# What is the angular momentum of earth?

May 8, 2018

The angular momentum due to the earth's rotation is $\approx 7.2 \times {10}^{33} \setminus {\text{Kg"\ "m"^2"s}}^{-} 1$
(this value is with respect to a co-moving observer)

#### Explanation:

We can estimate the angular momentum due to the earth's rotation by approximating the earth by a uniform sphere of

• mass $M = 6.0 \times {10}^{24} \setminus \text{Kg}$ and
• radius $R = 6.4 \times {10}^{6} \setminus \text{m}$

The moment of inertia of a uniform solid sphere about any axis passing through the center is

$I = \frac{2}{5} M {R}^{2}$

and so, for the earth it is

$I = \frac{2}{5} \times 6.0 \times {10}^{24} \times {\left(6.4 \times {10}^{6}\right)}^{2} \setminus {\text{Kg"\ "m}}^{2}$
$\quad = \approx 9.8 \times {10}^{37} \setminus {\text{Kg"\ "m}}^{2}$

The earth's angular velocity is

omega = (2 pi)/(1\ "day") = (2pi)/(24times 60times 60)\ "s"^-1~~7.3times 10^-5\ "s"^-1

So, the angular momentum of the earth's rotation (with respect to an observer co-moving with it) is

$L = I \omega \approx 7.2 \times {10}^{33} \setminus {\text{Kg"\ "m"^2"s}}^{-} 1$

Note that

• the angular momentum due to the revolution of the earth (with respect to the sun)is much larger than this.
• since the earth actually has a dense inner core, the actual moment of inertia is smaller than that estimated here.