Is Newton's 3rd law of motion not applicable to gravitational force?

Question

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.

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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.

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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?

Answer 1

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.

Answer 2

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).

Answer 3

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).

Answer 4

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.

Answer 5

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).

Answer 6

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.