[Total: 3 Average: 5/5] You must sign in to vote

This question has popped up many times. So here is an attempt to address it.

To answer this question at the **elementary** level, a number of assumption will be made, which will become obvious later on. Still, at this point, we will simply deal with objects with spherical symmetry and no complicated mass distribution. In other words, the mass of each object can be considered to be at their respective center of mass.

For two objects with masses, the gravitational force on mass 1 due to the gravitational force of mass 2 can be written as

F_{12} = \frac{Gm_1 m_2}{h^2}

where m_1, m_2 are the masses of object 1 and 2 respectively, G is the **universal gravitational constant**, and h is the distance between the two masses. Of course, this equation is symmetric, and the force on 2 due to 1 also has the identical form.

Let m_1 be any mass and m_2 = M, the mass of the earth. Then the force on object 1 due to the earth is

F_{1M} = \frac{Gm_1 M}{h^2}

But since this is the force on m_1, we can equate the dynamics of this object with

F_{1M} = \frac{Gm_1 M}{h^2} = m_1 a

Simplifying, we get

\frac{GM}{h^2} = a

which we define to be “g” at the surface of the earth, and is a constant if we always put different masses at the same location.

This means that no matter what m_1 is, the acceleration is always a constant g. It shows that the acceleration due to gravity is a constant.

Now, in deriving this, I’ve made an implicit assumption of one thing, that “M” really doesn’t get affected much by m_1. Recall that I said earlier that the gravitational force equation is symmetric. One can also talk about the force on the earth due to the mass m_1. So that mass certainly exert a force on M. However, this is where we have to consider a little bit about the center of mass of the M + m_1 system.

r_{cm} = \frac{m_1 r_1 + MR}{m_1 + M}

where r_1 is the location of the center of mass of m_1, and R is the location of the center of mass of M, i.e. the earth. If we put our origin at the center of mass of the earth, then R = 0. So let’s do that. The center of mass location is then simplified to

r_{cm} = \frac{m_1 r_1}{m_1 + M}

Now, for ordinary, terrestrial objects, the kind that Galileo (and you and I) would test in our intro physics classes, the mass of the object m_1 is much, much smaller than the mass of the earth, i.e. m_1 < M. In that case, we can simplify the center of mass equation to

r_{cm} = \frac{m_1 r_1}{M}

It means that the location of the center of mass of ordinary object + Earth system, for an object placed near or on the surface of the earth (i.e. r_1 ~ R_E), depends predominately on the ratio of m_1/M.

An ordinary object weights in the range of kg. The earth weighs on the order of 10^{24} kg, and the radius of the earth is of the order of 10^6m.

If you do that computation, you’ll get a very, very, small number for r_{cm}. It means that the center of mass of this system is practically the same as the center of mass of the Earth, which is situated at r=0.

What I’m trying to show here is in this scenario, the earth does not make any significant motion due to the force exerted by m_1. This means that for this case, it is perfectly fine to consider the earth as the fixed object, and only consider that it is the smaller mass that falls towards the earth. So when you consider that, then all the simplification that is done to allow us to deduce that the acceleration due to gravity of ALL objects falling on the surface of the earth to be a constant, independent of the mass of the object.