Question

The polar ice caps contain about 2.3 × 1019 kg of ice. This contributes essentially nothing to the moment of inertia of the earth because it is located at the poles, close to the axis of rotation. Estimate the change in the length of the day to be expected if the polar ice caps melt, distributing the water uniformly over the surface of the earth. Take mass of earth = 5.98 × 1024 kg and radius of earth = 6.37 × l06 m.

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{The}\mathrm{change}\mathrm{in}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{a}\mathrm{day}, \\ ∆\mathrm{T}={\mathrm{T}}_{\mathrm{after}}-{\mathrm{T}}_{\mathrm{before}}......\left(1\right) \\ \mathrm{Because}\mathrm{the}\mathrm{net}\mathrm{torque}\mathrm{acting}\mathrm{on}\mathrm{the}\mathrm{earth}\mathrm{during}\mathrm{this}\mathrm{process}\mathrm{is}\mathrm{zero}, \\ \mathrm{angular}\mathrm{momentum}\mathrm{is}\mathrm{conserved}. \\ ∆\mathrm{L}={\mathrm{L}}_{\mathrm{after}}-{\mathrm{L}}_{\mathrm{before}}=0 \\ \mathrm{Substituting}\mathrm{for}{\mathrm{L}}_{\mathrm{after}}\mathrm{and}{\mathrm{L}}_{\mathrm{before}}\mathrm{yeilds}, \\ \left({\mathrm{I}}_{\mathrm{sphere}}+{\mathrm{I}}_{\mathrm{shell}}\right){\mathrm{\omega }}_{\mathrm{after}}-{\mathrm{I}}_{\mathrm{sphere}}{\mathrm{\omega }}_{\mathrm{before}}=0 \\ \left({\mathrm{I}}_{\mathrm{sphere}}+{\mathrm{I}}_{\mathrm{shell}}\right)\frac{2\mathrm{\pi }}{{\mathrm{T}}_{\mathrm{after}}}-{\mathrm{I}}_{\mathrm{sphere}}\frac{2\mathrm{\pi }}{{\mathrm{T}}_{\mathrm{before}}}=0..........(\mathrm{because}\mathrm{\omega }=\frac{2\mathrm{\pi }}{\mathrm{T}}) \\ \frac{\left({\mathrm{I}}_{\mathrm{sphere}}+{\mathrm{I}}_{\mathrm{shell}}\right)}{{\mathrm{T}}_{\mathrm{after}}}-\frac{{\mathrm{I}}_{\mathrm{sphere}}}{{\mathrm{T}}_{\mathrm{before}}}=0 \\ {\mathrm{T}}_{\mathrm{after}}=\left(1+\frac{{\mathrm{I}}_{\mathrm{shell}}}{{\mathrm{I}}_{\mathrm{sphere}}}\right){\mathrm{T}}_{\mathrm{before}} \\ \mathrm{Sustitute}\mathrm{this}\mathrm{in}\mathrm{equation}\left(1\right), \\ ∆\mathrm{T}=\left(1+\frac{{\mathrm{I}}_{\mathrm{shell}}}{{\mathrm{I}}_{\mathrm{sphere}}}\right){\mathrm{T}}_{\mathrm{before}}-{\mathrm{T}}_{\mathrm{before}} \\ ∆\mathrm{T}=\left(\frac{{\mathrm{I}}_{\mathrm{shell}}}{{\mathrm{I}}_{\mathrm{sphere}}}\right){\mathrm{T}}_{\mathrm{before}} \\ \mathrm{where},{\mathrm{I}}_{\mathrm{shell}}=\mathrm{moment}\mathrm{of}\mathrm{inertia}\mathrm{of}\mathrm{shell} \\ {\mathrm{I}}_{\mathrm{sphere}}=\mathrm{moment}\mathrm{of}\mathrm{inertia}\mathrm{of}\mathrm{sphere} \\ \mathrm{Substitute}\mathrm{the}\mathrm{value}\mathrm{of}{\mathrm{I}}_{\mathrm{shell}}\mathrm{and}{\mathrm{I}}_{\mathrm{sphere}} \\ ∆\mathrm{T}=\left(\frac{{\displaystyle \frac{2}{3}}{\mathrm{mr}}^2}{{\displaystyle \frac{2}{5}}{\mathrm{M}}_{\mathrm{e}}{{\mathrm{R}}_{\mathrm{e}}}^2}\right){\mathrm{T}}_{\mathrm{before}}=\frac{5\mathrm{m}}{3{\mathrm{M}}_{\mathrm{e}}}{\mathrm{T}}_{\mathrm{before}} \\ \mathrm{Substitute}\mathrm{the}\mathrm{values}, \\ ∆\mathrm{T}=\frac{5\left(2.3\times {10}^{19}\right)}{3\left(5.98\times {10}^{24}\right)}\left(24\times 3600\right) \\ \mathrm{where},{\mathrm{T}}_{\mathrm{before}}=\mathrm{total}\mathrm{time}\mathrm{in}\mathrm{a}\mathrm{day}\mathrm{in}\mathrm{seconds} \\ ∆\mathrm{T}=0.55\mathrm{s} \end{gathered}$$