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I always intuitively understood how Newton's Third Law applies to simple contact forces because I could imagine how to reciprocate the force of one object on another, the recipient's molecular structure itself pushes back in an attempt to maintain itself so long as they are touching (as in the case of a normal force). However, now studying electricity and Coulomb's Law, I don't see how this concept applies (likewise, how it applies to the Law of Universal Gravitation). Mathematically speaking Coulomb's Law depicts the magnitude of the force between two charges irrespective of direction. It is beyond me how the force between two charges can be opposite and equal if one charge is five billion times larger than the other. Likewise, the notion that the force between Earth and the Sun is equal in magnitude makes no sense when the Sun has over 300K the mass of Earth. Would anyone be able to provide an explanation? Thank you.

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  • $\begingroup$ Have you heard about conservation of linear momentum? Do you know about Noether's theorem, and how it is connected to said conservation? $\endgroup$ Feb 24 at 19:55
  • $\begingroup$ Related: physics.stackexchange.com/a/252511/392 The equal and opposite forces have different effects on two bodies regardless of the source of the force or the distance between them. This effect depends on the concept of the reduced mass. True for planets and newton's cradle at the same time. $\endgroup$ Feb 24 at 20:54
  • $\begingroup$ It seems to me this question just arises due to unfamiliarity with non-contact forces in our every day life rather than something more concrete. Do this: grab two magnets, one in each hand and repel them. Did you feel an equal and opposite force? You can do the same thing with a huge magnet and a little magnet though it's trickier to feel because you will also feel the different weights in your hands. If you have a magnet be twice the weight of the other you can kind of lay each one down on the smooth surface and repel it with the other and observe the difference in acceleration. $\endgroup$
    – DKNguyen
    Feb 24 at 23:57
  • $\begingroup$ There are only 4 known forces: electromagnetic, gravitation, strong nuclear, and weak nuclear. Contact forces are actually electromagnetic repulsion between the electrons of two different surfaces, which actually makes them non-contact forces. $\endgroup$ Feb 25 at 0:10

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Mathematically, it's simply true: the forces will have equal magnitude. Taking gravity for example, the force acting on either the sun $M_\text{Sun}$ or the earth $m_\text{earth}$ is the same magnitude: $$|\vec{F}_\text{gravity}| = \frac{GM_\text{Sun} m_\text{earth}}{r^2}$$ where $r$ is the distance between the two. Something to point out though is that the acceleration the two bodies feel is different, since $\vec{F} = m \vec{a}$, we find that the accelerations: $$|\vec{a}_\text{Sun} | = \frac{G m_\text{Earth}}{r^2}, \quad |\vec{a}_\text{Earth} | = \frac{G M_\text{Sun}}{r^2}$$ different!

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  • $\begingroup$ I think I understand it mathematically, but is there an intuitive way I may be able to look at this? $\endgroup$ Feb 24 at 20:11
  • $\begingroup$ @EthanDandelion Sometimes when we encounter a new concept, we are able to apply intuition we already have. Other times we need to build new intuition because our old intuition is inadequate. For most people, Newton's 3rd law is very unintuitive and it takes some time to build new intuition with this law as the starting point. $\endgroup$
    – d_b
    Feb 24 at 22:26
  • $\begingroup$ @EthanDandelion Intuition is just fancy term for "you know...from experience". It's not like there is real physical contact going on when you push something with your hands either. Even when you touch something, there are repulsive forces at work at small distances that prevent the nuclei of atoms from physically making contact (whatever that really means). The only difference is the gap is too small for you to see but you're just so used to it you don't think about it. $\endgroup$
    – DKNguyen
    Feb 25 at 0:07
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I belive you missunderstood the case of contact forces, the one which you think it makes sense. It is the common missundertsanding that one of the forces in the Newton's third law comes as a response to the other one, which acts on the "recipient". This is a missconception due to the fact that Newton's third law is still called the law of "action and reaction". Actually, the two forces act simultaneously, there is no delay, no active partner and no recipient. What we have is an inter-action between two objects as equal partners. The two forces, together, describe this unique interaction. They start and end at the same time. If you think that the pair of forces describes an unique thing, the interaction, it may become more confortable to accept that they have the same magnitude. And your objection relative to the interaction Earth-Sun can be applied to contact forces too. A fly sitting on a table is milions (thousands of K) of times lighter than the table but the contact forces between the fly and the table are equal in magnitude. But you say that this makes sense, don't you?

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Coloumb and Cavendish made carefully experiments using torsion balances to calculate the forces for electrical and gravitational forces respectively.

In the case of charges for example, we can imagine that a charge $q$ in the balance causes a deflection $\delta \theta$ on the balance when another charge $2q$ was nearby at a distance $d$.

When he inverted the experiment, and put $2q$ in the balance and $q$ at the same distance, he must have measured the same force, otherwise he could not have published that $$F = \frac{kQq}{r^2}$$.

So, if the force is the same for both cases, it follows that the charge out of the balance should be doing always the same force on the balance. The outcome of the experiences, that leads to the formula, requires that action and reaction must be the same.

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You are having some wrong ideas about the concept of 'force'. You seem to think that a force is something were one object is the active part and the other the passive. This is incorrect. A force is something that happens as a result of the mutual interaction between the two objects. One object can not exert a force without the other object. In mathematical terms, this is expressed by the fact that the force law contains the product of the masses or charges of the two objects involved i.e. $m\cdot M$ or $q\cdot Q$. And it does obviously not make a difference if you write $M\cdot m$ or $Q\cdot q$ instead, the value of the force is the same.

Note also that strictly speaking there are actually no contact forces. If you push with your hand against an object, then the force you feel is exactly the Coulomb force that you have been quoting as an example of a non-contact force. It is only that the charges in your hand and those in the object come extremely close, but they are still separated by a small gap (about the size of an atom), kept apart by the Coulomb force. So contact forces are conceptually a contradiction in terms, maybe useful for some engineering problems, but potentially misleading when it comes to understanding physical principles.

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  • $\begingroup$ I think this explanation is getting to the core of my misunderstanding. Would you be able to provide a good mathematical definition for force so that I may better understand it? $\endgroup$ Feb 28 at 20:21
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It is beyond me how the force between two charges can be opposite and equal if one charge is five billion times larger than the other. Likewise, the notion that the force between Earth and the Sun is equal in magnitude makes no sense when the Sun has over 300K the mass of Earth. Would anyone be able to provide an explanation?

You are having the common difficulty in distinguishing between Newton's 3rd law, that every action (force) has an equal and opposite reaction (force), and Newton's 2nd law which deals with the effect (acceleration) of the equal and opposite force on the bodies individually. Per Newton's 2nd law

$$F_{net}=ma$$

For example, for a gravitational force $F$ that mass $m$ exerts on mass $M$ there is an equal and opposite force $-F$ that $M$ exerts on $m$, per Newton's 3rd law. If that force is the only force on the two bodies, then the magnitude of their respective accelerations will be, per Newton's 2nd law

$$a_{M}=\frac{F_{net}}{M}$$

$$a_{m}=\frac{F_{net}}{m}$$

If $M$ is much larger than $m$, like the Sun versus the Earth, then

$$a_{M}<<a_{m}$$

In the case of the Sun and Earth, if the mass of the Sun is 300,000 times that of the Earth, then the acceleration of the Sun due to the force exerted on it by the Earth would be 1/300,000 of the acceleration of the Earth due to the equal and opposite force the Sun exerts on the Earth.

This applies for non contact forces (gravity, electrostatic,..) and contact forces.

Hope this helps.

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  • $\begingroup$ I understand why acceleration is different but I still don't understand why force itself is reciprocated. What about the smaller object makes it capable of reciprocating the force exerted on it by the larger object? With contact forces I can see how this is just the molecular structure of the object but with field forces I am having some confusion. $\endgroup$ Feb 25 at 0:52
  • $\begingroup$ It sounds like your asking for proof of the third law. Is that it? $\endgroup$
    – Bob D
    Feb 25 at 8:00
  • $\begingroup$ @EthanDandelion See my new answer in response to your comment. $\endgroup$
    – Bob D
    Feb 25 at 18:34
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Assume that the Earth, mass $M_{\rm E}$, and the Sun, mass $M_{\rm S}$, can be treated as point masses at their respective centres of mass with separation $R$.

At a distance $R$ the magnitude of the gravitation field strength of the Sun is $g_{\rm S} = G\dfrac{M_{\rm S}}{R^2}$ and hence the magnitude of the gravitational force on the Earth is $M_{\rm E}\cdot g_{\rm S}=M_{\rm E}\cdot G\dfrac{M_{\rm S}}{R^2}$

At a distance $R$ the magnitude of the gravitation field strength of the Earth is $g_{\rm E} = G\dfrac{M_{\rm R}}{R^2}$ and hence the magnitude of the gravitational force on the Sun is $M_{\rm S}\cdot g_{\rm E}=M_{\rm S}\cdot G\dfrac{M_{\rm E}}{R^2}$

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This answer is in response to your following comment on my original answer:

I understand why acceleration is different but I still don't understand why force itself is reciprocated. What about the smaller object makes it capable of reciprocating the force exerted on it by the larger object? With contact forces I can see how this is just the molecular structure of the object but with field forces I am having some confusion.

The force is reciprocated because, otherwise, there would be a violation of the conservation of momentum law which states that, for an isolated system (a system not subjected to any net external force) momentum is conserved. This applies to both contact forces and forces due to the presence of fields.

Let two objects $m_1$ and $m_2$ be an isolated system. Let $m_1$ exert a gravitational force of $F_{12}$ on $m_2$ and $m_2$ exert a gravitational force $F_{21}$ on $m_1$. Per Newton's second law, the change in momentum of each object due to the gravitational force is:

$$F_{12}=\frac{dp_{2}}{dt}$$ $$F_{21}=\frac{dp_{1}}{dt}$$

For a total momentum change of

$$F_{12}+F_{21}=\frac{d(p_{2}+p_{1})}{dt}$$

Since the total momentum change for an isolated system of the two objects has to be zero for conservation of momentum:

$$F_{12}+F_{21}=0$$

$$F_{12}=-F_{21}$$

Which is the third law. Note that the result is independent of the sizes (masses) of the two gravitating objects.

Hope this helps.

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  • $\begingroup$ If F$_1$$_2$ is the force on m$_2$, then F$_1$$_2$ = dp$_2$/dt etc. The result is the same. $\endgroup$ Feb 25 at 18:56
  • $\begingroup$ @Not_Einstein I don't understand your comment $\endgroup$
    – Bob D
    Feb 25 at 19:01
  • $\begingroup$ @Not_Einstein Oops! Now I see it. I transposed my numbers. Will correct. Thanks $\endgroup$
    – Bob D
    Feb 25 at 20:51
  • $\begingroup$ Upon reading both of your answers I have concluded that my problem is with a fundamental misunderstanding of what force is so I'm going to see what I can do about that. $\endgroup$ Feb 28 at 20:23
  • $\begingroup$ @EthanDandelion No problem. Good luck $\endgroup$
    – Bob D
    Feb 28 at 20:25

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