A body weighed with a spring balance in a train at reset, shows a weight W₀. When the train begins to move with a velocity v around the equator from west to east and if the angular velocity of earth is ω then the weight recorded by the spring balance is

$W_0\left(1-\frac{2 v \omega}{g}\right)$

If W represents weight of the body when earth were not rotating, then weight of body, W₀ in train at rest is given by

$W_0=W-\frac{mv_{e}^2}{R}$           ........(1)

When the train moves from west to east with velocity v, the velocity of the body relative to earth gets added to the velocity of earth.

Therefore, the weight of the body,

$W_1=W-m \frac{\left(v_{e}+v\right)^2}{R}$      ........(2)

Subtracting (2) from (1)

$W_0-W_1=\frac{m}{R}\left[\left(v_{e}+v\right)^2-v_{e}^2\right]$

$=\frac{m}{R}\left[v_{e}^2+v^2+2 v_{e} v-v_{e}^2\right]$

$W_0-W_1=\frac{m}{R}\left(v^2+2 v_{e} v\right)$

Since $v^2<2 v_e v$ and $\frac{v_e}{R}=\omega$

$W_0-W_1=\frac{mg}{Rg}\left(2 v_{e} v\right) =\frac{mg}{R}(2 vR \omega)$

$W_0-W_1=W_0 2 v \frac{\omega}{g}$

$W_1=W_0\left(1-\frac{2 v \omega}{g}\right)$