The aircrafts land exactly 12 hours later at the diametrically opposite point. The Earth completes one full rotation ($2\pi R$) in 24 hours. Therefore, in 12 hours, the Earth rotates by distance $\pi R$.
Since the planes also cover the distance $\pi R$ (half circumference) in 12 hours, their speed relative to the Earth’s surface is: $$ v = \omega R $$

A pendulum clock’s time measurement depends on the effective gravity in its frame of reference: $T \propto 1/\sqrt{g_{eff}}$.
The standard “acceleration of free fall” $g$ at the equator already accounts for Earth’s rotation: $$ g = g_{true} – \frac{(\omega R)^2}{R} = g_{true} – \omega^2 R $$ Therefore, $g_{true} = g + \omega^2 R$.

For a plane moving with speed $v$ relative to Earth, the absolute speed $V_{abs}$ in the inertial frame determines the centrifugal force. $$ g_{eff} = g_{true} – \frac{V_{abs}^2}{R} $$

Plane 1 (Eastbound): Moves in the direction of rotation. $$ V_1 = \omega R + v $$ $$ g_1 = g_{true} – \frac{(\omega R + v)^2}{R} $$ Substitute $g_{true} = g + \omega^2 R$: $$ g_1 = (g + \omega^2 R) – (\omega^2 R + v^2/R + 2\omega v) = g – \frac{v^2}{R} – 2\omega v $$
Plane 2 (Westbound): Moves against rotation. $$ V_2 = |\omega R – v| $$ $$ g_2 = g_{true} – \frac{(\omega R – v)^2}{R} $$ $$ g_2 = (g + \omega^2 R) – (\omega^2 R + v^2/R – 2\omega v) = g – \frac{v^2}{R} + 2\omega v $$

The time measured $\tau$ is related to the flight time $t$ by $\tau = t \sqrt{\frac{g_{eff}}{g}}$ (assuming clocks are calibrated to $g$).
Using binomial expansion $\sqrt{1 \pm x} \approx 1 \pm x/2$:

For Plane 1: $$ \tau_1 = t \sqrt{1 – \frac{2\omega v + v^2/R}{g}} \approx t \left( 1 – \frac{\omega v}{g} – \frac{v^2}{2gR} \right) $$ For Plane 2: $$ \tau_2 = t \sqrt{1 + \frac{2\omega v – v^2/R}{g}} \approx t \left( 1 + \frac{\omega v}{g} – \frac{v^2}{2gR} \right) $$
Difference ($\Delta \tau = \tau_2 – \tau_1$): $$ \Delta \tau \approx t \left[ \left( 1 + \frac{\omega v}{g} – \frac{v^2}{2gR} \right) – \left( 1 – \frac{\omega v}{g} – \frac{v^2}{2gR} \right) \right] $$ The $v^2$ terms cancel out: $$ \Delta \tau = t \left( \frac{2\omega v}{g} \right) $$

Substitute $v = \omega R$: $$ \Delta \tau = t \frac{2\omega (\omega R)}{g} = t \frac{2\omega^2 R}{g} $$ The ratio of the difference to the time taken ($t$) is: $$ \frac{\Delta \tau}{t} = \frac{2\omega^2 R}{g} $$