Question

What would be length of a day if earth shrinks to half its radius?

Answer 1

We can consider two scenarios: one with a constant mass and the other with change in mass proportional to the change in volume.

In both cases, the same principle applies: the conservation of angular momentum. Angular momentum is

L=I×ωL=I×ω

where II is the inertia moment and ωω is the angular velocity. The angular momentum LLis conserved, meaning it must stay the same value, so if the inertia II decreases by some factor, the speed ωω has to increase by the same factor.

For a sphere (let’s assume the Earth is a perfect sphere): I=25mR2I=25mR2 (mm is the mass and RR is the radius)

So let’s find out how II changes in both scenarios.

Scenario 1: constant mass

If the radius is halved, that means the inertia moment is reduced by a factor 4 (because of the R2R2). So the angular speed ωω must be increased by a factor 4, which means a day would be 6 hours instead of 24.

Scenario 2: changing mass

if the mass changes proportionnally to the volume, it means the mass is reduced by a factor 8 (because the volume of a sphere is proportional to R3R3). So the inertia moment is reduced by a factor 4 because of the radius, and by a factor 8 because of the mass. So the total reduction is by a factor 32! which means the angular speed is increased by a factor 32, so a full day is only 45 minutes long !

Answer 2

It depends on whether or not Earth’s mass shrinks with its radius. Let’s assume that the mass stays the same, so that the Earth just compresses as its radius shrinks. We’ll also make the simplifying assumption that Earth is a sphere of uniform density.

So planet Earth would rotate about four times in 24 hours, reducing our day length to six hours.

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