# Force between a second and third body is such that they are rigidly bound together to form a composite body

I am currently studying the textbook Classical Mechanics, fifth edition, by Kibble and Berkshire.

In classical mechanics, we have that

$$m_1 \mathbf{a}_1 + m_2 \mathbf{a}_2 + m_3 \mathbf{a}_3 = \mathbf{0}$$

If we suppose that the force between a second and third body is such that they are rigidly bound together to form a composite body, their accelerations must be equal: $$\mathbf{a}_2 = \mathbf{a}_3$$. In that case, we get

$$m_1 \mathbf{a}_1 = -(m_2 + m_3) \mathbf{a}_2,$$

which shows that the mass of the composite body is just $$m_{23} = m_2 + m_3$$.

So let's say that $$m_{23}$$ is the mass of a human, and $$m_1$$ is the mass of the Earth. We get

$$(\text{Earth})\mathbf{a}_1 = -(\text{human})\mathbf{a}_2.$$

If I am interpreting this correctly, since masses are always positive, this means that we humans are decelerating towards the Earth at a deceleration of $$\mathbf{a}_2$$ and the Earth is accelerating towards us humans at an acceleration of $$\mathbf{a_1}$$ (or, equivalently, the Earth is decelerating towards at a deceleration of $$\mathbf{a}_1$$, and we humans are accelerating towards the Earth at an acceleration of $$\mathbf{a}_2$$)?

But this then raises another question: If we know the masses of the two bodies, then assuming that we don't already know either of the accelerations, how is it possible for us to then calculate the accelerations $$\mathbf{a}_1$$ and $$\mathbf{a}_2$$? And, because of relativity, if everything is relative (there is no "absolute measuring stick"), how is it even possible for us to ever calculate such acceleration values? (I suspect that the issue here is that I'm not fully understanding the concept of an inertial frame of reference?)

These are just some "novice" thoughts I had whilst studying the material. I would greatly appreciate it if people would please take the time to clarify this.

Deceleration is a word I urge you to avoid using in a physics context. Acceleration is a vector quantity, so I would describe $$M_{E}\mathbf a_1= -m_H\mathbf a_2$$ as saying that the human has acceleration $$\mathbf a_2$$ and the Earth has acceleration $$\mathbf a_1 = -\frac{m_H}{M_E}\mathbf a_2$$. Equivalently, you could say that the Earth has acceleration $$\mathbf a_1$$ and the human has acceleration $$\mathbf a_2 = -\frac{M_E}{m_h}\mathbf a_1$$. The relative minus sign between the two accelerations indicates that they are in opposite directions.

If we know the masses of the two bodies, then assuming that we don't already know either of the accelerations, how is it possible for us to then calculate the accelerations $$\mathbf a_1$$ and $$\mathbf a_2$$?

That's what Newton's 2nd law does. If we know the forces acting on the two objects, then we can calculate their accelerations. In this particular case, imagine that the human is somewhere above the surface of the Earth (in, say, the $$+\hat z$$ direction). Newton's law of gravitation tells us that the force on the person is

$$\mathbf F = G\frac{M_E m_h}{r^2} (-\hat z)$$

where $$r$$ is the distance between the human and the center of the Earth, and $$-\hat r$$ indicates that the force is directed downward (toward the center of the Earth). If this is the only force acting on the human, then since $$\mathbf F = m\mathbf a$$, we have that the human would accelerate with acceleration given by

$$\mathbf a_2 = G\frac{M_E}{r^2}(-\hat z)$$

and the Earth would have acceleration $$\mathbf a_1 = G\frac{m_H}{r^2} \hat z$$ i.e. upward, toward the human.

And, because of relativity, if everything is relative (there is no "absolute measuring stick"), how is it even possible for us to ever calculate such acceleration values?

Velocity is relative, but acceleration is not. This particular example bumps up against the equivalence principle (which roughly says that free-fall in gravitational field is locally indistinguishable from non-accelerated motion in free space), but generally speaking, two observers with constant relative velocities must observe the same physics, but two observers with relative accelerations need (and generically will) not.