Moment of Inertia: Solid Sphere

A solid sphere about an axis through its centre has $I = \tfrac{2}{5}mr^2$, and the coefficient of 0.4 sits below the disk's 0.5 for a straightforward geometric reason. In a disk, every piece of mass lies in one plane at its full distance from the axis. In a sphere, the material near the poles is close to the axis even though it is far from the centre, so a larger share of the mass sits near the line it is turning about, and the average $r^2$ comes out lower.

A billiard ball of 0.17 kg and 28.5 mm radius has $I = 0.4 \times 0.17 \times 0.0285^2 \approx 5.5\times10^{-5}$ kg·m². The same expression scales to a planet: Earth's mass is $5.97\times10^{24}$ kg and its radius $6.37\times10^6$ m, so a uniform Earth would have $I = 0.4MR^2 \approx 9.7\times10^{37}$ kg·m².

The measured value is about $8.0\times10^{37}$, which is $0.33MR^2$ rather than $0.40MR^2$, and that shortfall is one of the more elegant pieces of evidence in geophysics. A coefficient below the uniform-sphere value means the mass is concentrated toward the centre — the planet has a dense core. The same measurement, made from a spacecraft's tracking data, is how the internal structure of the Moon, Mars and the outer planets' moons is inferred without ever landing on them. Jupiter comes in near 0.25, more centrally concentrated still.

The mistake is applying this to a ball that is hollow. A thin spherical shell — a ping-pong ball, a basketball, an empty tank — has all its mass at the full radius and its moment of inertia is $\tfrac{2}{3}mr^2$, which is 0.667, not 0.4. That is a 67% difference, and it shows up immediately in a rolling race: a solid ball beats a hollow one down a ramp every time, because less of the available energy is diverted into spinning it. Anything with a thick wall lies between the two and needs the full $\tfrac{2}{5}m(r_2^5-r_1^5)/(r_2^3-r_1^3)$ treatment. Two other checks: the axis must pass through the centre — a sphere swung on the end of a rod is a parallel-axis problem, $I = \tfrac{2}{5}mr^2 + md^2$, and for a long rod the $md^2$ term dominates completely — and the sphere must be uniform, which is exactly what the Earth example demonstrates is often not the case.