Question

Calculate the moment of inertia of earth about its diameter. Consider the earth to be a sphere of radius 6400 km and mass ${10}^{25}kg$

• A. $1.64\times {10}^{38}kg{m}^2$
• B. $1.74\times {10}^{38}kg{m}^2$
• C. $1.84\times {10}^{38}kg{m}^2$
• D. $1.94\times {10}^{38}kg{m}^2$

Answer 1

The correct option is A $1.64\times {10}^{38}kg{m}^2$

$\begin{array}{c}\text{Given, mass of earth}\left(m\right)={10}^{25}kg\text{.}\\ \text{and, radius of earth}\left(r\right)=6400km=6.4\times {10}^{6}m\text{.}\\ \text{Thus, moment of inertia of earth above its diameter}=\frac{2}{5}{mr}^2\end{array}$

$\begin{array}{cc}& =\frac{2}{5}\times {10}^{25}\times {(6.4\times {10}^6)}^{2}kg-m^2\\ =& 16.384\times {10}^{37}kg-m^2\\ =& 1.64\times {10}^{38}{kgm}^2\end{array}$