Answer to Question #314616 in Physics for Anon

Question

A 60.0 kg space explorer just discovered a perfectly spherical planet. She found out that her apparent weight is 97% at the equator of a planet compared to that at the pole, which is 360N. If the rotational period of the planet is 14.0 h, what is the radius of the planet?

Answer 1

The centripetal force is 3% of the weight of space explorer. Hence

$F_c=\frac{mv^2}{R}=\frac{4\pi^2mR}{T^2}=0.03W$

The radius of a planet

$R=\frac{0.03WT^2}{4\pi^2m}$

$R=\frac{0.03\times 360\times (14.0\times 3600)^2}{4\times 3.14^2\times 60.0}=1.16\times 10^7\:\rm m$