated for almost a century. When one types "Gravitational force" one may expect to see a contemporary, Einsteinian equation for this force, which BTW, I placed a few years ago in Wikipedia but then it has been deleted when someone started to rearrange gravitation pages according to his/her poor understanding of gravitation.

How Wikipedia is going to help in understanding the contemporary physics if such a simple thing as ordinary gravitational force is called a "force of attraction" while we know for almost a century that nothing is attracted here to anything else. Just imagine how would Einstein feel if he knew that almost a century after he discovered where the gavitational force comes from, some guys in Wikipedia think that it comes from Newtonian "spooky action at distance" (from "attraction" according to Newtonian physics).

So please eiter restore my old post with Einsteinian equation for gravitational force or, if you don't like it for some reason, put your own version of it. But let it be Einsteinian and not Newtonian physics, which is nice for historical and many practical reasons but unfortuntely not true. I also hope that at certain point you don't return to geocentric planetary system as being good enough for the readers of Wikipedia and easier for you to make sense of. Jim 20:53, 5 November 2006 (UTC)

Let me put my 2 cents in. Newton's theory is not outdated. Most scientists are convinced that a graviton conveys force. In my opinion, Newton's theory is mathematically wrong, but conceptually correct. If Einsteinian gravity were explained in terms of vectors instead of warped space, very little of the theory would change. It is indeed possible that Einstein's theory is mathematically correct and conceptually flawed. It also might not hurt for you and the person who altered your page to be a little more open minded Donteras 06:31, 15 August 2007 (UTC)

The old "feather and hammer in a vacuum" chestnut...

Can somebody who knows the deal explain that to me?

I understand that two small objects different in mass, but attracted to the same massive object, will accelerate at the same rate, because the increased inertia counteracts the increased force (f=ma). However, shouldn't the massive object have a greater amount of acceleration in the case of the larger (small) object? It has the same mass in both cases, but in the second there is greater force. Therefore wouldn't the relative acceleration as seen by an observer on the massive object be greater for larger objects? This has been bugging me for years agnabbit! --Sophistifunk 04:53, 22 November 2006 (UTC)

You are of course right. If you take for small objects the Earth and the Moon and for the massive object the Sun then the acceleration of the Sun towards the common center of mass with the Earth will be greater than acceleration of the Sun towards the common center of mass with the Moon.

The force acting between the Sun and the Earth is $F_{se}=Gm_{s}m_{e}/r^{2}$ so the acceleration of the Sun towards their common center is $a_{se}=F_{se}/m_{s}=Gm_{e}/r^{2}$. The force acting between the Sun and the Moon is $F_{sm}=Gm_{s}m_{m}/r^{2}$ so the acceleration of the Sun towards their common center is $a_{sm}=F_{sm}/m_{s}=Gm_{m}/r^{2}$ and so since $m_{e}>m_{m}$ so is $a_{se}>a_{sm}$. Jim 17:26, 22 November 2006 (UTC)

The simple answer is that the reason the two small objects apparently fall at the same speed in vacuum is that mass of the Earth is so large compared to both the feather and the hammer, that the difference in mass between the two small objects is too small to produce an observable difference in acceleration. You are incorrect that the two small objects fall at exactly the same rate - the hammer does actually fall very, very slightly faster. It is not true that the increased mass of the hammer "cancels out" the increased gravitational attraction due to inertia. More massive objects accelerate faster toward gravitational partners (in this case "fall") faster than less massive objects in all cases, everything else being equal. It is also to important to consider Newton's Third Law (the "equal and opposite" one), which points out that the attraction between the earth and say, the hammer, is symmetrical. The force that the Earth experiences from the gravitational presence of the hammer is the same as the force that the hammer feels from the earth. Mass and inertia do come into play here. Remember that "force = mass * acceleration", Newton's Second Law. The force on the hammer and the Earth are the same, but the Earth has a much, much larger mass. As a result, the acceleration that the Earth experiences (in the reference frame of say, the Solar System) is much, much smaller than that of the hammer. But the Earth does move a tiny, tiny distance toward the hammer when you drop it, and the hammer moves a much larger distance toward the Earth. You are right that the Earth moves ever so slightly more when you drop a hammer than when you drop a feather. -- Beland 19:01, 17 March 2007 (UTC)

Wow. Let's see if I can explain this. Let's say the feather weighs 1g and the hammer weighs 20g. The gravitational force on the hammer is 20 times that of the feather (I'm assuming that we are dropping them on a massive body like the earth). The feather has 1g of inertia and the hammer has 20g of inertia (as per the equivalence of mass/inertia principal), so it will accelerate at 1/20th of the rate of the feather. This exactly compensates for the 20x force. If dropped with their bottoms at the same height and over level ground, they would hit the ground at the same time in a vacuum. Object shape and density make a huge difference when they suffer wind resistance. Donteras 06:53, 15 August 2007 (UTC)

wow

who the hell did ALL the formulas with r? ... it's f=(Gm1m2)/d^2 ... it's over DISTANCE squared... the only people i've ever seen use r were the slow students in my physics class so that they could remember to use a planets radius to find the acceleration due to gravity on that planet's surface, because you use the distance from each object's center of gravity. common convention says d is the variable used for distance... speed isn't s=r/t ... is it!??!?!?! NO ... even if at one time, you decided to determine the speed based off of a circumference of a circle, where you'd have s=2πr/t you wouldn't continue to use s=anything other than d/t afterwards would you?!?!?!? NO.... god... go read a physics textbook if you have to...

• It's distance between two point masses - which, since gravitation typical deals with objects rotating about one another, the use the value "r", since it implies a radial distance. However, it really, really, doesn't matter, and I don't understand why you're getting upset about it. If I say the equation of a parabola is $x=y^{2}$, where x is the vertical axis and y is the horizontal axis, am I really saying anything different? No, of course not. There are different conventions for distance in different places, and you should not let that throw you into a tizzy. --Haemo 08:23, 10 January 2007 (UTC)

this is the only response that i'll take with a grain of salt, as it is the only logical answer i've received here. perhaps the convention has changed. oh, and you're obviously not saying anything different by using $x=y^{2}$ just as you could say $g=h^{2}$ where g is the vertical axis and h is the horizontal, however i'd still have even more issues with that than i do with the equation at hand (as most anal mathematicians would as well). changing the letter for a variable in the equation of universal gravitation is one thing, changing the letter for the variable in a cartesian coordinate system graph is completely different, that's heresy my friend. since i accept your explanation (times do change eh?) i'll change the variables back.

• Perhaps to elaborate - if I say "distance" from the earth to the moon, there can be some confusion about what I am saying. Is it the distance from the earth, or from the center of the earth? Is it to the surface of the moon, or to the center of the moon? With d, it's not intuitive what is being said - if I told you the distance between two books on a shelf was 1m, would you assume that was from cover to cover, or from middle to middle. It's not clear. However, if we use r, the analogy with the radius of a circle is brought to mind - the radius of a circle is from the CENTER of the circle, always. This helps the symbol be a mnemonic for the formula. --Haemo 08:30, 10 January 2007 (UTC)

and? like i said, it was used by the slow students to remember that the distance is from the center of gravity. the equation as it stands (and i'll even quote mr feynman on this one :P) is f=g(m1m2)/d^2 (although Haemo makes a good point)

• Also, just FYI, in most college physics textbooks I have seen, s is reserved for displacement, not speed. See? Symbols can differ.

FYI I've seen the use of d, and Δx or Δy (when dealing with rectilinear displacement along either axis) all used for displacement, my physics textbook from college which i'm looking at right now uses d and a subscript depending on the direction, however note that the unit for displacement is (gasp) meters... which implies (gasp) distance, hence the use of d :P... i'm not sure what textbook you use that uses s, but it's not in the majority.

Thanks for editing them back, but just so you know, I made all of those comments. Also, the reason s is typically used for displacement is because when you do calculus, in Liebneiz's notation, you get annoying expressions like $dd/dt$ for velocity, if you use d for displacement. --Haemo 00:00, 11 January 2007 (UTC)

d? r? What's the difference? I've seen them both used. I prefer d, myself, but I'm not married to it. Donteras 17:51, 17 August 2007 (UTC)

Gravity should not be based on an unexplainable or magical force.

My reasoning has me convinced magical or miracles attraction should not be treated as a force but rather as a perception...

Force is calculated and explained via velocities and or as per Newtons Laws, and velocity is calculated and explained via repulsive events as per Newtons Laws..

Repulsion is self explanatory but when we refer to a perceived attraction we should note it is only possible via velocities converging upon another in the same direction. let me explain...

Let ">" equals a velocity to the right.

Let "<" equals a velocity to the left.

Let "-" equals a magnitude of force the more "--" the greater the magnitude of force and or velocity gained via Newtons Laws.

Let "o" equals that Newtons Laws are fully complied to via an equal or greater opposing force.

If we are to have a velocity to the left and hence a force exerting to the left we would need a footing or force in the opposite direction so to speak..

Put simply ask what velocities or force would we need to create a force to the left?

is this ">" direction with a velocity or force possible with out this "<" direction with a velocity or force possible?

No it is not! Force depends on an equal or greater force interaction -

As in <--o--> In this diagram a force or velocity to the left is satisfied with a force or velocity to right and vice Vs The above diagram depicts a repulsive reaction, lets now attempt to depict attraction now..

"-> <-" In this diagram the two velocities must have originated from somewhere - But how? Magic? or maybe some Gawd? via a miracle <LOL>

As one SHOULD now see, attractive perception can only be gained by some previous REPULSION'S) as in <-o-> <-o->

And guess what! that goes for Gravity and Electromotive Forces as well!

Thank you for your Attention...

Yours,

peter_j_schoen@msn.com

—The preceding unsigned comment was added by 203.122.123.9 (talk) 00:28, 14 January 2007 (UTC).

No offense, but the above appears to be pseudoscientific nonsense. -- Beland 18:31, 17 March 2007 (UTC)

Your theory would be better served if you don't change terms mid-theory. o is not an object and <-o-> is not a repulsion. <-o-> is action/reaction and can also be written as ->o<-. Donteras 06:58, 16 August 2007 (UTC)

History

It would be nice to have a history of the theory, including genesis, acceptance, refinement and expansion, application, theoretical obsolescence, and current role in education. -- Beland 19:03, 17 March 2007 (UTC)

Section about Einsteinian look at gravitation

Hi, I added at the end of the article a section explaining to high school stidents how Einstein's theory explains gravitational force so every high school student can easily imagine why things fall on the Erth rather than fly away from the Earth. Jim 18:57, 7 July 2007 (UTC)

With respect, this is way too long for an article on Newton's theory. I'll solicit more input, but I think this article should have no more than a brief comment on the fact that Newton's ideas have been superseded by relativity. -- Duae Quartunciae (t|c) 00:36, 20 July 2007 (UTC)

We all know that the gravitational force is always attractional in nature.but could you please tell me why is this gravitational force always attractibe, why not repulsive? What is the cause of attraction in this force? —The preceding unsigned comment was added by 59.88.212.125 (talk • contribs) 07:26, 13 July 2007 (UTC).

Conspiracy theories?

This part of the article is bothering me:

"Yet still in 2007 there is no official confirmation that the universe is stationary (a small wonder when papers delivering any evidence of it are consistently rejected by editors of scientific journals)."

Oh, I see, the official confirmation is being withheld by those evil editors who insist on silly things such as rigorous mathematical proofs and unheard-of things called "facts". They must all be part of some conspiracy.

I'd remove the bit between parenthesis, unless backed by some *really convincing* explanation. -- Rizzardi

The parenthical remark is certainly out of place; but actually this whole paragraph was really just biographical remarks on Einstein; not about relativity itself. I've removed it. The whole section should still be massively pruned. A link to GR, and a statement that this has superseded Newton, is entirely in order. An attempt to explain GR is out of place. The initial claim that Einstein gave a mechanism is not really true. He gave a description in terms of geometry, but not a mechanism. A mechanism can't really be claimed until we know more about gravitons and the linkage with particles and QM. -- Duae Quartunciae (t|c) 00:43, 20 July 2007 (UTC)

I believe Einstein did give something similar to a mechanism. Warped time changes frequencies, which is directly linked to energy. Seeing as this energy cannot simply go away, it causes an object to move. I'm a bit fuzzy on why down is the preferred direction. This would probably be a bad time to point out that QM is fantasy. The first experiment designed to prove quantum theory (the double slit experiment) actually proved it wasn't true. I'm at a loss to explain why the theory was not immediately dropped. Donteras 07:39, 15 August 2007 (UTC)

Pruning/moving the explanation of general relativity

I propose the following as a replacement for what is currently given as "Einstein's solution". I feel this section is much too long for a page on Newton's theory. I suggest:

This is not a criticism of the long passage I am proposing to remove. In fact, it seems to me to be a beautiful and intuitive account of general relativity that avoids getting bogged down in the difficult maths. If there are physics experts here present who can confirm my impression, I wonder if it would be useful incorporated into the page on general relativity, as a good introduction for the interested amateur or high school student. Here it is for the record:

Click show to see the removed text
I modified this to be collapsible. —Duae Quartunciae (talk · cont) 07:39, 16 August 2007 (UTC)
In 1915 Einstein solved the puzzle of how the gravitational force was generated. First Einstein realized that if the time runs slower closer to mass M then from the point of view of any observer the speed of light would seem slower in vicinity of that mass M by the same amount as the time slows down. Since according to the famous Einstein's equation the energy of a particle of mass m is $E=mc^{2}$, where c is speed of light, then the energy of any particle has to be smaller in vicinity of mass M. Since any particle is subject to a force towards the state of its lower energy it will move towards mass M, changing potential energy into kinetic energy, keeping its total energy intact. Furtheremore, if the particle is prevented from that movement by any obstacle it will push at the obstacle with inertial force $F=-dE/dx$, where dE is the change of particle's energy along the distance dx oriented towards mass M (as known from elementry mechanics). This way Einstein knew that the effect of the time running slower in vicinity of mass M produces an effect similar to attraction of particles in vicinity of mass M and the equation for this "attractive force" must be $F=-d(mc^{2})/dx=-2mc(dc/dx)$. So it was enough to assume that $dc/dx=-g/(2c)$ to obtain the Newtonian equation for "gravitational force" $(F=mg)$. It would be a different theory than Newton's, with no attractive force, just with the internal energy of the particle $(E=mc^{2})$ but giving the same results as Newton's theory. There was just one obstacle to this theory though: when one assumes on the principle of equivalence that a light ray is deflected in an accelerating rocket by the same amount as in a gravitational field of the same acceleration then the change of the observed speed of light in the rocket comes out as $dc/dx=-g/c$ which is twice more than needed for the results being the same as Newton's. So something didn't fit (most high school students should be able to figure out how much a light ray bends in an accelerating rocket). So for the time being Einstein's theory was good only for predicting bending of the light rays in the vicinity of masses if the time really runs slower in their vicinity but somewhat suspicious as a theory of gravitation since it predicted twice of what "Newtonian gravitational force" was supposed to be in vicinity of the Sun. After Eddington measured the real deflection of light rays in gravitational field of the Sun it turned out that it is twice more than Newtonian prediction, as if the gravitational force acting on light were twice as big as Newtonian prediction. Einstein immediately realized that except the time slowing down around masses there is also more space around them than it would be in Euclidean space and so the missing thing in his model with accelerating rocket was the curvature of space. It apparently added the same deflection to the light ray as the slowing of the time did. So the angle of deflection in his calculations of speed of light should be only half what he assumed from the model of accelerating rocket. Only half of it was due to the change of speed of light (to "gravitation" as Einstein called it then) and the other half was due to the curvature of space. So now everything fit and the curved space became an indispensable part of gravitation. So the full model of gravitation contains now curvatures of the whole spacetime with symmetric metric tensor and therefore with an expanding universe and when tested against movement of Mercury it delivers correct results. After the Mercury and the light deflection tests Einstein's theory of gravitation has been accepted by most physicists. Einstein's theory is accepted by physicists as a best model of gravitation, passing so far every test, which now are many more than just bending of light near the Sun and the precession of Mercury's orbit (the first two tests that showed that Einstein's theory can explain things that Newton's theory can't). For a fully blown explanation of Einstein's theory of gravitation see the theory of general relativity, as it is called now.

-- Duae Quartunciae (t|c) 07:04, 20 July 2007 (UTC)

Duae_Quartunciae - I suspect Jim Jast's explanation to be original research. One issue is that, in any inertial frame, the speed of light in vacuum will be measured to be the same number c. Therefore, in an inertial frame, dc/dx = 0. Of course, you can use other kinds of frames but it is necessary to specify what frame you are using before dc/dx can be meaningful, and this explanation does not do that. I notice that user:Pdn, who seems to have known his general relativity, asked JimJast here, on September 3, 2005, for a reference in which this stuff was published, but I can find no evidence that he received a satisfactory answer. The closest thing to an reply I see is this edit in which JimJast said "I talked to physics professors and astronomy professors and none is seeing a formal problem with this approach. Some believe it is a right approach. I'm looking for critical opinions though.," which seems to imply that the dc/dx = g/(2c) stuff was original research at that time. I also notice that, on September 2, 2005, here, JimJast referred to dc/dx = g/(2c) "my theoretical result", which again suggests that it was original research. Still, it is almost two years later, and perhaps in the meantime JimJast has published this stuff, or found a reference. However, I would be very leery of allowing this to remain in any article without a reference. Cardamon 09:53, 20 July 2007 (UTC)

Oh well. Pity. I confess I was not even looking particularly hard at the maths, but at the notion of an intuitive explanation for how curved space manifests in the same way as an attractive force. I took the formulae as giving an approximate intuition for helping relate apparent forces to curved spacetime. I was aware already that Einstein's initial predictions with respect to Venus were not accurate, due to not quite applying the fledging theory in all its details; but I was not well aware of the specifics. I took this explanation at face value as a way of getting an intuition for what occurred. I withdraw my suggestion.

In the meantime; what do you think of the proposed replacement section? Is that okay? -- Duae Quartunciae (t|c) 10:06, 20 July 2007 (UTC)

I think you should make it quite explicit that general relativity has completely supersceded Newtonian gravity on a theoretical basis, but that Newtonian gravity is still used in applications, being much easier to calculate. I'm afraid to make the edit myself though, having no sources, unfortunately. 151.152.101.44 23:08, 20 July 2007 (UTC)

OK, there has been plenty of time for comment, so I am going ahead with the edit. I have added a final one brief paragraph on continuing use of Newton's theory, which I think is uncontroversial. If my wording can be improved, go right ahead! -- Duae Quartunciae (talk · cont) 03:58, 22 July 2007 (UTC)

Removed incorrect paragraph

I removed the incorrectly phrased paragraph on acceleration addition. Gravitational acceleration is not additive, so the paragraph is incorrect. The only correct thing say is that :

-in the Earth frame, the probe mass is accelerated with a1=GM/r^2

-in the probe mass frame the Earth is accelerated with a2=Gm/r^2

There is no scientific support for adding accelerations calculated in two DIFFERENT frames. WWStone

As the reminder at the top of the edit page states, remember to sign your posts by typing four tildes (~~~~). Please do so in the future.

I believe you are misunderstanding the physics here. First, those accelerations are not calculated in two different frames, nor do they need to be. What makes you believe that?

Because whoever wrote the paragraph inadvertently added them WWStone 15:30, 30 September 2007 (UTC)

Second, you say "in the Earth frame" the probe mass is accelerated with a1=GM/r^2. The article said "the position of the mass from the earth accelerates according to a1+a2".

... which is clearly incorrect. I explained that WWStone 15:30, 30 September 2007 (UTC)

Clearly the second derivative of "the position of the mass from the earth", according to Newton's laws is a1+a2.

Clearly , this is wrong. If you want to make wiki a joke, go ahead. WWStone 15:30, 30 September 2007 (UTC)

Are you denying this? If so please recheck your math, and if you are still getting an error please post your math here.

I already did, I explained why the paragraph is wrong. If you don't understand this, this is too bad. WWStone 15:30, 30 September 2007 (UTC)

I am reverting your deletion as you currently are in the minority. Please do not change it again while we discuss your understanding on this subject (if you wish to continue discussing it). GrapeSmuckers 10:45, 30 September 2007 (UTC)

In minority? There is one of you and one of me. Do you have more people on your side? If you think the paragraph is correct, can you produce a reference? No physics book that respects itself would show such a thing. Where did you get it from? WWStone 15:30, 30 September 2007 (UTC)

Please do not reply by breaking up a comment like that, for if we were to continue replying like that (inserting piece-wise a comment to each statement) the conversation could not be easily followed by anyone. Second, I noticed you reformated part of another user's reply (mine). It is frowned upon to do so. Please do not do so in the future.

In response to my direct request to show your math if you still agree you just state you already explained. But you have shown no math, and only made statements or your beliefs with no cogent explanations of them. Despite your refusal to respond to requests for math or some kind of backup of your statements, I will assume good faith on your part (that you actually believe this and are not trolling). So I will try to explain mathematically the most basic of your errors as a starting point. If you still refuse to respond to simple requests to explain yourself I will no longer assume you are actually trying to communicate here, and end this conversation.

I'll begin by addressing your belief that those accelerations are necessarily calculated in two different frames.

As in the paragraph, let us consider two masses $m_{1},m_{2}$ and their positions. I will denote their positions from an inertial coordinate system origin as ${\vec {r_{1}}}$ and ${\vec {r_{2}}}$ repectively.

As mentioned in the article: ${\vec {F}}_{12}=-G{m_{1}m_{2} \over {\vert {\vec {r}}_{12}\vert }^{2}}\,\mathbf {\hat {r}} _{12}$ Where ${\vec {F}}_{12}$ is the force applied on object 2 due to object 1, and ${\vec {r}}_{12}={\vec {r}}_{2}-{\vec {r}}_{1}$ the vector from object 1 to object 2.

Newton's second law tells us that the acceleration caused by this force is F=ma, and thus the acceleration on object one in this inertial coordinate system will be: ${\vec {a}}_{1}=-G{m_{2} \over {\vert {\vec {r}}_{21}\vert }^{2}}\,\mathbf {\hat {r}} _{21}$

Similarly a_2 can be calculated.

Note that this calculation in no way needed to specify which inertial frame this was calculated in. Newton's laws are invarient to the choice of the inertial frame. Therefore, despite your insistence, these calculations do not require two different inertial frames. All these calculations can be done from one inertial frame, and ANY inertial frame. So once again I request you to please recheck your math for there are errors and if you are still getting the errors, please post your math here.

You ask why you are in the minority here -- Because some other poster made that paragraph, and I second it. Also, because you are arguing against Newtonian mechanics (and refusing to answer questions about your reasoning), which will automatically put you in the minority against the standard teaching of non-relativistic mechanics.

I will be reverting your deletions again. Please stop deleting this. I have tried to clearly lay out and explain some of your beginning confusion regarding Newtonian mechanics. Until you have a better hold on this subject it is not appropriate for you to be editting this article. I will gladly try to help explain this material to you in the interest of building consensus, but your first step will require a willingness to learn... and continually deleting a section of the article is not a good sign of this. GrapeSmuckers 02:42, 1 October 2007 (UTC)

Fine, you made a large paragraph where you are repeating uncontested stuff. I asked you to show what justifies the adding of the two accelerations, you demonstrated nothing and you couldn't produve any reference (for good reason, no physics book worth a penny would show such an idiocy). The paragraph was made by User:Paul_venter on Aug 31 2006. The erroneous has been sitting unnoticed for more than a year. Checking your posts as well as the author's posts, I did not see any posts that have anything to do with physics outside the reversal of the erroneous post in discussion. The accelerations of the two bodies do not add up, there is no justification why they should add up. I notice that you were unable to produce any reference as I asked you, you simply repeated the original post. There is no math for me to check since I have not added any math, I just removed the ridiculous paragraph that you keep putting back in. It is a known fact that the two bodies gravitate about each other and that the accelerations do not add naively, as shown by the paragraph in debate. See : http://farside.ph.utexas.edu/teaching/336k/lectures/node107.html and http://farside.ph.utexas.edu/teaching/336k/lectures/node58.html. If we are intent on treating the two bodies as a single body, then that can be done correctly by employing the notion of "reduced mass" (see http://farside.ph.utexas.edu/teaching/336k/lectures/node57.html#e7.6) but that is a completely different story from the naive and incorrect paragraph currently present in the main article. WWStone 04:27, 1 October 2007 (UTC)

You write: "Fine, you made a large paragraph where you are repeating uncontested stuff." So does this mean you now retract your previous statement complaining that we're dealing with quantitites from two different frames? If so, this is progress. Altough I find it disturbing that you seem to be trying to pretend now that you never made those statements.

So now that we hopefully got that problem fixed (even if just swept under the rug), we can move onto your next problem. The paragraph is discussing the acceleration of "the position of the mass from the earth". It is using the mass as object 1 and the earth as object 2, so "the position of the mass from the earth" is ${\vec {r}}_{12}={\vec {r}}_{2}-{\vec {r}}_{1}$ and the acceleration of this due to gravity is ${\vec {a}}_{12}={\frac {d^{2}}{dt^{2}}}{\vec {r}}_{12}={\frac {d^{2}}{dt^{2}}}{\vec {r}}_{2}-{\frac {d^{2}}{dt^{2}}}{\vec {r}}_{1}=-a_{2}{\hat {r}}_{12}+a_{1}{\hat {r}}_{21}=(a_{1}+a_{2}){\hat {r}}_{21}$. In other words, the magnitudes are such that $a_{12}=a_{1}+a_{2}$ just as the paragraph stated.

What I find very strange is that your sources actually support the very paragraph you are trying to delete. Did you even read those sources? Look at Eq. 376, 377 on [2], it states exactly what you are trying to delete.

This is rediculous. It makes it truly difficult for me to believe you are earnest here and not just trolling for a fight. Please prove me wrong by dropping this line or argument. GrapeSmuckers 09:09, 1 October 2007 (UTC)

As explained, this is the acceleration of the "reduced mass", associated with a fictitious body. The way the paragraph is constructed, it shows a mindless addition of the two accelerations. If you read the reference on "Binary Stars" as you claim, you would have seen that there is no support for adding the scalar accelerations as the wiki page shows. The two stars gravitate about their COM. I should delete the formula a1+a2=GM/r^2 but I will let ir stand, as testimony to the ineptitude of what is put into wiki pages. No, I am not trolling, I just pointed out the correct way of dealing with the problem. It would require rewriting the paragraph the way I did it or deleting it.

BTW: the last sentence in the paragraph is patently wrong (because it is derived from the acceleration of the "reduced mass", which is NOT the acceleration of any of the two bodies). This gives rise to the ridiculous sentence "If m1 is negligible compared to m2, small masses would have approximately the same acceleration. However, for appreciably large m1, the combined acceleration, should be considered." . I propose we delete it since it contradicts the well known fact that all bodies fall with the same acceleration GM/r^2. Or they no longer do so , according to this wonderful wiki page ? :-)

WWStone 14:52, 1 October 2007 (UTC)

You seem unable to admit your mistakes like a normal human being, and until you do this discussion is apparrently futile. So I will try to wrap up this discussion by summarizing the argument so far:

• We are not discussing quantities from two different frames. (Unlike you claimed in your original remark. I explained this to you in detail including some math.)
• The "position of the mass from the earth" is ${\vec {r}}_{12}={\vec {r}}_{2}-{\vec {r}}_{1}$. Thus the acceleration of the "position of the mass from the earth" is indeed just the simple linear combination of the acceleration of object 1 and object 2. (I also showed this explicitly to you with math.)
• There was no need to bring reduced mass into the discussion at this point, as my math above shows, although that method does give the same result (which you argued against) as it should.
• The last sentence in that paragraph followed directly and simply from the equation before it. When m1 << m2, m2 will determine the acceleration with neglible impact from m1. Again instead of showing math (despite requests to do so), you just restate you incorrect position with no explanation and delete it yet again.
• And to wrap it all up, even though I've shown you mathematically that the acceleration equation in question is correct, and furthermore that even your own sources agree with it, you still refuse to accept you are wrong stating "I should delete the formula a1+a2=GM/r^2 but I will let ir stand, as testimony to the ineptitude of what is put into wiki pages."

You clearly are not willing to learn as it has been explained to you several times now in much detail. And still you refuse to admit your mistakes. I will be reverting your changes, and I expect you to stop trolling by continuing to delete that section of the article. Any further attempts on your part to push your incorrect view on this article will be summarily undone. GrapeSmuckers 22:54, 1 October 2007 (UTC)

You are simply unable to understand that taking the second order derivative of r12 is a meaningless exercise. The references I broght in talk about two bodies as they gravitate about each other, they do NOT show anything remotely close to the braindead acceleration addition. At this point, I am going to ask you one more time to produce a reference to ANY textbook that shows the derivation a1+a2=GM/R^2. Falure to do so will result into a request for arbitration. Wiki is not the place for crackpots to do their "original research". Produce a reference, please. WWStone 23:06, 1 October 2007 (UTC)

Wow, what a surprise -- you removed the source link for that equation. Now that you see your source (from a university physics professor's lecture notes) disagrees with you, it is no longer an acceptable source? You clearly are trolling for arguments.

You want a text book source? Because the whole point here is that in general both masses need to be considered, because there is a force on both accelerating them together. So the "reference" point itself is accelerating. I will refer you to the chapter on non-inertial reference frames in Classical Dynamics of Particles and Systems, 4th ed., Marion and Thorton. pg 386

$m{\vec {a}}_{r}={\vec {F}}-m{\ddot {\vec {R}}}$

Where ${\vec {a}}_{r}$ is the relative acceleration of a mass m from a position R (let this be earth if you wish), ${\vec {F}}$ is the force on mass m according to any inertial frame, and ${\ddot {\vec {R}}}$ is the acceleration of R according to any inertial frame. Changing the notation to match what we've been using here:

$m{\vec {a}}_{21}=m{\vec {a}}_{1}-m{\vec {a}}_{2}$

What do you know, it agrees with your source, and the common sense math explanation above. ${\vec {a}}_{21}={\vec {a}}_{1}-{\vec {a}}_{2}$ and if just looking at the magnitudes $a_{21}=a_{1}+a_{2}$. The fact that you even call this original research, when I wasn't the one that originally added it, and you even found a source yourself backing this up ... why it just boggles the mind.

I will ask you only once more, show your math or admit your error and agree with the math shown above ... I have shown math, given explantions, given sources, and gone to great lengths to explain this to you. If you reply arguing against this again without showing math proving the simple reasoning above wrong, I will be reporting you as a troll on the administrators notice page. GrapeSmuckers 06:17, 2 October 2007 (UTC)

You are a lot more stupid than expected (more stubborn as well). It is very simple really: in the two body problem the two bodies GRAVITATE about each other. You can see the equations of motion and the trajectory obtained by integrating the equation of motion in the reference I gave you. They GRAVITATE, capisci? They do not follow the idiotic sentence: "Therefore the position of one mass from the second mass gravitationally accelerates according to[1] with the acceleration a1+a2". This is the idiocy in the wiki paragraph that you keep supporting, I tried to expunge it but you keep bringing it back. Contrary to the new stupidity that you have just added, the two masses DON'T "fall faster into each other", they GRAVITATE about each other. The correct and complete equations are given a little below (379,380) on the same page I referenced earlier. You only had to read. The whole point is that you either give the whole story (that would require showing the gravitating equations of the two bodies)or nothing. You cannot put down selective equations in a brain-dead paragraph.As to your threats, I looked at your posts, you are mainly posting in userTalk pages,pretending to be some sort of authority, nothing of substance. Are you some sort of policeman? Don't tell me that you have a BS in physics :-) WWStone 13:50, 2 October 2007 (UTC)

GrapeSmuckers, I suggest you stop feeding the troll. WWStone has changed his stance greatly since you started, and that is as close as you're going to get to an admission of fault or any consensus. Search Adrian Sfarti on google to see who you are dealing with. Good luck. FriendlyWarning 16:07, 2 October 2007 (UTC)

FriendlyWarning, thank you for your warning, but I'm not 100% convinced they are the same people. Skimming through the Adrian Sfarti / Ati3414 / Moroder links starting from google (and it is quite fascinating), Adrian Sfarti is clearly combative, but comes across as a crackpot (believes in time dilation but not length contraction, believes in E=mc^2 but claims adding or subtracting energy from a box in its rest frame won't change the mass, and then that whole thing with insisting all uranium isotopes had the same abundance during the formation of the earth in order to push his "research" on the age of the earth on wikipedia, etc.) whereas our friend WWStone here will not answer direct questions or admit fault, and has some physics misunderstanding, he doesn't come across as a crackpot to me. I haven't looked at WWStone's other contributions yet though. I'll look into it later and file a sockpuppet report if necessary. If you can provide more proof, that would be helpful.

WWStone, releasing a mass from a fixed point above another mass ... yes indeed they will accelerate towards each other on a collision course. And this acceleration of the position of one mass from the other depends on BOTH masses in general, as has been showed to you time and time again (you have yet to show your math proving otherwise). So yes, a much heavier object released in this fashion WILL fall into the asteroid faster. If you disagree with this, again please show your math as several sources and the simple math above show you are clearly wrong.

And as for your retort "They GRAVITATE", gravitate just means to attract each other, or move due to the forces of gravity (I've also seen it used to mean emit gravitational waves, but that doesn't fit here at all). So all you're saying is, 'NO, they move according to the force of gravity'. How you can possibly believe this is a rational retort to describing the motion of masses according to the force of gravity is mind boggling.

So, to put this on a more productive note, do you agree that the math and statements I am making are technically correct? And you are just complaining that they are not described well in the article? If so, this is a much more productive comment. If not, again, please show your math proving wrong the sources and my simple math explanations above. GrapeSmuckers 03:10, 3 October 2007 (UTC)

Lead sentence

Is this a law or a theory? --— Gadget850 (Ed) ^talk - 04:00, 30 December 2007 (UTC)

Quick answer: both. Quick, but not that quick answer: everything in science is theory - even everything which is (today) considered to be "correct". Why? Since falsifiability is a crucial element in science and the philosophy of science. And law, well, that's just a short name for a mathematical assumption which can predict (more or less accurately) physical observable results of the underlying theory. Make sense? Newton's law, historically named so, is still in use, even if Einstein's theory of general relativity is considered to be "the one". Today, that is, I must add... --Dna-Dennis (talk) 05:05, 27 January 2008 (UTC)

Can you bend Gravity?

I am writing my thesis and was wondering if gravity could bend. There was no mention of it in 'Newton's Laws of Gravitation.' —Preceding unsigned comment added by 75.144.248.221 (talk) 20:18, 19 September 2008 (UTC)

I'm not 100% certain I understand what you mean, but you might mean can gravity bend space? According to Einstein's theory of general relativity the answer is a definite "yes"; Einstein's theory (contrary to Newton's) explains the movement of for instance a small body A through a gravitational field caused by a massive body B as a simple consequence of the fact that mass B bends the spacetime continuum, with no forces involved. In this theory, even light (electromagnetic radiation) is bent. Newton, on the other hand, explains gravity as a consequence of forces between different masses, which cause them to independently attract eachother. These two theories predict the reality almost equally correct on an astronomical scale, but the underlying theories are more or less contradictory to each other (they are also centuries apart in history:) ). Regards, --Dna-Dennis (talk) 23:52, 5 December 2008 (UTC)

Eh, I probably should point out that Einstein's theory has proven to be more accurate than Newton's, when scientists have used more modern means of measurements. So, Einstein's theory is today widely considered to be the superior theory of gravity. For more info, see Tests of general relativity. --Dna-Dennis (talk) 00:02, 6 December 2008 (UTC)

Is there any who can answer to the following.

Critique 1

Suppose two masses m₁ = m₂ = 1 kg, diameter of m₁ = m₂ = 0.5 m Both masses are in space and the centre to center distance between them is 1 metre = r , then

F= Gm₁m₂/r² = G = 6.67*10^-11 Newton.

But Gravitational accelerations or forces of attractions of both masses balance each other and therefore value of gravitational force F should be zero between those two masses, if not, then Which one is gravitating/ or falling mass. And how aforementioned masses attract each other by F= Gm₁m₂/r² = G = 6.67*10^-11 Newton, when there is no gravitating or falling mass.

Make it more simple

let say earth is an sphere. Now if we put imaginary earth on earth or sun on sun and now apply F= GMm/R²to the masses, where g = GM/d². M= mass of earth, m= mass of imaginary earth, centre to centre distance between two masses = diameter of earth. Now again which one is falling mass or gravitating mass where gravitational acceleration of each mass cancel each other.

Critique 2

Sorry, I messed up with calculation in my previous critique 2 therefore I want to represent it again in different way. Consider an apple (or sphere) is at any height “h” and starts falling with g1 towards the surface of the earth. Both apple and earth attract each other with a force equal in magnitude (F1=F2) but opposite in direction. The g1 of apple (or sphere) is 9.8 m/sec/sec towards earth while The g2 of earth is 1.63 x 10^-24 m/sec/sec towards apple (or sphere) Since g1>g2 therefore apple (or sphere) move (fall) faster towards earth and both bodies comes to rest when they touch each other. At rest neither apple (or sphere) is accelerating (at any rate m/sec/sec) further towards earth nor earth towards apple (or sphere). The velocity of any object is zero at its rest position and since acceleration is the rate of change of velocity therefore, technically/ theoretically at rest, acceleration of any object is also equal to zero. Therefore at rest it is wrong to say that the value of g is always 9.8 m/sec/sec for w = mg. Therefore, when both apple (or sphere) and earth rest on each other the value both g1 and g2 should also equal to zero because of no further rate of change of velocity towards each other. Therefore F1=g1m=g2m=0=F2; Just like we don’t feel the weight of earth nor earth feel ours. Further, If two objects are under influence of gravitation either at rest or uniform motion then there is contradiction between First law of motion and Universal law of gravitation.

Critique 3

Equation was derived on the assumption that earth is a homogeneous sphere while in reality it is composed of many different materials with variant densities. So the actual center of gravity of earth is somewhere else but not at its center. So this means thing should fall on the ground at an angle to the normal except at two location where the real center of gravity of the earth is closest and farthest to its surface. So why an object is falling straight on the ground surface not an angle???. If answer is micro gravity then do we have to change the value of R in equation g = GM/R². If we change R then value of g will be diffrent than 9.8 m/s/s.

Critique 4

Here is the summary of attraction Forces of Sun - Moon, Earth - Moon and Sun - Earth during Total Eclipse when the Moon is between Sun and Earth.

F = GMm/r²

Sun - Moon - Earth S-M = 4.1984 x 10²⁰ M - E = 2.2 x 10^20 Net force on the Moon = 4.1984 x 10²⁰ Minus 2.2 x 10²⁰ = 1.998 x 10²⁰ towards Sun At this point why Earth force the moon to revolve around its centre when the net force on the moon is much much greater towards the sun ? Explain please. Check the calculation pls. If we consider the Sun - Earth Force S - E= 3.67 x 10²² , then Net Force on the Moon = 1.998 x 10²⁰ Plus 3.67 x 10²² = 3.68 x 10²² towards Sun.

Please also note that force of attraction F = GMm/r² between sun and moon in any case (perigee, apogee, average) is much greater than between moon and earth F = GMm/r² (perigee, apogee, average). So technically it should revolve around the sun in separate orbit and not around earth. So why moon revolves around earth?

Here is also the detail of escape velocity of moon V_em wrt to both sun and earth when it is in between earth and sun in total eclipse.

Sun------------(V _{em, wrt s}=2GM _s/R_sm)^1/2 -----------Moon-------(V _{em, wrt e}=2GM _e/R_em)^1/2--------Earth

V _{em, wrt s}=(2GM _s/R_sm)^1/2 >>> Vem, wrt e=(2GM _e/R_em)^1/2

V _{em, wrt s}>>>> V _{em, wrt e}

Please check all calculation.

Critique 5

Let “P” is a point or an origin of two circles of radius r₁=1 meter and r₂= 2 meter. Consider these two circle as spheres (empty from inside) or consider these circles as two bangles in space. Now apply Newton’s law of gravitation i.e. F=GMm/R² to those two masses and neglect all other local attractions.

As gravitational force of attraction between these two bangles is infinity (center to center distance b/t masses is zero)

Now how much force is required to separate aforementioned masses, infinity or less? If less then what about the Newton’s law? Myktk (talk) 04:38, 7 August 2008 (UTC)Zarmewa

Critique 6:

Galileo had concluded hundreds of years before - All objects released together fall at the same rate regardless of mass. i.e. g = GM/R² and had proved on the lunar surface by falling feather and hammer at the same time in the absence of air.

As g= GM/R² and does not depend upon the falling mass therefore why the masses of atoms / molecules (mass of any kind) of air/ gases do not fall at the same rate on the surface of earth along with other masses or obey the Newton law of gravitation.e.g a blimp or ballon filled with helium gas. Further movements of air molecule in atomosphere depend upon temperature, pressure and Newton did not mention any temperature or pressure in his law of gravitation. 96.52.178.55 (talk) 08:09, 20 February 2009 (UTC) zarmewa khattak

Hi! I'm not sure I follow your reasoning completely, but I will try to respond anyway.

You said : (Critique 1:) Now again which one is falling mass or gravitating mass where gravitational acceleration of each mass cancel each other.

Which one is falling/gravitating? Well, both, of course.

No, the forces do not cancel each other out - this is only if you look at the gravitational system as a whole. From the perspective of each mass, each mass experiences a force F, and will therefore undergo an acceleration a=F/m (according to Newton's second law F= m x a, see Newton's laws of motion).

Critique 2 and 3 I don't quite understand, but they might have to do with what I said above.

You said: (Critique 4:) So technically (the Moon) it should revolve around the sun in separate orbit and not around earth. So why moon revolves around earth?

But you are doing a significant mistake here; you assume that the masses have no initial velocity, and thus no Momentum (p=m x v). Earth, the Moon (and the Sun) all have velocities, and the Moon has been in a stabile orbit around the Earth for a very, very long time. All these objects have thus a significant Momentum, and it is thus quite hard to force them off their course in space. The Moon is simply too close to the Earth and too far from the Sun to be pulled "out of orbit".

Critique 5 I'm not 100% sure I get your point, but strictly mathematically, yes, the force would be infinitely large. But! The distance between two different objects can in reality never be zero; they then become per definition the same object, or rather a new object (compare with Nuclear fusion). But I must also mention that your reasoning comes close to the rules for the creation of black holes, see Schwarzschild radius.

Critique 6 seems to question Newton's law by applying the concept of temperature. Why this? The one has nothing to do with the other. Temperature in physics is a collective concept which relates to the average of the velocities of all particles, and therefore temperature is in this case more closely related to Momentum (p=m x v).

All in all, Newton's Law of Universal Gravitation describes the influence of gravitation, i.e. mass attracts mass, nothing more, nothing less. It does not describe Momentum, temperature and everything else, for this you need to consider other physics, including Newton's laws of motion. Regards, --Dna-Dennis (talk) 00:28, 18 March 2009 (UTC)

A good try but not satisfactory

1- Imaigine a rope pulled by two persons with equal forces but opposite in direction. Thus F1=Gm₁m₂/D²= Gm₁m₂/D²=F2. Also m₁ try to accelarate m₂ towards m₁ but at the same time m₂ try to accelarate m₁ towards m₂ with same amount but opposite in direction. I forgot to write the diameters of those two masses viz 0.5 m.

2- Centripital and centrifugal (gravitational) forces are the reseans for orbital velocities of moon and earth around sun therefore moon could have got different velocity around sun upon its creation.

3- There is infinity force b/t aforementioned masses as per law of gravitation. Another simple example is sperical ballon or plastic swimming pool ball filled with water or air. There are two masses, one is water or air and the other is ballon. Center of both masses are coinicides.

4- A ballon filled with water and partially (small amount) with air will come up to the surface if it is put at the bottom of swimming pool and so temperatute is not the right justfication. Adio 96.52.178.55 (talk) 06:24, 20 March 2009 (UTC) zarmewa

I have to say the same; A good try but not satisfactory... far from.

1. a rope pulled by two persons with equal forces - this is a very bad example; with a rope, the two persons (and forces) are physically connected, as the rope is made up of atoms. Gravitational force does not need any physical connection; it works perfectly well in vacuum. Thus, as I said, the forces do not cancel each other out. Basic physics.

2 So why moon revolves around earth? was the original question. Last time I looked, yes, the Moon was still orbiting the Earth. And last time I calculated (in school a very long time ago), the reality was consistent with Newton's laws of motion and other laws of physics. And why focus on The Moon, there are many moons in the Solar System, orbiting other planets (e.g. Triton, Titan, Titania and a sh*tload of others). I bet there are even moons around moons. They also follow the laws of physics. Clarification: The Moon does orbit the Sun, it just orbits the Earth as well. If you wonder why the moon orbits the Earth, you probably should study Astrophysics and Astronomy. Newton's theory does not, and can not explain why the Moon was caught by Earth in the first place, this is the role of Astronomy. AFAIK - and I'm far from sure - the Moon is believed to be a byproduct of some planetary collision a very long time ago. Check the article The Moon - the info is probably there.

3 There is infinity force b/t aforementioned masses as per law of gravitation. Again, only mathematical, theoretical infinity. And (simple) mathematics almost never perfectly describes reality, compare with Chaos Theory. In reality, we don't know yet, as very small distances can not be perfectly examined due to quantum effects, see Uncertainty principle. There may very well be (and probably are) other effects when we deal with very small distances. And the effect of gravitation will in either case be very hard to distinguish, as it is very weak compared to other forces, e.g. the Electromagnetic force.

4. A ballon filled with water and partially (small amount) with air will come up to the surface if it is put at the bottom of swimming pool... So what? This has nothing whatsoever to do with gravitation. It's about pressure, and is covered by completely different physics. Why apply gravitation to it in the first place? It's just confusing to mix things together this way.

As I said, Newton's Law of Universal Gravitation describes the influence of gravitation, i.e. mass attracts mass, nothing more, nothing less. It never was, and never will be a Grand unification theory. Anyway, if you are dying to criticise gravitational theories, why don't you start with Einstein's General Relativity? After all, this is the currently superior theory. Good luck, btw. And, if you're dying to discuss physics and try to attack a gravitational theory which has been used successfully for hundreds of years, and still is, I recommend doing it at http://www.physicsforums.com/index.php. Good luck, again. You will have a lot of explaining to do... --Dna-Dennis (talk) 07:51, 20 March 2009 (UTC)