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Question on Gravity and the old "all objects free fall at the same rate regardless of their mass"

I "drop" three objects, one at a time, towards the earth

  • First object has less mass than the earth

  • Second object has mass equal to the earth

  • Third object has greater mass than the earth

in all cases as Gravity "pulls" the two together. The greater mass object wins this simple "tug of war" due to inertia and is moved relatively less. The lighter mass object moves relatively more. You could plot a curve (heavier moves less) with the object of equal mass at center.

So what am I missing in the semantics of "all objects free fall at the same rate regardless of their mass" is popularly taught and I think the average person understands they are being told?

AccidentalFourierTransform
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Hume Bates
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2 Answers2

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The word "fall" is hiding some assumptions in your statement. You have some object with position $x$, while surface of the Earth sits at position $X$. Does "the rate of falling" mean $\frac{d^2 x}{dt^2}$ or $\frac{d^2(x-X)}{dt^2}$? The first is the rate of acceleration of the object through an inertial reference frame. The second is the acceleration of the separation of Earth and the object. The statement is fully correct in the first case but breaks down for large objects in the second case.

The first clearly does not depend on the mass of the object: $\ddot{x} = -\frac{1}{m}\frac{GMm}{(x-X)^2}$. But the second does! $$ \ddot{x}-\ddot{X} = -\frac{GMm}{(x-X)^2}\left(\frac{1}{m}+\frac{1}{M}\right)=-\frac{G(m+M)}{(x-X)^2}$$

Of course, for $M\gg m$, the mass of the object is negligible, which is why we usually ignore this issue and just say "everything falls at the same rate." And since most people are only going to imagine sufficiently small objects being dropped I think it is not very misleading to say.

Note that this happens because when you drop the object the Earth also accelerates so the surface of the Earth is no longer an inertial reference frame.

Luke Pritchett
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Let's do the math "properly". Assume there are two point masses, $m_1$ and $m_2$. The question "how quickly do they fall" can be rephrased as "how long is it until the distance between them is reduced from $D$ to $d$, where $d<D$ is the distance where they touch (I am using point masses - but obviously they are of finite size so they will hit when their centers are still some distance apart).

This is a problem that is most easily solved in the center of mass frame. The two points will move towards the center of mass with an acceleration that is inversely proportional to their mass (the heavier object will accelerate less). The distance to the center of mass is also proportional to the inverse of the mass.

Distance that object 1 has to fall, $d_1 = \frac{D m_2}{m_1+m_2}$; and $d_2 = \frac{Dm_1}{m_1+m_2}$. The total acceleration (of the distance between them) can be computed as the sum of the accelerations of $m_1$ and $m_2$ at a given moment in time. Now since they experience the same force, we can say

$$\frac{d^2x_1}{dt^2} = \frac{F}{m_1}\\ \frac{d^2x_2}{dt^2} = \frac{F}{m_2}$$

And their sum

$$a = F\frac{m_1+m_2}{m_1 m_2}$$

When $F=\frac{Gm_1 m_2}{r^2}$ this tells us that the rate at which the masses accelerate towards each other is given by

$$a = \frac{G(m_1+m_2)}{R^2}$$

As long as $m_1>>m_2$, this means the acceleration depends only on the distance; but when that assumption is no longer true, the masses do "fall towards each other", and we conclude that "really heavy things fall faster" (because the earth will "fall up" to meet them).

Another way of looking at this: the "effective $g$" (let's call it $g'$) when the mass of the object is $m$, falling towards an object with mass $M$, is $$g'=\frac{M+m}{M} g$$

Floris
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