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Newton's law of gravitation states:

Every particle attracts every other particle in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between the centers.

and it can be mathematically expressed as $$F=G\ \frac{m_1\ m_2}{r^2}$$
Newton's 3rd law of motion states:

When two bodies interact, they apply forces to one another that are equal in magnitude and opposite in direction.


Consider a scenario where there's an object of mass 1 kg near Earth's surface.

Let's assume that Newton's 3rd law of motion isn't acting for now.

As per the law of gravitation, the Earth pulls the object with a force of approx. 9.8 N and the object also pulls Earth with a force of 9.8 N. It isn't that only Earth pulls the object, both the object and Earth are pulling each other with a force of same magnitude that is calculated from the above-stated formula.
So, the object and Earth each are experiencing a force of 9.8 N from each other.

Now, let's think that Newton's 3rd law of motion starts acting.

As the object was pulling Earth with a force, Earth now in turn pulls the object with the same magnitude of force. Thus, the object now experiences 9.8 N (gravitational force) + 9.8 N (reaction force from the object pulling Earth) = 19.6 N (net force experienced) and similarly, Earth also experiences 19.6 N of net force from the object.
So, when the 3rd law of motion is in action, the object and Earth each should experience 19.6 N of force from each other.


In reality, this is not what we observe.

We see that an object of 1 kg accelerates at only 9.8 m/s^2 near Earth's surface and not 19.6 m/s^2 as it should if the 3rd law of motion was acting. That means that the object experiences only 9.8 N of force from Earth and this matches the situation before the 3rd law of motion was acting in our scenario.

Does that mean that Newton's 3rd law of motion doesn't apply to gravitational force?
Am I thinking something wrong?

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So, the total force experienced by the Earth and the object by each other is approx. 19.6 N.

I guess you are free to consider that a sum of the magnitudes of the two different forces, but it is unclear to me how that would help you. The forces still act on different bodies. The earth has a force of 9.8N on it, and the object has a force of 9.8N.

We observe that an object of 1 kg accelerates at only 9.8 m/s^2 near Earth's surface and not at 19.6 m/s^2; this means that the force experienced by a 1 kg object is 9.8 N near Earth's surface.

Yes. In fact that's exactly how you began the scenario. We can compute the force on one object and find that to be the magnitude.

You don't get to add up the magnitudes of two forces on two different objects in two different directions and think that magnitude applies to one object in one direction.

Let's assume that Newton's 3rd law of motion isn't acting for now.

That's a tough ask. We're used to forces arising from a coupled interaction. Both sides of the couple experience this interaction as a force.

So, the object and Earth each are experiencing a force of 9.8 N from each other.

That's exactly what we'd expect normally. How does this create a situation where Newton's law "isn't acting"?

Newton's 3rd law doesn't describe some additional force that pops into existence. It just says that if something creates a force (like gravity between two objects), that force is created on both (in opposite directions). That it's not possible to create a "one-way" force.

The gravitational force of 9.8N on both objects is consistent with the 3rd law.

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    $\begingroup$ I would like to put a huge like-thumb 👍 on your last paragraph: "You don't get to add up the magnitudes of two forces on two different objects in two different directions and think that magnitude applies to one object in one direction." $\endgroup$
    – md2perpe
    Nov 25 at 17:58
  • $\begingroup$ I guess you've misunderstood my question. I've made my question clearer. Please check. $\endgroup$
    – silica19
    Nov 26 at 6:47
  • $\begingroup$ I've added a bit. $\endgroup$
    – BowlOfRed
    Nov 26 at 7:44
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    $\begingroup$ "reaction" forces aren't different kinds of forces. They're just regular forces that Newton's 3rd says will always exist. The force is due to gravitation. It also happens to be paired with another gravitational force. You seem to be saying that this is different for gravitation, but it is identical to other situations. If I had a spring pulling two objects, the forces would be equal and opposite but both caused by the spring. One is not a magical "reaction" force. $\endgroup$
    – BowlOfRed
    Nov 26 at 18:10
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    $\begingroup$ That sounds perfect! $\endgroup$
    – BowlOfRed
    Nov 30 at 16:31
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I think your confusion comes from 'double counting' the effect of the third law. Newton's law of gravitation states that all objects attract each other with the force $$F = G \frac{m_1 m_2}{r^2} \, .$$ Because this force is symmetric in $m_1$ and $m_2$, it means that both objects 1 and 2 attract each other with the same force. The gravitational law is therefore consistent with Newton's third law of motion.

The statement of the third law of motion is not to miraculously copy forces from one body to the other, it simply constrains the form of physically possible force laws. For example, the third law tells you that a hypothetical gravitational law, such as "the heavier object attracts the lighter one, but the lighter object does not attract the heavier one" is not physical (even though it appears to be true in day-to-day life).

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  • $\begingroup$ Newton's 3rd law of motion involves an action-reaction pair of forces. What is the action force and the reaction force in this case? $\endgroup$
    – silica19
    Nov 26 at 18:01
  • $\begingroup$ @silica19 That is an arbitrary distinction. You could say that particle 1 attracts particle 2, and that particle 2 "reacts" by attracting particle 1. Or you could view it the other way around. The third law states that both viewpoints are equivalent. $\endgroup$
    – leapsheep
    Nov 26 at 20:50
  • $\begingroup$ I lacked to understand that the two forces involved as action-reaction are not 'action' and 'reaction'; they are both forces acting simultaneously on their own. For example, if 2 balls are colliding, it isn't that 1st ball collides with the 2nd, it is both balls colliding with each other. But I have another question: Imagine 2 persons in deep space in front of each other that push each other simultaneously with 10 N of force using their hands. In this case, what will be the net force experienced by each person? Also, how is this condition different from the gravitational force between objects? $\endgroup$
    – silica19
    Nov 27 at 12:52
  • $\begingroup$ Both of them will feel $10 N$. As soon as they come in contact, both of them will be repelled by the same force, regardless if one or both of them intend to push the other. $\endgroup$
    – leapsheep
    Nov 28 at 20:00
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    $\begingroup$ Yes, I think you got it! $\endgroup$
    – leapsheep
    Dec 1 at 10:16
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9.8 N of force causes a 1 kg object to accelerate at 9.8 m/s2.

But the earth has a mass of about $5.97\times10^{24}\ {\rm kg}$. So 9.8 N causes the Earth to accelerate toward the object at only about $1.64\times10^{-24}\ {\rm m/s^2}$.

Thus the relative acceleration you observe standing on the earth and watching the object fall toward it is only 9.800000000000000000000164 m/s2 (but of course it's not really exactly that because the figure of 9.8 m/s2 was never accurate to so many significant figures).

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There's nothing special about gravity.

The force on a mass $m_1$ at $\vec{r}_1$ due to a mass $m_2$ at $\vec{r}_2$ is $\frac{-Gm_1m_2}{|\vec{r}_1-\vec{r}_2|^3}(\vec{r}_1-\vec{r}_2)$. (The power in the denominator is $3$ rather than $2$, because the vector outside the fraction isn't a unit vector.) Similarly, the force on $m_2$ due to $m_1$ is $\frac{-Gm_2m_1}{|\vec{r}_2-\vec{r}_1|^3}(\vec{r}_2-\vec{r}_1)=\frac{Gm_1m_2}{|\vec{r}_1-\vec{r}_2|^3}(\vec{r}_1-\vec{r}_2)$. This is $-1$ times the former force, as per Newton's third law.

To address your combining-accelerations ambition, let's put gravity aside for a moment. Suppose a body of mass $m_1$ experiences a force $\vec{F}$ due to a body of mass $m_2$, for whatever reason . Then $m_1$ has acceleration $\vec{F}/m_1$. By Newton's third law, $m_2$ experiences a force $-\vec{F}$, giving it the acceleration $-\vec{F}/m_2$. The relative acceleration is $\vec{F}/m_1+\vec{F}/m_2=\vec{F}/\mu$ with $\mu:=\frac{m_1m_2}{m_1+m_2}$ the reduced mass of the two bodies.

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According to the law of gravitation, the object and the Earth will apply a approximate force of 9.8 N to each other.

Yes

...according to Newton's 3rd law of motion, as the object pulls the Earth, the Earth also pulls the object with the same magnitude of force (9.8 N) and vice-versa.

Yes.

So, the total force experienced by the Earth and the object by each other is approx. 19.6 N.

This is not a helpful way to think about the situation; it is not even wrong.

But in actual, this is not the case.

It is the case that each feels a 9.8N force due to the other

We observe that an object of 1 kg accelerates at only 9.8 m/s^2 near Earth's surface and not at 19.6 m/s^2;

Yes. A 9.8N force on a 1kg object leads to a 9.8m/s^2 acceleration.

this means that the force experienced by a 1 kg object is 9.8 N near Earth's surface.

Yes.

Does this mean that Newton's 3rd law of motion is not applicable to gravitational force?

No.

Am I thinking something wrong?

Yes.

The force on the 1kg object is 9.8N. The acceleration of the 1kg object is 9.8m/s^2.

The force on the 5.972 × 10^24 kg earth is 9.8N. The acceleration of the 5.972 × 10^24 kg earth is 1.6744809 x 10^-25 m/s^2 (which is so small you can't notice it).

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  • $\begingroup$ I guess you've misunderstood my question. I've made my question clearer. Please check. $\endgroup$
    – silica19
    Nov 26 at 6:47
  • $\begingroup$ The question has been re-written a bit, but does not seem to have been made any clearer. $\endgroup$
    – hft
    Nov 26 at 20:01
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No body feels both forces. Each body (Earth and object) only feels the force exerted on it and not the force that it itself exerts.

It is a crucial part of Newton's 3rd law to realise that the two forces that make up the force-pair are not exerted on the same body. So there is no issue with the observation you make about the gained acceleration, and Newton's 3rd surely still does apply for gravitational forces as well.

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  • $\begingroup$ I guess you've misunderstood my question. I've made my question clearer. Please check. $\endgroup$
    – silica19
    Nov 26 at 6:48
  • $\begingroup$ @silica19 "Thus, the object now experiences 9.8 N (gravitational force) + 9.8 N (reaction force from the object pulling Earth) = 19.6 N (net force experienced" This sentence is incorrect. When the object pulls in earth via gravity and earth also pulls in the object via gravity, then these two forces are the force action/reaction pair. So there are no further reaction forces to include via Newton's 3rd law. They are already all accounted for. $\endgroup$
    – Steeven
    Nov 26 at 8:11
  • $\begingroup$ When the object pulls in earth via gravity and earth also pulls in the object via gravity, then these two forces are the force action/reaction pair. Why do you call these forces an action-reaction pair when they're the forces generated by those objects themselves. According to the law of gravitation, both objects should pull each other with their own forces. $\endgroup$
    – silica19
    Nov 26 at 9:43
  • $\begingroup$ @silica19 No, according to Newton's law of gravitation a gravitational force is exerted on both objects (it is equal but opposite) due to both objects being present. Not just due to one object. As the formula shows, both masses are involved. Thus, the gravitational force is not an object-specific force but a system-specific force. It only exists between two objects, never just due to one of them. And it acts on both with equal but opposite forces exactly as Newton's 3rs law expects. $\endgroup$
    – Steeven
    Nov 26 at 9:54
  • $\begingroup$ I'm having a hard time understanding how you conclude that the forces applied by the object and Earth on each other are related to Newton's 3rd law of motion. I understand how gravitational force is a system-specific force but I don't understand how these forces act. I'm not able to observe the action and reaction pair in these forces because as you said, it isn't that only Earth pulls the object and the object also pulls the Earth as a reaction force, the forces that they apply on each other are gravitational forces. Please explain how are they related to Newton's 3rd law of motion. $\endgroup$
    – silica19
    Nov 26 at 12:30

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