I tried this question on the Physics SE in November and didn't get a response. Let's see how it does over here!

My copy of Feynman's "Six Not-So-Easy Pieces" has an interesting introduction by Roger Penrose. In that introduction (copyright 1997 according to the copyright page), Penrose complains that Feynman's "simplified account of the Einstein field equation of general relativity did need a qualification that he did not quite give." Feynman's intuitive discussion rests on relating the "radius excess" of a sphere to a constant times the enclosed gravitational mass $M$: for a sphere of measured radius $r_{\mathrm{meas}}$ and surface area $A$ enclosing matter with average mass density $\rho$ smoothly distributed throughout the sphere,
$$\sqrt{\frac{A}{4π}}−r_{\mathrm{meas}}=\frac{G}{3c^2}\cdot M,$$
wherein $G$ is Newton's gravitational constant, $c$ is the speed of light in vacuum, and $M=4π\rho r^3/3$. I don't know what $r$ is supposed to be, but it's presumably $\sqrt{\frac{A}{4\pi}}$. Feynman gratifyingly points out that $\frac{G}{3c^2}\approx 2.5\times 10^{−28}$ meters per kilogram (for Earth, this corresponds to a radius excess of about $1.5$ mm). Feynman is also careful to point out that this is a statement about average curvature.

Penrose's criticism is: "the 'active' mass which is the source of gravity is not simply the same as the energy (according to Einstein's $E=mc^2$); instead, this source is the energy plus the sum of the pressures". Damned if I know what that means -- whose pressure on what?

So, taking into account Penrose's criticism but maintaining Feynman's intuitive style, what is the active mass $M$?

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