Zorch, an archenemy of superman, decides to slow earth's rotation to once per 27.4 h by exerting an opposing force at and parallel to the equator. superman is not immediately concerned, because he knows zorch can only exert a force of 3.98 ✕ 107 n (a little greater than a saturn v rocket's thrust). how long in seconds must zorch push with this force to accomplish his goal? (this period gives superman time to devote to other villains.)

If we assume that Earth is a solid sphere, then I = (2/5)mr² = (2/5) * 5.98e24kg * (6.371e6m)² = 9.71e37 kg·m² 
torque τ = Iα, but also τ = F*r = 3.98e7N * 6.371e6m = 2.54e14 N·m Since τ = τ, 2.54e14 N·m = 9.17e37kg·m² * α α = 3.61e-23 rad/s² 
ωo = 2πrads / (24h*3600s/h) = 7.27e-5 rad/s ω1 = 2πrads / (28h*3600s/h) = 6.23e-5 rad/s 
Plug in numbers t = 2/5 * (5.98 * 10^24 kg) * (6.37*10^6 m) * (7.27 * 10^-5 rad/s - 6.23 * 10^-5 rad/s) / (3.98*10^7 N) = 3.98*10^22 s = 1.26*10^17 yrs