Hourglass on the Moon Someone asked me this, and I was surprised to find I couldn't answer it: suppose I have an hourglass / egg timer that times two minutes in Earth's gravity. If I used it on the Moon, how long would it take for all the sand to fall?
The reason I can't answer it is that I don't know exactly what physical processes cause the sand to fall at such a nicely reproducible rate under Earth's gravity. So I suppose an equivalent question is, how exactly does an hourglass work, at the microscopic level?
 A: What is the relationship between hourglass flowrate and local gravity?
As in the excellent answer to a related question (hourglass flowrate vs. sand grain size) and this published paper, the mass flowrate $Q$ through an hourglass is dependent on local gravity as
$Q\ \propto\ \sqrt{g}$
This is derived through dimensional analysis, as follows. (Quoting from the answer by Georg Sievelson linked above.)

Let us consider a cylinder of diameter $D$, with a circular hole punched on the bottom side with radius $a$. We fill the cylinder with a height $H$ of sand. If we look at the speed of sand grains going out the bucket, we observe (experimentally) that it does not depend of the height of sand $H$, if $H$ is big enough (compared to the diameter $D$ - because the constraint saturates). We are left with two parameters : the diameter of the hole $a$ and the gravity field $g$ that makes it fall, so the output speed $v$ has to be proportional to $\sqrt{g a}$. The flow rate is the speed times the section, thus it is $Q \propto v \, a^2$, so it is of order $Q \propto g^{1/2} a^{5/2}$ (this is the Beverloo law).

Assuming you have the same hourglass on the Earth and the Moon, then the hourglass has the same mass of sand to move $m$ but different mass flowrates $Q$ so it will take a different time $t$. Since $Q=\frac{m}{t}$ and $Q\ \propto\ \sqrt{g}$, simple algebra shows:
$\frac{t_{Moon}}{t_{Earth}} = \sqrt{\frac{g_{Earth}}{g_{Moon}}}$
Since $\frac{g_{Earth}}{g_{Moon}}=6$ then:
$t_{Moon} = t_{Earth} \sqrt{6}$
So your $2$ minute egg timer here on Earth becomes a $\approx 5$ minute egg timer on the Moon. Thankfully we have more reliable methods of timekeeping in space!
A: I'm probably wrong, but I think it would take about 6 times as long, 
assuming the moon's surface gravity is 1/6th that of Earth: 


*

*According to 
http://www.physics.umd.edu/deptinfo/facilities/lecdem/services/demos/demosc5/c5-41.htm, 
"[d]uring the steady-state sand fall the extra 
force of sand hitting the bottom very nearly cancels the loss of 
weight of the sand in the air" 

*Since the force of sand hitting the bottom is only 1/6th as much 
on the moon, there can only be 1/6th as much sand in the air at any 
given time. 

*Thus, it takes 6 times as long to drain the sand. 
This only works because the speed (and thus force) at which sand falls is 
unrelated to the amount of sand in the top bulb. In other words, if 
two hourglasses had the same stem width, but one had 50 pounds of sand 
in the top and the other had only 1 pound, sand would fall at the same 
rate and with the same force. Of course, the 50 pound hourglass would take 50 times longer to empty, but that's only because it contains 50 times as much sand.
This applies even when the amount in the top bulb is nearly 0, at which point gravity is the only force acting on the sand.
This is confirmed at http://www.technologyreview.com/view/418993/the-mystery-of-sand-flow-through-an-hourglass/ 
with a note that the commonly accepted reason for this oddity may be incorrect. 
http://en.wikipedia.org/wiki/Hourglass#Practical_uses also notes "The rate of flow of the sand is independent of the depth in the upper reservoir" sourced by the European journal of physics
