Question

Find the moment of inertia of earth assuming that it is a sphere of radius 6400 km and
uniform density 5520 kg/m3
about an axis of rotation passing through its centre.

Answer 1

Moment of inertia is $\bold{1.38 \times 10^{32} kg - m^2}$

Explanation:

Given,

Radius of sphere , R = 6400 km = $6.4\times 10^6$ m

Density,$\rho$ = 5520$kg/m^3$

Moment of inertia,

we are taking earth as a sphere, so moment of inertia of the sphere

I = $\frac{2}{3} \times M \times r^2$

∵ $M = volume \times {density}$

$M = \frac{4}{3} \times 3.14 \times (6.4\times 10^6)^3 \times 5520$

$M = 5.09\times 10^{18}$ kg

Then,

$I = \frac{2}{3}\times 5.09\times 10^{18} \times (6.4\times 10^6)^2$$kg -m^2$

$I = 1.38 \times 10^{32} kg - m^2$