Answer:
[tex]\Delta t = 3.95 \times 10^{18} seconds[/tex]
Explanation:
Torque due to applied force along its surface is given as
[tex]\tau = r \times F[/tex]
here we know that
r = radius of earth = [tex]6.37 \times 10^6 m[/tex]
Now we have
[tex]\tau = (6.37 \times 10^6)(4.00 \times 10^7)[/tex]
[tex]\tau = 2.548 \times 10^{14} Nm[/tex]
now we know that
initial angular speed of Earth is
[tex]\omega_i = \frac{2\pi}{24\times 3600}[/tex]
final angular speed will be
[tex]\omega_f = \frac{2\pi}{28\times 3600}[/tex]
so now we have
[tex]\tau \Delta t = I(\omega_i - \omega_f)[/tex]
here we have moment of inertia of Earth is given as
[tex]I = \frac{2}{5} mR^2[/tex]
[tex]I = \frac{2}{5}(5.98 \times 10^{24})(6.37 \times 10^6)^2[/tex]
[tex]I = 9.7 \times 10^{37} kg m^2[/tex]
now we have
[tex](2.548 \times 10^{14})\Delta t = (9.7 \times 10^{37})(\frac{2\pi}{24\times 3600} - \frac{2\pi}{28\times 3600})[/tex]
[tex]\Delta t = 3.95 \times 10^{18} seconds[/tex]
