formula

$T[y]=\int^{400km}_{0km}\sqrt{\frac{1+y'^2}{2gy}}dx\\ =\int^{2\pi}_{0}\sqrt{\frac{1+(\frac{dy}{d\theta}/\frac{dx}{d\theta})^2}{2gy(\theta)}}\frac{dx}{d\theta}d\theta\\ =\sqrt{\frac{A}{g}}2\pi$