At the equator, a person stands on a spring balance. By what fraction is the reading on the balance less than the person's true weight?

Step-by-step solution

True weight of the person, W = mg Apparent weight of the person at the equator due to Earth's rotation, W' = m(g - ^2 R) Decrease in weight, W = W - W' = m ^2 R Fractional decrease in weight is given by: W W = m ^2 R mg = ^2 R g We know that the angular velocity of Earth is = 2 T , where T = 24 3600 s . Radius of Earth, R 6.4 10^6 m and g = 9.8 m/s ^2 . Substituting the values: W W = 4 ^2 R T^2 g = 4 ^2 6.4 10^6 (24 3600)^2 9.8 Using ^2 9.87 , we get: W W 252.67 10^6 73.15 10^9 3.45 10⁻³ This value is closest to 3.