What is the radius of the planet?

Question

A 60.0 kg space explorer discovered a perfectly spherical planet. She found out that her apparent weight is 97% at the equator of the 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

In terms of centripetal force F_W - F_N = F_C The net force on the explorer is such that he goes around in a circle about the equator with F_C having to be mv²/R. Since we are given period, we might as well use v = 2πR/T to write the F_C as 4π²mR/T². The apparent weight is the normal force a scale would measure and F_W is the weight from just gravitational pull (like at the pole)

F_C = F_W - F_N = .03(360N) = 4π²(60 kg)R/(14 hr* 3600 s/hr)²

solve for R in meters.