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The Eötvös Experiment

Is there any way to test Newton's conjecture that "inertial mass" (the quantitative measure of an objects resistance to acceleration by an applied force) is different from "gravitational mass" (the factor determining the weight of said object)? Certainly. But first we must make the proposition more explicit:

• Inertial mass $m_{\scriptscriptstyle I}$ is an additive property of matter. That is, two identical objects, when combined, will have twice the inertial mass of either one by itself.⁵
• When subjected to a given force $\vec{\bf F}$  [a vector quantity, since it certainly has both magnitude and direction], an object will be accelerated in the direction of $\vec{\bf F}$ at a rate $\vec{\bf a}$ which is inversely proportional⁴ to its inertial mass $m_{\scriptscriptstyle I}$. Mathematically,
  

  $$\vec{\bf a} \; \propto \; {\vec{\bf F} \over m_{\scriptscriptstyle I}}.$$ (1)

  
  

• Gravitational mass $m_{\scriptscriptstyle G}$ is also an additive property of matter.
• The force of gravity $\vec{\bf W}$ pulling an object "down" toward the centre of the Earth (i.e. its weight) is proportional to its gravitational mass $m_{\scriptscriptstyle G}$. Let's write the constant of proportionality "g" so that $W = g \, m_{\scriptscriptstyle G}$ (where $W \equiv \vert\vec{\bf W}\vert$ is the magnitude of the weight, which is usually all we need, knowing as we do which way is "down") - or, in full vector notation,
  

  $$\vec{\bf W} = - g \, m_{\scriptscriptstyle G} \, \hat{\bf r}$$ (2)

  
  
  (where $\hat{\bf r}$ is the unit vector pointing from the centre of the Earth to the object in question).

The combination of the last two postulates is easy to check using a simple balance. However, it is not so easy to separately check these two propositions. See why? Fortunately, we don't have to.

If we put together the two equations $\vec{\bf a} \; \propto \; \vec{\bf F}/m_{\scriptscriptstyle I}$ and $\vec{\bf W} = - g \, m_{\scriptscriptstyle G} \, \hat{\bf r}, \;$ noting that, in the case of the force of gravity itself, $\vec{\bf F} \equiv \vec{\bf W}$, we get

$$\vec{\bf a} \; \propto \; - \hat{\bf r} \, g \, {m_{\scriptscriptstyle G} \over m_{\scriptscriptstyle I}}$$ (3)

- i.e. the acceleration due to gravity is in the $-\hat{\bf r}$ direction (towards the centre of the Earth), and is proportional to the ratio of the gravitational mass to the inertial mass. So... if the gravitational mass is proportional to the inertial mass, then all objects should experience the same acceleration when falling due to the force of gravity, at least in the absence of any other forces like air friction. Wait! Isn't this just what Galileo was always trying to tell us? Yep. But was he right?

Clearly the answer hangs on the proportionality of $m_{\scriptscriptstyle G}$ and $m_{\scriptscriptstyle I}$. As we shall see, any nontrivial constant of proportionality can be absorbed into the definition of the units of force; thus instead of $\vec{\bf a} \; \propto \; \vec{\bf F}/m_{\scriptscriptstyle I}$ we can write $\vec{\bf a} = \vec{\bf F}/m_{\scriptscriptstyle I}$ and the question becomes, "Are inertial mass and gravitational mass the same thing?" The experimental test is of course to actually drop a variety of objects in an evacuated chamber where there truly is no air friction (nor, we hope, any other more subtle types of friction) and measure their accelerations as accurately as possible. This was done by Eötvös to an advertised accuracy of 10^-9 (one part per billion - often written 1 ppb) who found satisfactory agreement with Galileo's "law."⁶ Henceforth I will therefore drop the _G and _I subscripts on mass and assume there is only one kind, mass, which I will write m.