Hyperloop: Dropping coins

"The acceleration of gravity is 9.8 m/s/s" means that this is the acceleration when it is stationary. When it is moving, though, some of the gravitational acceleration is centripetal, and only the leftover acceleration is radial.

I will call the time it takes to fall $T$ and I will call the radial acceleration when the Hyperloop is moving $a$

Since it falls the same (radial) height whether it is moving or not, we know:

$9.8T^2=a(\frac{1}{3})^2$

And therefore:

$T=\frac{1}{3}\sqrt{\frac{a}{9.8}}$

As I hinted at before, $a$ is 9.8 minus the centripetal acceleration, which is:

$a=9.8-\frac{300^2}{6370000}$

So now you can calculate $T$ , and then take $\frac{1}{3}-T$ and you will get the correct answer.

When moving in an approximate circle around the Earth, some of the gravity acting on the coin goes to supplying the centripetal force, thus making the "effective gravitational acceleration" less (the same principle applies for why astronauts are weightless aboard the ISS). We know that:

$a' \ = \ g \ - \ \frac{v^2}{R}$

While at rest, the acceleration will simply be $a_0 \ = \ g$ . The distance the coins drop is the same. We know that:

$d \ = \ \frac{1}{2} a t^2 \ \Rightarrow \ t \ = \ \sqrt{\frac{2d}{a}}$

$\frac{t_0}{t'} \ = \ \frac{\sqrt{\frac{2d}{a_0}}}{\sqrt{\frac{2d}{a'}}} \ \Rightarrow \ t_0 \ = \ t' \ \sqrt{\frac{a'}{a_0}} \ \Rightarrow \ \Delta t \ = \ t' \ - \ t_0 \ = \ t' \Big(1 \ - \ \sqrt{\frac{a'}{a_0}} \Big) \ = \ \frac{1}{3} \ \Big( 1 \ - \ \sqrt{\frac{g \ - \ \frac{v^2}{R}}{a_0}} \Big) \ = \ \frac{1}{3} \ \Big( 1 \ - \ \sqrt{\frac{9.8 \ - \ \frac{300^2}{6370000}}{9.8}} \Big)$

$\Delta t \ \approx \ 0.00024 s$