Given Newton's Third law, why don't we have to include acceleration or energy lost to Earth in calculations?

Question

This is probably an elementary question, but I am trying to help my nephew answer a worry. His question is: given Newton's third law, when a ball is thrown up in the air, we would push the earth a little downward. But whatever acceleration imparted and distance traveled during the throw the earth moved, why don't we have to take this into consideration?

I know from experience it is because we assume the earth didn't move. But I want to give him a better reason. I think the following, can someone confirm it is right we can assume it doesn't move because:

1. There are thousands of things happening at once, therefore all these actions/reactions tend to cancel out.
2. Earth is too massive to make a difference.
3. Our force might only compress part of earth, not earth as a whole.
4. Even if it did move or accelerate when the ball was thrown, when they collide as it comes back to ground, technically each would stop one another.

Therefore, we can safely ignore in all practical problems. This correct? Darn it, should have kept my highschool physics book given all this remote learning.

Answer 1

I think all of your practical considerations are correct. However, we can seriously simplify the situation. Let's imagine a perfectly rigid earth and ball (all parts move together). Then to throw a ball in the air requires a force, which will be applied equally to the earth, as in Newton's third law.

This force is capable of launching the ball a metre or so into the air because it has a low mass. The earth on the other hand has an incredibly large mass! Nevertheless the earth will move in the opposite direction but by a tiny amount because of the large mass and the small force. This movement would be imperceptible. Since $F = ma$, the relative difference in the acceleration felt by each is in the ratio of the masses. Assuming a 1kg ball, the relative acceleration of the earth will be ~$5\times 10^{24}$ times smaller!!

Now consider the point where the ball is in the air and about to fall back to earth. There is a gravitational pull between the ball an earth, which again are equal and opposite. As the ball falls back to earth, the earth will also fall towards the ball (but again by a ridiculously small amount).

Even in this simple model you can see that the effect on the earth of someone throwing a ball is not measurable and so we can consider the earth as immovable.

Answer 2

In my view, all those reasons are not suffice.

"There are thousands of things happening at once, therefore all these actions/reactions tend to cancel out" and "Our force might only compress part of earth, not earth as a whole“

For above two reasons I would like to point out that even if you stop all those things happening all around the Earth and consider Earth to be perfectly rigid, then also you may not consider the Earth's acceleration.

"Earth is too massive to make a difference"

It is the point - Earth is so massive. Even if throwing the ball gives the Earth velocity of 10^-20 m/s, then it's momentum would be of order 10^5 kg-m/'s and it's Kinetic Energy would be even much more-of order 10^35 Joules . So that is surely going to make a difference.

For the forth reason, you must think a bit, it is not so logical why it should be the reason for not considering Earth's acceleration. I am not demotivating you, just trying to point out where the things went wrong.

What I think is the true reason is the choice of your system. If we choose our ball as our system, then we consider only the energy (and all other quantities) of the ball. There is no need( or you can say it is the condition for newton's laws to work. Normally we take the ball as the system

Now, if instead we take (the ball+Earth) as our system, then we must take into account Earth's acceleration. Hey! Earth moved when you throwed the ball up, such that the centre of mass of the (Earth+ball) system remains at rest.