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