Question

How long would one day be if the Earth were to rotate at a speed great enough for objects at the equator to have apparent weights equal to one fifth of their actual weights? Clearly the days will be shorter, but I am unsure of how to come to the answer.

Answer 1

The apperent weight
F_a = F_g  - F_c
Where F_a is the apparent force ,F_g is the gravitational force and F_cis the centripetal force on the objects due to the rotation of earth
The apparent force is (1/5)^th of the original force (F_g)
F_g / 5 = F_g  - F_c
(4 / 5) F_g  = F_c
(4 / 5) m g = m v² / R
Where R is the radius of the earth
(4 / 5) m g = m v² / R
(4/ 5) m g = m ² R
   = sqrt (4 g / 5 R)
  = sqrt (4 x 9.81 / 5 x 6.371 x 10⁶ )
  = 1.11 x 10^-3 rad/s
The time period is
T = 2 /
T = (2 x 3.14) / 1.11 x 10^-3
T = 5657.66 s
T = 5657.66 / (60 x 60)
T = 1.57 hrs