Problem — Angular momentum of Earth's spin

Problem

Estimate the angular momentum of Earth’s spin ($M\approx 6\cdot 10^{24}\;\mathrm{kg}$, $R\approx 6{,}4\cdot 10^{6}\;\mathrm{m}$), treating it as a uniform sphere with $I = \tfrac{2}{5}MR^2$.

Solution

The spin occurs about the Earth’s axis with period $T = 1\;\text{day} \approx 8{,}64\cdot 10^4\;\mathrm{s}$, so $\omega = \dfrac{2\pi}{T} \approx 7{,}3\cdot 10^{-5}\;\mathrm{rad/s}$. $I = \tfrac{2}{5}MR^2 = \tfrac{2}{5}\cdot 6\cdot 10^{24}\cdot (6{,}4\cdot 10^{6})^2 \approx 9{,}8\cdot 10^{37}\;\mathrm{kg\,m^2}$ $L = I\,\omega = 9{,}8\cdot 10^{37}\cdot 7{,}3\cdot 10^{-5}$ $\ev{L \sim 7\cdot 10^{33}\;\mathrm{kg\,m^2/s}}$ This is about ten thousand times smaller than Earth’s orbital angular momentum ($\sim 3\cdot 10^{40}$): Earth “weighs” much more in its orbit around the Sun than in its spin about itself.

Links

Topics: Rotational dynamics Concepts: Angular momentum · Moment of inertia Skills: Fermi estimation