Consider Earth as a uniform sphere of diameter 12.7  10⁶ m, with a mass of 5.98  10³⁰ kg, and a rotational period of 1 day. If Earth suddenly had a diameter of half this value without any loss of mass, calculate the new period of rotation. A) 0.5 day B) 2.0 days C) 0.25 day D) 4.0 days E) 1.4 days

Question

Quizplus · Accepted Answer

The new diameter of Earth is 6.35 · 10^6 m. 
The moment of inertia for a uniform sphere can be calculated as I = (2/5)MR^2, where M is the mass and R is the radius/diameter. 
Before: I1 = (2/5)(5.98 · 10^24 kg)·(6.35 · 10^6 m)^2 = 9.44 · 10^37 kg·m^2
After: I2 = (2/5)(5.98 · 10^24 kg)·(3.175 · 10^6 m)^2 = 2.36 · 10^37 kg·m^2
The conservation of angular momentum (L = Iω) tells us that the product of the moment of inertia and angular velocity is constant. Therefore, I1ω1 = I2ω2. Solving for ω2 gives:
ω2 = (I1/I2)ω1 = (9.44 · 10^37 kg·m^2) / (2.36 · 10^37 kg·m^2) · (2π rad/day) = 8π rad/day
The new rotational period is therefore T2 = 2π/ω2 = 0.25 day, which corresponds to answer choice C.

Consider Earth as a Uniform Sphere of Diameter 12