Hyperloop: Dropping coins

EDITED 27-Aug Initially I had $t$ and $t_1$ the wrong way around, and added extra explanation.

We are trying to find out the effect of the pod's centripetal acceleration on the coin falling time. The assumption is made that the pod is moving parallel to the earth, so it experiences an apparent centrifugal force of $\frac{mv^2}{r}$ .

So the effective gravity in the pod will be $g_1 = g - \frac{300^2}{6370000}$ .

Let $t$ be the time the coin takes to fall under gravity $g$ , and $t_1 = \frac{1}{3}$ be the time the coin takes to fall under the apparent gravity $g_1$ .

The distance is the same in both cases, so

$\frac{1}{2}gt^2 = \frac{1}{2}g_1{(\frac{1}{3})}^2$

Rearranging this gives :

$t^2 = {(\frac{1}{3})}^2 \frac{g_1}{g}$ so $t = \frac{1}{3}\sqrt{\frac{g_1}{g}}$ so $t_1 - t = \frac{1}{3}(1 - \sqrt{\frac{g_1}{g}})$

Numerically we get $g_1 = 9.78587...$ , $\frac{g_1}{g} = 0.998558...$ , and $t_1 - t = 2.4037116... *10^{-4}$

I know that centripetal acceleration is equal to v^2/r, but why are you adding the acceleration to gravity to that? Can you also tell me the motion equation?

In this problem are unfortunately neglected the rotational speed of the Earth (see http://en.wikipedia.org/wiki/Gravitational_acceleration) and the "course" of the pod, which are both basically needed to solve the question.

Skipping the computation I did, I think I could say, for example, that:

supposing, for simplicity, the Hyperloop lies on the equator ,

• and the pod is directed eastwards , the answer would be 0.001 sec

• and the pod is directed westwards the answer would be - 0.0005 sec.

It is worth noting that this differences are progressively higher with higher latitudes values.