The gravity train is 'just' simple harmonic motion - the attractive force is proportional to the distance from the centre and the 'constants involved just involve the density of the particular planet. You can either accept that the sums work or slog through it. What fascinated me about that problem was that it reveals that a piece of dust would have the same time period moving through a hole in a large granite rock as long as there were nothing else around and the rock was not tumbling (of course)

Anyway, I have had an interesting hour or so - which I should have spent doing something useful - working out some answers.

For the diagonal path.

The acceleration, a, is g/√2 and the distance to travel is l√2.

The equation of motion gives s = at^{2}/2

So l√2 = gt^{2} /(2√2)

Giving

t = 2√(l/g)

For the circular path it is easiest to consider the angle from the centre of the circle – starting at θ = 0 and going up to 90^{o}

At any point on the path, the tangential speed, v, is l dθ/dt and the 'imposed' tangential acceleration is

g cos(θ) so you can write what is happening to the mass at that point

The acceleration will be dv/dt,

or ld_{2}θ/dt^{2}

and you can write the equation as

g cos(θ) = ld_{2}θ/dt^{2}

I was messing around for ages before I realised that this is the diff equation for a simple PENDULUM –durrrr.

The period of a pendulum with small displacements (swings) is given by 2π√(l/g) and we are dealing with a quarter of a period so the time would be

t = (π/2)√(l/g) for a small displacement. (Edit: that's a pi - not an n - it isn't clear on my browser)

Or 1.57√(l/g)

BUT, for large displacements, the sums are much harder and I had to look it up. You can’t work it out simply and you have to get into “elliptic functions of the first kind – aaaargh! Wiki tells us that, for a 90^{o} swing, there is an 18% increase in time, giving

t = 1.85√(l/g)

Please check for accuracy but I think it's ok and shows that the straight path takes longer.

I then looked at what happens if the mass falls vertically, then, with the acquired speed (after a tight curve) it travels the horizontal distance.

The vertical drop takes time t_{v} = √2 √(l/g) (same as the circular path, apparently - and a bit counter intuitive) and its final speed is √2gl, so it takes an extra time

t_{h} = l/√2gl or √(l/2g)

Total time

t = t_{v} + t_{h}

t = √(2l/g) + √(l/2g)

or √(l/g) (√2 + √(1/2))

so

t = 2.12√(l/g)

Which, is greater than for the circular path.

Somewhere in there there must be an optimum.

There must be some Maths wizards out there who could find it?