Height of Water in Straw vs. Time

Imagine you had a very long straw, one end is in the ocean (sea

level) and the other end is in space. Air pressure at sea level

is a standard atmospheric pressure and space is a perfect vacuum.



My end goal is to be able to plot "height of water in straw"

vs. time.



Here is what I've tried and maybe someone can point out where I'm

going wrong.



net force = atmospheric pressure - force of gravity on mass of water in straw



Atmospheric Pressure

Standard atmospheric pressure is ~101 kg / (m * s^2) per unit area

Let's call A the cross-sectional area of the straw.

atmospheric pressure = 101 * A



Force of gravity on mass of water in straw

Force of gravity on water depends on how much water is in the

straw (above sea level). Let's call h the current height of the

water in the straw.



mass-of-water-in-straw(h) = volume * density = h * A (density of water = 1)

force_of_gravity(h) = mass-of-water-in-straw(h) * g = h * A * g



F = 101 * A - h * A * g

So acceleration (F / m) = (101 - g * h) / h

Written another way: a(t) = (101 - g * h(t)) / h(t)



Ok, great, so I think h as a function of time should be the solution to this equation:

h(t) = integral_0_t (integral_0_t a(t) dt) dt



Unfortunately, I don't know how to solve that... so I tried to

numerically approximate it. Let's just assume constant

acceleration over short periods of time dt, then we can use much

simpler equations and step through time in small dt steps.



Simpler equations:

a(t) = (101 - g * h(t-dt)) / h(t-dt)

v(t) = v(t-dt) + a(t) * dt

h(t) = h(t-dt) + v(t) * dt

with starting conditions h(0) = 0 and v(0) = 0 and a(0) = 0



When I try to calculate my initial acceleration

a(1), it's infinity (or rather, undefined) since the force of air

pressure is pushing on no mass (no water is above sea level

at the start of the problem)! There is obviously some flaw

in that logic, but I'm not sure what it is.



Any help is apprecated, thank you.