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?

• This is a fantastic question... – Earth is Donut Nov 22 '14 at 15:35
• Since an hourglass depends on friction between the particles (and BTW I rather doubt the value of particles/second is strictly constant in an hourglass!), this could get messy in a hurry. An hourglass brought along by Philae, e.g., probably never drops any grains. – Carl Witthoft Nov 22 '14 at 17:50
• @CarlWitthoft you're right about it probably not being constant - I've changed it to "reproducible". – Nathaniel Nov 22 '14 at 22:44
• @Nathaniel Believe it or not, the rate is nearly constant as well! (see my answer) – user854 Nov 23 '14 at 2:14
• @CarlWitthoft As a four year old I used to love to watch the sand in my gran's egg timer, and I strongly suspect you're right. I would suggest the following experiment. At first, the sand heaps up to a cone until the vertex angle of repose as its vertex angle. Thereafter, you see little "landslides" from time to time. It would be a reasonable bet that each of these landslides would represent similar volumes of sand, so you could time the landslide events and check whether the rate is constant after the angle of repose is reached. This may at least tell you something about weight vs flow rate. – Selene Routley Nov 23 '14 at 2:43

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!

• Nice answer, but I would imagine one could refine it by including general relativity effects, i.e. the flow of time is not the same on Earth and on the Moon, leading to additional corrections. Still +1 – Frédéric Nov 25 '14 at 20:01
• I hope you don't mind that I edited in the derivation of the $\sqrt g$ law from the other answer, since that's what I wanted to know. – Nathaniel Nov 26 '14 at 2:02
• @Frédéric the correction due to general relativity will be of the order 0.02 seconds per year of measured time (quoting from Wikipedia because I'm lazy) - far smaller than the error involved in measuring time with an egg timer, so I think this is a perfectly acceptable answer! – Nathaniel Nov 26 '14 at 2:08

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

• Those quoted sources do not IMHO address the problem at handm tho' the conveyor belt test comes close. My bet is that gravitational force, which acts from "below", as opposed to the conveyor belt's force which acts from "above", leads to different behavior when the force is small compared with the inter-particle friction/binding forces. – Carl Witthoft Nov 23 '14 at 2:30
• "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. " Doesn't follow. What is canceled is the loss of weight which is proportional to both $g$ and the mass flow rate. But the impact impulse is also proportional to both $g$ and the mass flow rate. I don't think that you can conclude anything from that. – dmckee --- ex-moderator kitten Nov 23 '14 at 3:07
• @dmckee You are correct. The missing piece of the puzzle is that sand falls slower on the moon. Combining that with the facts above should yield an answer, though the answer may depend on the length of the stem. – user854 Nov 23 '14 at 3:15