Railroad Prank

Let the radius of the arc be $r$ and the angle extended by the bent track at the center of the circle be $\theta$ . Then we note that:

$\begin{cases} r \theta = 1001 & \implies \theta = \dfrac {1001}r \\ \sin \dfrac \theta 2 = \dfrac {500}r & \implies \sin \dfrac {1001}{2r} = \dfrac {500}r \end{cases}$

Since $\theta$ is very small, we can estimate $\sin \frac \theta 2$ using Maclaurin series and we get $\dfrac {1001}{2r} - \dfrac 16 \left(\dfrac {1001}{2r}\right)^3 \approx \dfrac {500}r$ as suggested by @Michael Mendrin . $\implies r = \sqrt{\dfrac {1001^3}{6\times 2^3 (500.5-500)}} \approx \boxed{6464.66}$ .

Solve for $r$ in the following equation

$r\left(\dfrac{1001}{2r}-\dfrac{1}{6}\left( \dfrac{1001}{2r} \right)^3 \right) = \dfrac{1000}{2}$

and get $r \approx 6464.66...$

Bonus question: how high does the arc reach above the ground?