Confusion Regarding Reaction Forces and Newton's Third Law of Motion

Question

So I was confused about reaction forces for a LONG TIME and I have finally been able to clear some of my confusions, but the concept is still not totally clear to me. Can someone confirm whether my above conclusions are correct or not, and explain my mistakes if there are any?

Suppose a heavier object $A$(with mass $m_a$) capable of applying force by itself is applying a force $F_a$ on a lighter object $B$(with mass $m_b$). So, the equal and opposite reaction force $F_b$ provided by the body $B$ is acting on the body $A$.

Now, since $$F_a=F_b$$ And $$m_a>m_b$$ And considering acceleration as $a$, $$F=ma$$ We can see that $$a_a<a_b$$

Now if $B$ applies a force on $A$ with force $F_b$, then similarly, $a_a<a_b$, and sometimes the acceleration $a_a$ of the heavier body is so less that it is considered $≈0$, i.e., $lim_{a_a→0⁺}(a_a)$ as in the case of us applying a force on the Earth, or pushing a wall(I understood this from the answers under a similar question which I had asked earlier).

Now consider the situation where a guy(with mass $m_p$) is standing on the earth(with mass $m_e$), and $m_e>>m_p$. The person applies a force $mg$ on earth, and earth applies and equal and opposite force $mg$ on the guy too.

Now since both apply the same force on each other,

$$a_p>a_e$$ which is absolutely understandable. At first I thought that why don't we just get launched into the air when we stand, but then I quickly realized that its because we DO get launched into the air with some acceleration when the force is instantaneous, e.g., when we jump, or when we drop a body on the Earth's surface from some height. But this got me confused again, considering the fact that the dropped body applies a force $mg$ on the Earth, which applies and equal force on the body, so why doesn't the body move up with an equal velocity

But now I am confused about the physics behind a force which is acting constantly on a body, e.g., any body or person simply standing on the Earth's surface.

My official question is:

Is it that as soon as the body acquires an acceleration upwards, it applies the same acceleration downwards and so the net acceleration is $0$? And the case is not the same when we apply a force instantaneously? Considering the fact that the dropped body applies a force $mg$ on the Earth, which applies and equal force on the body, so why doesn't the body move up with an equal acceleration? Is it because of energy loss as heat, sound, etc., when the object makes contact with the ground, it applies less force than $mg$, so the reaction force is also $<mg$, and hence less acceleration? If that's true, then does every object dropped on the Earth's surface technically bounce, but the surface and material of the object and the surface governs the amount of energy lost, which causes the difference in force applied and in turn, upward acceleration?

Answer 1

Take the case of a person standing on the Earth.

A downward force $F_d=m_pg$ acts on him and thus on the Earth.

However, $N3L$ dictates that the Earth exerts a reaction force on the person, of equal magnitude (scalar) but opposite direction: $F_{u}=-m_p g$.

The net force (at least in the vertical direction, $y$) acting on the person is:

$$\sum F_y=m_pg-m_p g=0$$

So by $N2L$, the acceleration (in the $y$ direction) is zero.

Considering the fact that the dropped body applies a force $mg$ on the Earth, which applies an equal force on the body, so why doesn't the body move up with an equal acceleration?

The Earth doesn't experience a force, the falling body applies an impulse to the Earth, which is very different from a force.