# What is the moment of inertia of the earth?

## Given that the radius of the earth is $6.38 \times {10}^{6} \text{m}$ and the mass of the earth is $5.98 \times {10}^{24} \text{kg}$.

Aug 15, 2017

$I \approx 9.736 \times {10}^{37} {\text{kgm}}^{2}$

#### Explanation:

If we think of the earth as a solid sphere rotating about its center, the moment of inertia is given by:

$\textcolor{\mathrm{da} r k b l u e}{I = \frac{2}{5} M {R}^{2}}$

where $M$ is the mass of the earth and $R$ is its radius

We are given the following information:

• $\mapsto M = 5.98 \times {10}^{24} \text{kg}$
• $\mapsto R = 6.38 \times {10}^{6} \text{m}$

Substituting these values into the equation above:

$I = \frac{2}{5} {\left(5.98 \times {10}^{24} \text{kg")(6.38xx10^6"m}\right)}^{2}$

$\implies = \frac{2}{5} \left(5.98 \times {10}^{24} {\text{kg")(4.07044xx10^13"m}}^{2}\right)$

$\implies \frac{2}{5} \left(2.4341 \times {10}^{38} {\text{kgm}}^{2}\right)$

$\implies \textcolor{\mathrm{da} r k b l u e}{= 9.736 \times {10}^{37} {\text{kgm}}^{2}}$