# Rate of Air Loss Through a Hole in a Spaceship

## Homework Statement

We want to calculate the rate of air loss from a space vehicle (module) if a meteoroid punches a hole in i
t. Assume the module is sealed off from other modules. It is shaped like a cylinder roughly 4 m in diameter and 10 m long. The hole's area is a cm2. The hole is punched in one of the end faces of the cylinder.

Show that the rate of loss of air molecules through the hole is

$$\left| \frac{dN}{dt} \right| \approx \frac{a \left| v_x \right|_{avg}}{2V} N$$

2. Homework Equations [/B]

## The Attempt at a Solution

Here's my thinking on this. Please point out any reasoning flaws.

I take a representative slice of the cylindrical module, of width dx, and ask how many molecules dN' pass through this slice in a time dt. Since the molecules are assumed to be very many in number and their motion is assumed to be randomly distributed, the fraction of all the molecules in this slice can be taken as the same as the fraction of the volume of the whole module taken up by this slice. So:

$$\frac{dN'}{N} = \frac{Av_x dt}{V}$$

where vx is the average x-velocity of a molecule. But I know that I will only want half of this, since the escape hole is only on the one side. So I divide by 2:

$$\frac{dN'}{N} = \frac{Av_x dt}{2V}$$

Now I don't want all of these molecules, either. I only want the fraction that will pass through the hole. The expression I have so far is how many would be lost if the entire cylinder end were open, i.e. just one big hole. So I fix this by multiplying by the fraction a/A, where A is the area of the entire end of the cylinder.

So my final result is
$$\frac{dN'}{dt} = \frac{a v_x N}{2V}$$

And since N' are the only molecules that can be lost (since molecules passing through other slices aren't lost) I can say dN' = dN. Which gives me the result I want.

My only reservation here is that I haven't dealt carefully with the averaged velocity. [/B]