Question

If the earth revolved rapidly enough, the weight of objects at the equator would be zero. What would be the length of the day in that case (in h)?

Answer 1

Circumference of Earth c = 40075.16 km
r = 6371 km
M = 5.98 x 10^24 kg

For m = 0 at center
Force of gravity = Force of Centripetal acceleration
mg = mv^2/r
g = v^2/r
v = sqr(gr)
dx/dt = sqr(gr)
x = c = sqr(gr) x [t] <-- Definite integral between t and 0
t = c/sqr(gr)
t = [40075.16 x 10^3 / sqr(9.8 x 6371 x 10^3)] s.
t = 5071.75 s = 1.408 hours