Inverse square law not accurate for non-point masses?
$begingroup$
It would seem that the inverse square law, popularized by Isaac Newton, is only true for masses which have no dimensionality and are a single point. In other words, it is only an ideal, not a practical reality. For example, imagine we have a small satellite, X, orbiting a much larger body:
X will be attracted to all of the parts of the planet which it orbits. If we imagine the mass of element A to be equal to that of C and A+C=B, then we can see that by the inverse square law, the force between X and A will be proportional to 1/9, the force between X and B will be proportional to 2/16=1/8, and the force between X and C will be proportional to 1/25. Thus, the total force which A and C exert on X is proportional to 34/225 which is more than 1/8.
We can extrapolate this same basic argument to every pair of symmetric elements of mass which comprise the planet. Therefore, the force between X and the planet is different than if all the mass of the planet had been located at B.
From this example, it would appear that the inverse square law of gravitation is not accurate in real scenarios, and that the closer a satellite is to its host, the less accurate the law is.
Is there some more accurate equation that more accurately represents this state of affairs, or is brute calculation by integration of finite elements necessary?
forces newtonian-gravity
asked Apr 21 at 5:39
Ambrose SwaseyAmbrose Swasey
1,59921645
$endgroup$
add a comment |
$begingroup$
It would seem that the inverse square law, popularized by Isaac Newton, is only true for masses which have no dimensionality and are a single point. In other words, it is only an ideal, not a practical reality. For example, imagine we have a small satellite, X, orbiting a much larger body:
X will be attracted to all of the parts of the planet which it orbits. If we imagine the mass of element A to be equal to that of C and A+C=B, then we can see that by the inverse square law, the force between X and A will be proportional to 1/9, the force between X and B will be proportional to 2/16=1/8, and the force between X and C will be proportional to 1/25. Thus, the total force which A and C exert on X is proportional to 34/225 which is more than 1/8.
We can extrapolate this same basic argument to every pair of symmetric elements of mass which comprise the planet. Therefore, the force between X and the planet is different than if all the mass of the planet had been located at B.
From this example, it would appear that the inverse square law of gravitation is not accurate in real scenarios, and that the closer a satellite is to its host, the less accurate the law is.
Is there some more accurate equation that more accurately represents this state of affairs, or is brute calculation by integration of finite elements necessary?
forces newtonian-gravity
asked Apr 21 at 5:39
Ambrose SwaseyAmbrose Swasey
1,59921645
$endgroup$
$begingroup$
The Earth is a point mass in comparison to the mass of the Sun, so to an extremely high degree of accuracy you can model the planets as being point masses, which is what Newton did in the Principia. Also, Newton did not merely 'popularise' the law: he demonstrated its mathematical coherence and derived it from the law of universal gravitation. Hooke did nothing to demonstrate it mathematically or universality and only speculated on its existence in the Solar System, don't be fooled by comments by Stephen Hawking as he did not always know what he was talking about.
$endgroup$
– Tom
Apr 21 at 15:36
$begingroup$
Since non-point masses can be considered as collections of a bunch of tiny point masses, the inverse-square law is still applied when you add up the inverse-square contributions of all of the tiny point masses. This is how the gravitational field from extended objects is calculated (though usually we work in the limit where the non-point mass is made up of infinitely many point masses of infinitesimal mass; dealing with adding them all up is precisely what calculus was invented to deal with).
$endgroup$
– probably_someone
Apr 21 at 17:06
$begingroup$
There's more mass in your sphere located near d=5 than near d=3.
$endgroup$
– Dmitry Grigoryev
Apr 21 at 18:48
add a comment |
$begingroup$
It would seem that the inverse square law, popularized by Isaac Newton, is only true for masses which have no dimensionality and are a single point. In other words, it is only an ideal, not a practical reality. For example, imagine we have a small satellite, X, orbiting a much larger body:
X will be attracted to all of the parts of the planet which it orbits. If we imagine the mass of element A to be equal to that of C and A+C=B, then we can see that by the inverse square law, the force between X and A will be proportional to 1/9, the force between X and B will be proportional to 2/16=1/8, and the force between X and C will be proportional to 1/25. Thus, the total force which A and C exert on X is proportional to 34/225 which is more than 1/8.
We can extrapolate this same basic argument to every pair of symmetric elements of mass which comprise the planet. Therefore, the force between X and the planet is different than if all the mass of the planet had been located at B.
From this example, it would appear that the inverse square law of gravitation is not accurate in real scenarios, and that the closer a satellite is to its host, the less accurate the law is.
Is there some more accurate equation that more accurately represents this state of affairs, or is brute calculation by integration of finite elements necessary?
forces newtonian-gravity
asked Apr 21 at 5:39
Ambrose SwaseyAmbrose Swasey
1,59921645
$endgroup$
It would seem that the inverse square law, popularized by Isaac Newton, is only true for masses which have no dimensionality and are a single point. In other words, it is only an ideal, not a practical reality. For example, imagine we have a small satellite, X, orbiting a much larger body:
X will be attracted to all of the parts of the planet which it orbits. If we imagine the mass of element A to be equal to that of C and A+C=B, then we can see that by the inverse square law, the force between X and A will be proportional to 1/9, the force between X and B will be proportional to 2/16=1/8, and the force between X and C will be proportional to 1/25. Thus, the total force which A and C exert on X is proportional to 34/225 which is more than 1/8.
We can extrapolate this same basic argument to every pair of symmetric elements of mass which comprise the planet. Therefore, the force between X and the planet is different than if all the mass of the planet had been located at B.
From this example, it would appear that the inverse square law of gravitation is not accurate in real scenarios, and that the closer a satellite is to its host, the less accurate the law is.
Is there some more accurate equation that more accurately represents this state of affairs, or is brute calculation by integration of finite elements necessary?
forces newtonian-gravity
forces newtonian-gravity
asked Apr 21 at 5:39
Ambrose SwaseyAmbrose Swasey
1,59921645
asked Apr 21 at 5:39
Ambrose SwaseyAmbrose Swasey
1,59921645
edited Apr 21 at 15:48
Ambrose Swasey
asked Apr 21 at 5:39
Ambrose SwaseyAmbrose Swasey
1,59921645
asked Apr 21 at 5:39
Ambrose SwaseyAmbrose Swasey
1,59921645
asked Apr 21 at 5:39
Ambrose SwaseyAmbrose Swasey
1,59921645
1,59921645
$begingroup$
The Earth is a point mass in comparison to the mass of the Sun, so to an extremely high degree of accuracy you can model the planets as being point masses, which is what Newton did in the Principia. Also, Newton did not merely 'popularise' the law: he demonstrated its mathematical coherence and derived it from the law of universal gravitation. Hooke did nothing to demonstrate it mathematically or universality and only speculated on its existence in the Solar System, don't be fooled by comments by Stephen Hawking as he did not always know what he was talking about.
$endgroup$
– Tom
Apr 21 at 15:36
$begingroup$
Since non-point masses can be considered as collections of a bunch of tiny point masses, the inverse-square law is still applied when you add up the inverse-square contributions of all of the tiny point masses. This is how the gravitational field from extended objects is calculated (though usually we work in the limit where the non-point mass is made up of infinitely many point masses of infinitesimal mass; dealing with adding them all up is precisely what calculus was invented to deal with).
$endgroup$
– probably_someone
Apr 21 at 17:06
$begingroup$
There's more mass in your sphere located near d=5 than near d=3.
$endgroup$
– Dmitry Grigoryev
Apr 21 at 18:48
add a comment |
$begingroup$
The Earth is a point mass in comparison to the mass of the Sun, so to an extremely high degree of accuracy you can model the planets as being point masses, which is what Newton did in the Principia. Also, Newton did not merely 'popularise' the law: he demonstrated its mathematical coherence and derived it from the law of universal gravitation. Hooke did nothing to demonstrate it mathematically or universality and only speculated on its existence in the Solar System, don't be fooled by comments by Stephen Hawking as he did not always know what he was talking about.
$endgroup$
– Tom
Apr 21 at 15:36
$begingroup$
Since non-point masses can be considered as collections of a bunch of tiny point masses, the inverse-square law is still applied when you add up the inverse-square contributions of all of the tiny point masses. This is how the gravitational field from extended objects is calculated (though usually we work in the limit where the non-point mass is made up of infinitely many point masses of infinitesimal mass; dealing with adding them all up is precisely what calculus was invented to deal with).
$endgroup$
– probably_someone
Apr 21 at 17:06
$begingroup$
There's more mass in your sphere located near d=5 than near d=3.
$endgroup$
– Dmitry Grigoryev
Apr 21 at 18:48
$begingroup$
The Earth is a point mass in comparison to the mass of the Sun, so to an extremely high degree of accuracy you can model the planets as being point masses, which is what Newton did in the Principia. Also, Newton did not merely 'popularise' the law: he demonstrated its mathematical coherence and derived it from the law of universal gravitation. Hooke did nothing to demonstrate it mathematically or universality and only speculated on its existence in the Solar System, don't be fooled by comments by Stephen Hawking as he did not always know what he was talking about.
$endgroup$
– Tom
Apr 21 at 15:36
$begingroup$
The Earth is a point mass in comparison to the mass of the Sun, so to an extremely high degree of accuracy you can model the planets as being point masses, which is what Newton did in the Principia. Also, Newton did not merely 'popularise' the law: he demonstrated its mathematical coherence and derived it from the law of universal gravitation. Hooke did nothing to demonstrate it mathematically or universality and only speculated on its existence in the Solar System, don't be fooled by comments by Stephen Hawking as he did not always know what he was talking about.
$endgroup$
– Tom
Apr 21 at 15:36
$begingroup$
Since non-point masses can be considered as collections of a bunch of tiny point masses, the inverse-square law is still applied when you add up the inverse-square contributions of all of the tiny point masses. This is how the gravitational field from extended objects is calculated (though usually we work in the limit where the non-point mass is made up of infinitely many point masses of infinitesimal mass; dealing with adding them all up is precisely what calculus was invented to deal with).
$endgroup$
– probably_someone
Apr 21 at 17:06
$begingroup$
Since non-point masses can be considered as collections of a bunch of tiny point masses, the inverse-square law is still applied when you add up the inverse-square contributions of all of the tiny point masses. This is how the gravitational field from extended objects is calculated (though usually we work in the limit where the non-point mass is made up of infinitely many point masses of infinitesimal mass; dealing with adding them all up is precisely what calculus was invented to deal with).
$endgroup$
– probably_someone
Apr 21 at 17:06
$begingroup$
There's more mass in your sphere located near d=5 than near d=3.
$endgroup$
– Dmitry Grigoryev
Apr 21 at 18:48
$begingroup$
There's more mass in your sphere located near d=5 than near d=3.
$endgroup$
– Dmitry Grigoryev
Apr 21 at 18:48
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
It is not quite true to say that the inverse-square law is "not accurate for non-point masses". The more correct answer is to say that it is not accurate for non-spherically symmetric masses.
The inverse-square law depends, essentially, on the fact that the source "looks" the same to the object being attracted thereby, no matter where it is at in relation. If it can "see" a bump nmming it from one direction than from another, it will feel more force from that bump than otherwise, even while being at the same distance from the center of mass, hence violating the inverse-square law. E.g. a bilobate asteroidal/cometary object like the recently-discovered "Ultima Thule" with one lobe nning at you, versus being at the same distance but "seeing" it sideways.
For a spherically-symmetric mass, you can actually show that Newton's law holds in its usual form. This can be done through a gravitational analogue of Gauss's law for electric fields:
$$ointint_S mathbf{g} cdot dmathbf{l} = -4pi GM_mathrm{enc}$$
where $S$ is a closed surface enclosing the mass (can't write that funny circle thing around the integral here properly, sorry) and $M_mathrm{enc}$ the enclosed mass. Alternatively, you can just integrate Newton's law, but that's not as nice.
It's also important to point out that the spherical symmetry must not just be in the gross shape of the mass, but also in its density distribution. In fact, the gravitational field of the Earth is non-spherically symmetric and so non-inverse square for near-Earth objects chiefly because of such inhomogeneities within it than anything else - a fact that shows up if you look at a "geoid", effectively a map of the shape of the Earth's gravity field, used in cartography to set a precise meaning to the term "sea level". This, despite that Earth's gross shape is as good if not a better sphere than many "spherical" objects you can hold in your hands.
(Note: another, but so far purely hypothetical or fantastic, possibility, but arguably more dramatic, way to violate it would be if there existed negative gravitational masses [likely violating Einstein's theory]: these would permit the creation of 'gravitational dipoles' [i.e. a non-removable gravitational dipole moment] that would be analogous to electric dipoles and so would have an inverse-cube law.)
ADD (see comments): As a note, this result is also a case of the famous shell theorem and was proven by the founder of classical mechanics, Sir Isaac Newton, himself. The above is a more modern approach to deriving the result.
ADD 2: I notice that an explanation has been sought, not only for the correct mode of proof of the inverse-square behavior for an extended spherically symmetric mass, but also for why the OP's attempted proof thereof is erroneous. The basic reason is that the essential characteristic - the inverse-square law - of the spherically-symmetric behavior cannot be analyzed in terms of the behaviors of its parts individually: it only appears when you add them all together, holistically, at least up to the point of attaining full spherical symmetry. The two mass elements selected from the sphere being symmetrical about its center is not enough. Because while they, together, have some symmetry (namely $mathrm{O}(2) times mathbb{Z}_2$), they do not have spherical symmetry. Thus, alone, they will never be entirely replaceable by a point concentration as you try to do and you can only be misled as to the whole behavior thinking about them alone, individually. This logical error actually has a name: it's called the fallacy of the composite - assuming that a composite system necessarily has, or "inherits", the same properties as its parts. It doesn't, or, if it does, that has to be proved, and not simply deduced from the fact of its composition thereby.
(Well actually, you can break it up, but you have to "peel it like an onion" into spherical shells. I don't include this because it is, in effect, "trivial": the 'good part' - i.e. the fundamental difference between spherical symmetry and non-spherical - is already there once you do that, as it has to be, and thus has no real explaining or proving power, I'd say. At best, it could be the final step in a proof, but not the most meaty step.)
answered Apr 21 at 6:03
The_SympathizerThe_Sympathizer
5,7431230
$endgroup$
$begingroup$
FWIW, the lack of the "funny circle thing" is a known issue in MathJax.
$endgroup$
– Ilmari Karonen
Apr 21 at 9:42
1
$begingroup$
Ok, what is the error in my analysis then?
$endgroup$
– Ambrose Swasey
Apr 21 at 12:44
4
$begingroup$
@AmbroseSwasey You've only properly considered the points A, B, and C. If you sum over all the points carefully, it averages out correctly. For example, if you consider the points D and E, so that DE is perpendicular to AB, then both D and E contribute less gravitational field than B does, partially balancing out the fact that (as you pointed out) A and C on average contribute more than B does.
$endgroup$
– knzhou
Apr 21 at 14:52
$begingroup$
@AmbroseSwasey Of course, doing this whole argument explicitly would be extremely difficult. Calculus was invented to make this kind of computation relatively easy and straightforward.
$endgroup$
– knzhou
Apr 21 at 14:55
7
$begingroup$
Are "nmming" and "nning" typos?
$endgroup$
– npostavs
Apr 21 at 15:44
|
show 9 more comments
$begingroup$
The inverse square law is exact as long as general relativistic corrections can be neglected. For any mass distribution the force can be found from summing up the inverse square law contribution from its parts, that is by convolution of the density distribution with the inverse square law.
answered Apr 21 at 10:26
my2ctsmy2cts
6,6342722
$endgroup$
add a comment |
$begingroup$
No, the inverse square law is not accurate for non-point masses. It isn't even accurate for spherically symmetric non-point masses. Take a look at the Wikipedia article on gravitational potential. The plot on the right shows the force of gravity for the Earth. It isn't an inverse-square rule plot. The force of gravity reduces as you climb up and away from the surface of the Earth, and it reduces as you tunnel down towards the centre. If you were in some gedanken chamber at the centre of the Earth, you'd float around weightless.
CCASA image by AllenMcC, see Wikipedia Commons
Generally speaking, I'd say the inverse square rule is good enough most of the time when we're talking about celestial mechanics. It probably isn't good enough for when we're talking about co-orbiting neutron stars. Maybe some astronomer can tell you something about that.
Edit: note the comments below.
answered Apr 21 at 9:06
John DuffieldJohn Duffield
7,43152547
$endgroup$
3
$begingroup$
In Newtonian gravity, the inverse-square law is accurate for spherically symmetric objects. It's just that, as The_Symphathizer points out in their answer, the Earth is not actually spherically symmetric. (In the case of "tunneling toward the center", you count the mass within the sphere corresponding to your current depth.)
$endgroup$
– Peter Erwin
Apr 21 at 9:16
12
$begingroup$
More to the point, the inverse square law is accurate for spherically symmetric masses outside the mass itself. Inside the massive body, only the part of the mass closer to the center than the observer counts (which is quite a remarkable result in itself).
$endgroup$
– Ilmari Karonen
Apr 21 at 9:47
$begingroup$
@Peter Erwin : noted. Yes, it's accurate when you allow for the distribution of matter. I was thinking of just the force of gravity for the Earth. Ilmari Karonen: also noted.
$endgroup$
– John Duffield
Apr 21 at 10:02
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "151"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f474092%2finverse-square-law-not-accurate-for-non-point-masses%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is not quite true to say that the inverse-square law is "not accurate for non-point masses". The more correct answer is to say that it is not accurate for non-spherically symmetric masses.
The inverse-square law depends, essentially, on the fact that the source "looks" the same to the object being attracted thereby, no matter where it is at in relation. If it can "see" a bump nmming it from one direction than from another, it will feel more force from that bump than otherwise, even while being at the same distance from the center of mass, hence violating the inverse-square law. E.g. a bilobate asteroidal/cometary object like the recently-discovered "Ultima Thule" with one lobe nning at you, versus being at the same distance but "seeing" it sideways.
For a spherically-symmetric mass, you can actually show that Newton's law holds in its usual form. This can be done through a gravitational analogue of Gauss's law for electric fields:
$$ointint_S mathbf{g} cdot dmathbf{l} = -4pi GM_mathrm{enc}$$
where $S$ is a closed surface enclosing the mass (can't write that funny circle thing around the integral here properly, sorry) and $M_mathrm{enc}$ the enclosed mass. Alternatively, you can just integrate Newton's law, but that's not as nice.
It's also important to point out that the spherical symmetry must not just be in the gross shape of the mass, but also in its density distribution. In fact, the gravitational field of the Earth is non-spherically symmetric and so non-inverse square for near-Earth objects chiefly because of such inhomogeneities within it than anything else - a fact that shows up if you look at a "geoid", effectively a map of the shape of the Earth's gravity field, used in cartography to set a precise meaning to the term "sea level". This, despite that Earth's gross shape is as good if not a better sphere than many "spherical" objects you can hold in your hands.
(Note: another, but so far purely hypothetical or fantastic, possibility, but arguably more dramatic, way to violate it would be if there existed negative gravitational masses [likely violating Einstein's theory]: these would permit the creation of 'gravitational dipoles' [i.e. a non-removable gravitational dipole moment] that would be analogous to electric dipoles and so would have an inverse-cube law.)
ADD (see comments): As a note, this result is also a case of the famous shell theorem and was proven by the founder of classical mechanics, Sir Isaac Newton, himself. The above is a more modern approach to deriving the result.
ADD 2: I notice that an explanation has been sought, not only for the correct mode of proof of the inverse-square behavior for an extended spherically symmetric mass, but also for why the OP's attempted proof thereof is erroneous. The basic reason is that the essential characteristic - the inverse-square law - of the spherically-symmetric behavior cannot be analyzed in terms of the behaviors of its parts individually: it only appears when you add them all together, holistically, at least up to the point of attaining full spherical symmetry. The two mass elements selected from the sphere being symmetrical about its center is not enough. Because while they, together, have some symmetry (namely $mathrm{O}(2) times mathbb{Z}_2$), they do not have spherical symmetry. Thus, alone, they will never be entirely replaceable by a point concentration as you try to do and you can only be misled as to the whole behavior thinking about them alone, individually. This logical error actually has a name: it's called the fallacy of the composite - assuming that a composite system necessarily has, or "inherits", the same properties as its parts. It doesn't, or, if it does, that has to be proved, and not simply deduced from the fact of its composition thereby.
(Well actually, you can break it up, but you have to "peel it like an onion" into spherical shells. I don't include this because it is, in effect, "trivial": the 'good part' - i.e. the fundamental difference between spherical symmetry and non-spherical - is already there once you do that, as it has to be, and thus has no real explaining or proving power, I'd say. At best, it could be the final step in a proof, but not the most meaty step.)
answered Apr 21 at 6:03
The_SympathizerThe_Sympathizer
5,7431230
$endgroup$
$begingroup$
FWIW, the lack of the "funny circle thing" is a known issue in MathJax.
$endgroup$
– Ilmari Karonen
Apr 21 at 9:42
1
$begingroup$
Ok, what is the error in my analysis then?
$endgroup$
– Ambrose Swasey
Apr 21 at 12:44
4
$begingroup$
@AmbroseSwasey You've only properly considered the points A, B, and C. If you sum over all the points carefully, it averages out correctly. For example, if you consider the points D and E, so that DE is perpendicular to AB, then both D and E contribute less gravitational field than B does, partially balancing out the fact that (as you pointed out) A and C on average contribute more than B does.
$endgroup$
– knzhou
Apr 21 at 14:52
$begingroup$
@AmbroseSwasey Of course, doing this whole argument explicitly would be extremely difficult. Calculus was invented to make this kind of computation relatively easy and straightforward.
$endgroup$
– knzhou
Apr 21 at 14:55
7
$begingroup$
Are "nmming" and "nning" typos?
$endgroup$
– npostavs
Apr 21 at 15:44
|
show 9 more comments
$begingroup$
It is not quite true to say that the inverse-square law is "not accurate for non-point masses". The more correct answer is to say that it is not accurate for non-spherically symmetric masses.
The inverse-square law depends, essentially, on the fact that the source "looks" the same to the object being attracted thereby, no matter where it is at in relation. If it can "see" a bump nmming it from one direction than from another, it will feel more force from that bump than otherwise, even while being at the same distance from the center of mass, hence violating the inverse-square law. E.g. a bilobate asteroidal/cometary object like the recently-discovered "Ultima Thule" with one lobe nning at you, versus being at the same distance but "seeing" it sideways.
For a spherically-symmetric mass, you can actually show that Newton's law holds in its usual form. This can be done through a gravitational analogue of Gauss's law for electric fields:
$$ointint_S mathbf{g} cdot dmathbf{l} = -4pi GM_mathrm{enc}$$
where $S$ is a closed surface enclosing the mass (can't write that funny circle thing around the integral here properly, sorry) and $M_mathrm{enc}$ the enclosed mass. Alternatively, you can just integrate Newton's law, but that's not as nice.
It's also important to point out that the spherical symmetry must not just be in the gross shape of the mass, but also in its density distribution. In fact, the gravitational field of the Earth is non-spherically symmetric and so non-inverse square for near-Earth objects chiefly because of such inhomogeneities within it than anything else - a fact that shows up if you look at a "geoid", effectively a map of the shape of the Earth's gravity field, used in cartography to set a precise meaning to the term "sea level". This, despite that Earth's gross shape is as good if not a better sphere than many "spherical" objects you can hold in your hands.
(Note: another, but so far purely hypothetical or fantastic, possibility, but arguably more dramatic, way to violate it would be if there existed negative gravitational masses [likely violating Einstein's theory]: these would permit the creation of 'gravitational dipoles' [i.e. a non-removable gravitational dipole moment] that would be analogous to electric dipoles and so would have an inverse-cube law.)
ADD (see comments): As a note, this result is also a case of the famous shell theorem and was proven by the founder of classical mechanics, Sir Isaac Newton, himself. The above is a more modern approach to deriving the result.
ADD 2: I notice that an explanation has been sought, not only for the correct mode of proof of the inverse-square behavior for an extended spherically symmetric mass, but also for why the OP's attempted proof thereof is erroneous. The basic reason is that the essential characteristic - the inverse-square law - of the spherically-symmetric behavior cannot be analyzed in terms of the behaviors of its parts individually: it only appears when you add them all together, holistically, at least up to the point of attaining full spherical symmetry. The two mass elements selected from the sphere being symmetrical about its center is not enough. Because while they, together, have some symmetry (namely $mathrm{O}(2) times mathbb{Z}_2$), they do not have spherical symmetry. Thus, alone, they will never be entirely replaceable by a point concentration as you try to do and you can only be misled as to the whole behavior thinking about them alone, individually. This logical error actually has a name: it's called the fallacy of the composite - assuming that a composite system necessarily has, or "inherits", the same properties as its parts. It doesn't, or, if it does, that has to be proved, and not simply deduced from the fact of its composition thereby.
(Well actually, you can break it up, but you have to "peel it like an onion" into spherical shells. I don't include this because it is, in effect, "trivial": the 'good part' - i.e. the fundamental difference between spherical symmetry and non-spherical - is already there once you do that, as it has to be, and thus has no real explaining or proving power, I'd say. At best, it could be the final step in a proof, but not the most meaty step.)
answered Apr 21 at 6:03
The_SympathizerThe_Sympathizer
5,7431230
$endgroup$
$begingroup$
FWIW, the lack of the "funny circle thing" is a known issue in MathJax.
$endgroup$
– Ilmari Karonen
Apr 21 at 9:42
1
$begingroup$
Ok, what is the error in my analysis then?
$endgroup$
– Ambrose Swasey
Apr 21 at 12:44
4
$begingroup$
@AmbroseSwasey You've only properly considered the points A, B, and C. If you sum over all the points carefully, it averages out correctly. For example, if you consider the points D and E, so that DE is perpendicular to AB, then both D and E contribute less gravitational field than B does, partially balancing out the fact that (as you pointed out) A and C on average contribute more than B does.
$endgroup$
– knzhou
Apr 21 at 14:52
$begingroup$
@AmbroseSwasey Of course, doing this whole argument explicitly would be extremely difficult. Calculus was invented to make this kind of computation relatively easy and straightforward.
$endgroup$
– knzhou
Apr 21 at 14:55
7
$begingroup$
Are "nmming" and "nning" typos?
$endgroup$
– npostavs
Apr 21 at 15:44
|
show 9 more comments
$begingroup$
It is not quite true to say that the inverse-square law is "not accurate for non-point masses". The more correct answer is to say that it is not accurate for non-spherically symmetric masses.
The inverse-square law depends, essentially, on the fact that the source "looks" the same to the object being attracted thereby, no matter where it is at in relation. If it can "see" a bump nmming it from one direction than from another, it will feel more force from that bump than otherwise, even while being at the same distance from the center of mass, hence violating the inverse-square law. E.g. a bilobate asteroidal/cometary object like the recently-discovered "Ultima Thule" with one lobe nning at you, versus being at the same distance but "seeing" it sideways.
For a spherically-symmetric mass, you can actually show that Newton's law holds in its usual form. This can be done through a gravitational analogue of Gauss's law for electric fields:
$$ointint_S mathbf{g} cdot dmathbf{l} = -4pi GM_mathrm{enc}$$
where $S$ is a closed surface enclosing the mass (can't write that funny circle thing around the integral here properly, sorry) and $M_mathrm{enc}$ the enclosed mass. Alternatively, you can just integrate Newton's law, but that's not as nice.
It's also important to point out that the spherical symmetry must not just be in the gross shape of the mass, but also in its density distribution. In fact, the gravitational field of the Earth is non-spherically symmetric and so non-inverse square for near-Earth objects chiefly because of such inhomogeneities within it than anything else - a fact that shows up if you look at a "geoid", effectively a map of the shape of the Earth's gravity field, used in cartography to set a precise meaning to the term "sea level". This, despite that Earth's gross shape is as good if not a better sphere than many "spherical" objects you can hold in your hands.
(Note: another, but so far purely hypothetical or fantastic, possibility, but arguably more dramatic, way to violate it would be if there existed negative gravitational masses [likely violating Einstein's theory]: these would permit the creation of 'gravitational dipoles' [i.e. a non-removable gravitational dipole moment] that would be analogous to electric dipoles and so would have an inverse-cube law.)
ADD (see comments): As a note, this result is also a case of the famous shell theorem and was proven by the founder of classical mechanics, Sir Isaac Newton, himself. The above is a more modern approach to deriving the result.
ADD 2: I notice that an explanation has been sought, not only for the correct mode of proof of the inverse-square behavior for an extended spherically symmetric mass, but also for why the OP's attempted proof thereof is erroneous. The basic reason is that the essential characteristic - the inverse-square law - of the spherically-symmetric behavior cannot be analyzed in terms of the behaviors of its parts individually: it only appears when you add them all together, holistically, at least up to the point of attaining full spherical symmetry. The two mass elements selected from the sphere being symmetrical about its center is not enough. Because while they, together, have some symmetry (namely $mathrm{O}(2) times mathbb{Z}_2$), they do not have spherical symmetry. Thus, alone, they will never be entirely replaceable by a point concentration as you try to do and you can only be misled as to the whole behavior thinking about them alone, individually. This logical error actually has a name: it's called the fallacy of the composite - assuming that a composite system necessarily has, or "inherits", the same properties as its parts. It doesn't, or, if it does, that has to be proved, and not simply deduced from the fact of its composition thereby.
(Well actually, you can break it up, but you have to "peel it like an onion" into spherical shells. I don't include this because it is, in effect, "trivial": the 'good part' - i.e. the fundamental difference between spherical symmetry and non-spherical - is already there once you do that, as it has to be, and thus has no real explaining or proving power, I'd say. At best, it could be the final step in a proof, but not the most meaty step.)
answered Apr 21 at 6:03
The_SympathizerThe_Sympathizer
5,7431230
$endgroup$
It is not quite true to say that the inverse-square law is "not accurate for non-point masses". The more correct answer is to say that it is not accurate for non-spherically symmetric masses.
The inverse-square law depends, essentially, on the fact that the source "looks" the same to the object being attracted thereby, no matter where it is at in relation. If it can "see" a bump nmming it from one direction than from another, it will feel more force from that bump than otherwise, even while being at the same distance from the center of mass, hence violating the inverse-square law. E.g. a bilobate asteroidal/cometary object like the recently-discovered "Ultima Thule" with one lobe nning at you, versus being at the same distance but "seeing" it sideways.
For a spherically-symmetric mass, you can actually show that Newton's law holds in its usual form. This can be done through a gravitational analogue of Gauss's law for electric fields:
$$ointint_S mathbf{g} cdot dmathbf{l} = -4pi GM_mathrm{enc}$$
where $S$ is a closed surface enclosing the mass (can't write that funny circle thing around the integral here properly, sorry) and $M_mathrm{enc}$ the enclosed mass. Alternatively, you can just integrate Newton's law, but that's not as nice.
It's also important to point out that the spherical symmetry must not just be in the gross shape of the mass, but also in its density distribution. In fact, the gravitational field of the Earth is non-spherically symmetric and so non-inverse square for near-Earth objects chiefly because of such inhomogeneities within it than anything else - a fact that shows up if you look at a "geoid", effectively a map of the shape of the Earth's gravity field, used in cartography to set a precise meaning to the term "sea level". This, despite that Earth's gross shape is as good if not a better sphere than many "spherical" objects you can hold in your hands.
(Note: another, but so far purely hypothetical or fantastic, possibility, but arguably more dramatic, way to violate it would be if there existed negative gravitational masses [likely violating Einstein's theory]: these would permit the creation of 'gravitational dipoles' [i.e. a non-removable gravitational dipole moment] that would be analogous to electric dipoles and so would have an inverse-cube law.)
ADD (see comments): As a note, this result is also a case of the famous shell theorem and was proven by the founder of classical mechanics, Sir Isaac Newton, himself. The above is a more modern approach to deriving the result.
ADD 2: I notice that an explanation has been sought, not only for the correct mode of proof of the inverse-square behavior for an extended spherically symmetric mass, but also for why the OP's attempted proof thereof is erroneous. The basic reason is that the essential characteristic - the inverse-square law - of the spherically-symmetric behavior cannot be analyzed in terms of the behaviors of its parts individually: it only appears when you add them all together, holistically, at least up to the point of attaining full spherical symmetry. The two mass elements selected from the sphere being symmetrical about its center is not enough. Because while they, together, have some symmetry (namely $mathrm{O}(2) times mathbb{Z}_2$), they do not have spherical symmetry. Thus, alone, they will never be entirely replaceable by a point concentration as you try to do and you can only be misled as to the whole behavior thinking about them alone, individually. This logical error actually has a name: it's called the fallacy of the composite - assuming that a composite system necessarily has, or "inherits", the same properties as its parts. It doesn't, or, if it does, that has to be proved, and not simply deduced from the fact of its composition thereby.
(Well actually, you can break it up, but you have to "peel it like an onion" into spherical shells. I don't include this because it is, in effect, "trivial": the 'good part' - i.e. the fundamental difference between spherical symmetry and non-spherical - is already there once you do that, as it has to be, and thus has no real explaining or proving power, I'd say. At best, it could be the final step in a proof, but not the most meaty step.)
answered Apr 21 at 6:03
The_SympathizerThe_Sympathizer
5,7431230
edited Apr 22 at 4:12
answered Apr 21 at 6:03
The_SympathizerThe_Sympathizer
5,7431230
answered Apr 21 at 6:03
The_SympathizerThe_Sympathizer
5,7431230
answered Apr 21 at 6:03
The_SympathizerThe_Sympathizer
5,7431230
5,7431230
$begingroup$
FWIW, the lack of the "funny circle thing" is a known issue in MathJax.
$endgroup$
– Ilmari Karonen
Apr 21 at 9:42
1
$begingroup$
Ok, what is the error in my analysis then?
$endgroup$
– Ambrose Swasey
Apr 21 at 12:44
4
$begingroup$
@AmbroseSwasey You've only properly considered the points A, B, and C. If you sum over all the points carefully, it averages out correctly. For example, if you consider the points D and E, so that DE is perpendicular to AB, then both D and E contribute less gravitational field than B does, partially balancing out the fact that (as you pointed out) A and C on average contribute more than B does.
$endgroup$
– knzhou
Apr 21 at 14:52
$begingroup$
@AmbroseSwasey Of course, doing this whole argument explicitly would be extremely difficult. Calculus was invented to make this kind of computation relatively easy and straightforward.
$endgroup$
– knzhou
Apr 21 at 14:55
7
$begingroup$
Are "nmming" and "nning" typos?
$endgroup$
– npostavs
Apr 21 at 15:44
|
show 9 more comments
$begingroup$
FWIW, the lack of the "funny circle thing" is a known issue in MathJax.
$endgroup$
– Ilmari Karonen
Apr 21 at 9:42
1
$begingroup$
Ok, what is the error in my analysis then?
$endgroup$
– Ambrose Swasey
Apr 21 at 12:44
4
$begingroup$
@AmbroseSwasey You've only properly considered the points A, B, and C. If you sum over all the points carefully, it averages out correctly. For example, if you consider the points D and E, so that DE is perpendicular to AB, then both D and E contribute less gravitational field than B does, partially balancing out the fact that (as you pointed out) A and C on average contribute more than B does.
$endgroup$
– knzhou
Apr 21 at 14:52
$begingroup$
@AmbroseSwasey Of course, doing this whole argument explicitly would be extremely difficult. Calculus was invented to make this kind of computation relatively easy and straightforward.
$endgroup$
– knzhou
Apr 21 at 14:55
7
$begingroup$
Are "nmming" and "nning" typos?
$endgroup$
– npostavs
Apr 21 at 15:44
$begingroup$
FWIW, the lack of the "funny circle thing" is a known issue in MathJax.
$endgroup$
– Ilmari Karonen
Apr 21 at 9:42
$begingroup$
FWIW, the lack of the "funny circle thing" is a known issue in MathJax.
$endgroup$
– Ilmari Karonen
Apr 21 at 9:42
1
1
$begingroup$
Ok, what is the error in my analysis then?
$endgroup$
– Ambrose Swasey
Apr 21 at 12:44
$begingroup$
Ok, what is the error in my analysis then?
$endgroup$
– Ambrose Swasey
Apr 21 at 12:44
4
4
$begingroup$
@AmbroseSwasey You've only properly considered the points A, B, and C. If you sum over all the points carefully, it averages out correctly. For example, if you consider the points D and E, so that DE is perpendicular to AB, then both D and E contribute less gravitational field than B does, partially balancing out the fact that (as you pointed out) A and C on average contribute more than B does.
$endgroup$
– knzhou
Apr 21 at 14:52
$begingroup$
@AmbroseSwasey You've only properly considered the points A, B, and C. If you sum over all the points carefully, it averages out correctly. For example, if you consider the points D and E, so that DE is perpendicular to AB, then both D and E contribute less gravitational field than B does, partially balancing out the fact that (as you pointed out) A and C on average contribute more than B does.
$endgroup$
– knzhou
Apr 21 at 14:52
$begingroup$
@AmbroseSwasey Of course, doing this whole argument explicitly would be extremely difficult. Calculus was invented to make this kind of computation relatively easy and straightforward.
$endgroup$
– knzhou
Apr 21 at 14:55
$begingroup$
@AmbroseSwasey Of course, doing this whole argument explicitly would be extremely difficult. Calculus was invented to make this kind of computation relatively easy and straightforward.
$endgroup$
– knzhou
Apr 21 at 14:55
7
7
$begingroup$
Are "nmming" and "nning" typos?
$endgroup$
– npostavs
Apr 21 at 15:44
$begingroup$
Are "nmming" and "nning" typos?
$endgroup$
– npostavs
Apr 21 at 15:44
|
show 9 more comments
$begingroup$
The inverse square law is exact as long as general relativistic corrections can be neglected. For any mass distribution the force can be found from summing up the inverse square law contribution from its parts, that is by convolution of the density distribution with the inverse square law.
answered Apr 21 at 10:26
my2ctsmy2cts
6,6342722
$endgroup$
add a comment |
$begingroup$
The inverse square law is exact as long as general relativistic corrections can be neglected. For any mass distribution the force can be found from summing up the inverse square law contribution from its parts, that is by convolution of the density distribution with the inverse square law.
answered Apr 21 at 10:26
my2ctsmy2cts
6,6342722
$endgroup$
add a comment |
$begingroup$
The inverse square law is exact as long as general relativistic corrections can be neglected. For any mass distribution the force can be found from summing up the inverse square law contribution from its parts, that is by convolution of the density distribution with the inverse square law.
answered Apr 21 at 10:26
my2ctsmy2cts
6,6342722
$endgroup$
The inverse square law is exact as long as general relativistic corrections can be neglected. For any mass distribution the force can be found from summing up the inverse square law contribution from its parts, that is by convolution of the density distribution with the inverse square law.
answered Apr 21 at 10:26
my2ctsmy2cts
6,6342722
answered Apr 21 at 10:26
my2ctsmy2cts
6,6342722
answered Apr 21 at 10:26
my2ctsmy2cts
6,6342722
answered Apr 21 at 10:26
my2ctsmy2cts
6,6342722
6,6342722
add a comment |
add a comment |
$begingroup$
No, the inverse square law is not accurate for non-point masses. It isn't even accurate for spherically symmetric non-point masses. Take a look at the Wikipedia article on gravitational potential. The plot on the right shows the force of gravity for the Earth. It isn't an inverse-square rule plot. The force of gravity reduces as you climb up and away from the surface of the Earth, and it reduces as you tunnel down towards the centre. If you were in some gedanken chamber at the centre of the Earth, you'd float around weightless.
CCASA image by AllenMcC, see Wikipedia Commons
Generally speaking, I'd say the inverse square rule is good enough most of the time when we're talking about celestial mechanics. It probably isn't good enough for when we're talking about co-orbiting neutron stars. Maybe some astronomer can tell you something about that.
Edit: note the comments below.
answered Apr 21 at 9:06
John DuffieldJohn Duffield
7,43152547
$endgroup$
3
$begingroup$
In Newtonian gravity, the inverse-square law is accurate for spherically symmetric objects. It's just that, as The_Symphathizer points out in their answer, the Earth is not actually spherically symmetric. (In the case of "tunneling toward the center", you count the mass within the sphere corresponding to your current depth.)
$endgroup$
– Peter Erwin
Apr 21 at 9:16
12
$begingroup$
More to the point, the inverse square law is accurate for spherically symmetric masses outside the mass itself. Inside the massive body, only the part of the mass closer to the center than the observer counts (which is quite a remarkable result in itself).
$endgroup$
– Ilmari Karonen
Apr 21 at 9:47
$begingroup$
@Peter Erwin : noted. Yes, it's accurate when you allow for the distribution of matter. I was thinking of just the force of gravity for the Earth. Ilmari Karonen: also noted.
$endgroup$
– John Duffield
Apr 21 at 10:02
add a comment |
$begingroup$
No, the inverse square law is not accurate for non-point masses. It isn't even accurate for spherically symmetric non-point masses. Take a look at the Wikipedia article on gravitational potential. The plot on the right shows the force of gravity for the Earth. It isn't an inverse-square rule plot. The force of gravity reduces as you climb up and away from the surface of the Earth, and it reduces as you tunnel down towards the centre. If you were in some gedanken chamber at the centre of the Earth, you'd float around weightless.
CCASA image by AllenMcC, see Wikipedia Commons
Generally speaking, I'd say the inverse square rule is good enough most of the time when we're talking about celestial mechanics. It probably isn't good enough for when we're talking about co-orbiting neutron stars. Maybe some astronomer can tell you something about that.
Edit: note the comments below.
answered Apr 21 at 9:06
John DuffieldJohn Duffield
7,43152547
$endgroup$
3
$begingroup$
In Newtonian gravity, the inverse-square law is accurate for spherically symmetric objects. It's just that, as The_Symphathizer points out in their answer, the Earth is not actually spherically symmetric. (In the case of "tunneling toward the center", you count the mass within the sphere corresponding to your current depth.)
$endgroup$
– Peter Erwin
Apr 21 at 9:16
12
$begingroup$
More to the point, the inverse square law is accurate for spherically symmetric masses outside the mass itself. Inside the massive body, only the part of the mass closer to the center than the observer counts (which is quite a remarkable result in itself).
$endgroup$
– Ilmari Karonen
Apr 21 at 9:47
$begingroup$
@Peter Erwin : noted. Yes, it's accurate when you allow for the distribution of matter. I was thinking of just the force of gravity for the Earth. Ilmari Karonen: also noted.
$endgroup$
– John Duffield
Apr 21 at 10:02
add a comment |
$begingroup$
No, the inverse square law is not accurate for non-point masses. It isn't even accurate for spherically symmetric non-point masses. Take a look at the Wikipedia article on gravitational potential. The plot on the right shows the force of gravity for the Earth. It isn't an inverse-square rule plot. The force of gravity reduces as you climb up and away from the surface of the Earth, and it reduces as you tunnel down towards the centre. If you were in some gedanken chamber at the centre of the Earth, you'd float around weightless.
CCASA image by AllenMcC, see Wikipedia Commons
Generally speaking, I'd say the inverse square rule is good enough most of the time when we're talking about celestial mechanics. It probably isn't good enough for when we're talking about co-orbiting neutron stars. Maybe some astronomer can tell you something about that.
Edit: note the comments below.
answered Apr 21 at 9:06
John DuffieldJohn Duffield
7,43152547
$endgroup$
No, the inverse square law is not accurate for non-point masses. It isn't even accurate for spherically symmetric non-point masses. Take a look at the Wikipedia article on gravitational potential. The plot on the right shows the force of gravity for the Earth. It isn't an inverse-square rule plot. The force of gravity reduces as you climb up and away from the surface of the Earth, and it reduces as you tunnel down towards the centre. If you were in some gedanken chamber at the centre of the Earth, you'd float around weightless.
CCASA image by AllenMcC, see Wikipedia Commons
Generally speaking, I'd say the inverse square rule is good enough most of the time when we're talking about celestial mechanics. It probably isn't good enough for when we're talking about co-orbiting neutron stars. Maybe some astronomer can tell you something about that.
Edit: note the comments below.
answered Apr 21 at 9:06
John DuffieldJohn Duffield
7,43152547
edited Apr 21 at 10:02
answered Apr 21 at 9:06
John DuffieldJohn Duffield
7,43152547
answered Apr 21 at 9:06
John DuffieldJohn Duffield
7,43152547
answered Apr 21 at 9:06
John DuffieldJohn Duffield
7,43152547
7,43152547
3
$begingroup$
In Newtonian gravity, the inverse-square law is accurate for spherically symmetric objects. It's just that, as The_Symphathizer points out in their answer, the Earth is not actually spherically symmetric. (In the case of "tunneling toward the center", you count the mass within the sphere corresponding to your current depth.)
$endgroup$
– Peter Erwin
Apr 21 at 9:16
12
$begingroup$
More to the point, the inverse square law is accurate for spherically symmetric masses outside the mass itself. Inside the massive body, only the part of the mass closer to the center than the observer counts (which is quite a remarkable result in itself).
$endgroup$
– Ilmari Karonen
Apr 21 at 9:47
$begingroup$
@Peter Erwin : noted. Yes, it's accurate when you allow for the distribution of matter. I was thinking of just the force of gravity for the Earth. Ilmari Karonen: also noted.
$endgroup$
– John Duffield
Apr 21 at 10:02
add a comment |
3
$begingroup$
In Newtonian gravity, the inverse-square law is accurate for spherically symmetric objects. It's just that, as The_Symphathizer points out in their answer, the Earth is not actually spherically symmetric. (In the case of "tunneling toward the center", you count the mass within the sphere corresponding to your current depth.)
$endgroup$
– Peter Erwin
Apr 21 at 9:16
12
$begingroup$
More to the point, the inverse square law is accurate for spherically symmetric masses outside the mass itself. Inside the massive body, only the part of the mass closer to the center than the observer counts (which is quite a remarkable result in itself).
$endgroup$
– Ilmari Karonen
Apr 21 at 9:47
$begingroup$
@Peter Erwin : noted. Yes, it's accurate when you allow for the distribution of matter. I was thinking of just the force of gravity for the Earth. Ilmari Karonen: also noted.
$endgroup$
– John Duffield
Apr 21 at 10:02
3
3
$begingroup$
In Newtonian gravity, the inverse-square law is accurate for spherically symmetric objects. It's just that, as The_Symphathizer points out in their answer, the Earth is not actually spherically symmetric. (In the case of "tunneling toward the center", you count the mass within the sphere corresponding to your current depth.)
$endgroup$
– Peter Erwin
Apr 21 at 9:16
$begingroup$
In Newtonian gravity, the inverse-square law is accurate for spherically symmetric objects. It's just that, as The_Symphathizer points out in their answer, the Earth is not actually spherically symmetric. (In the case of "tunneling toward the center", you count the mass within the sphere corresponding to your current depth.)
$endgroup$
– Peter Erwin
Apr 21 at 9:16
12
12
$begingroup$
More to the point, the inverse square law is accurate for spherically symmetric masses outside the mass itself. Inside the massive body, only the part of the mass closer to the center than the observer counts (which is quite a remarkable result in itself).
$endgroup$
– Ilmari Karonen
Apr 21 at 9:47
$begingroup$
More to the point, the inverse square law is accurate for spherically symmetric masses outside the mass itself. Inside the massive body, only the part of the mass closer to the center than the observer counts (which is quite a remarkable result in itself).
$endgroup$
– Ilmari Karonen
Apr 21 at 9:47
$begingroup$
@Peter Erwin : noted. Yes, it's accurate when you allow for the distribution of matter. I was thinking of just the force of gravity for the Earth. Ilmari Karonen: also noted.
$endgroup$
– John Duffield
Apr 21 at 10:02
$begingroup$
@Peter Erwin : noted. Yes, it's accurate when you allow for the distribution of matter. I was thinking of just the force of gravity for the Earth. Ilmari Karonen: also noted.
$endgroup$
– John Duffield
Apr 21 at 10:02
add a comment |
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f474092%2finverse-square-law-not-accurate-for-non-point-masses%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
The Earth is a point mass in comparison to the mass of the Sun, so to an extremely high degree of accuracy you can model the planets as being point masses, which is what Newton did in the Principia. Also, Newton did not merely 'popularise' the law: he demonstrated its mathematical coherence and derived it from the law of universal gravitation. Hooke did nothing to demonstrate it mathematically or universality and only speculated on its existence in the Solar System, don't be fooled by comments by Stephen Hawking as he did not always know what he was talking about.
$endgroup$
– Tom
Apr 21 at 15:36
$begingroup$
Since non-point masses can be considered as collections of a bunch of tiny point masses, the inverse-square law is still applied when you add up the inverse-square contributions of all of the tiny point masses. This is how the gravitational field from extended objects is calculated (though usually we work in the limit where the non-point mass is made up of infinitely many point masses of infinitesimal mass; dealing with adding them all up is precisely what calculus was invented to deal with).
$endgroup$
– probably_someone
Apr 21 at 17:06
$begingroup$
There's more mass in your sphere located near d=5 than near d=3.
$endgroup$
– Dmitry Grigoryev
Apr 21 at 18:48