Answer
$2.6\times10^{19} \ Nm$
Work Step by Step
a) Using the radius of the earth and the equation for the moment of inertia of a solid sphere, we obtain:
$ I =\frac{2}{5}mr^2 = \frac{2}{5}(5.97\times10^{24})(6.37\times10^6)^2=9.69\times10^{27}$
b) Using unit conversions, we know that this rotation corresponds to an angular acceleration of $2.68\times10^{-19} \ rads/s^2$
Thus, we find the torque using the rotational version of Newton's second law:
$\tau = I\alpha=( 9.69\times10^{27})(2.68\times10^{-19} \ rads/s^2)=2.6\times10^{19} \ Nm$
