I love snow! There is so much physics involved with snow. You can study thermodynamics, kinetics, optics, fluid dynamics, crystallography, solid state physics and statistical mechanics and still not know enough about snow. Quantum mechanics, chemistry…

There are some quick Fermi problem associated with snow:

--- Start Fermi Problem ---

- Estimate how much snow falls over the Earth each year.

- About how many snow flakes is this?

- How many protons are in this amount of snow?

- Estimate how much energy it would take to melt all that snow.

- Estimate how much energy it would take to vaporize all that snow.

- How many snow men could be made with all that snow?

- If all that snow was turned into snowballs, and a great world war with snowballs were launched involving the entire planets human population, about how much energy would the people throwing the snowballs expend (neglect throwing each snowball more than once).
- How long would the war last (assuming the snow does not melt first) ?

--- End Fermi Problem ---

War is hell, so what are the snowball’s chances?

Honestly, I dont know where to start (estimating the amount of snowfall)

ReplyDeleteTry bracketing the answer. Come on Samik, you know how to do these! ;)

ReplyDeleteHi!

ReplyDelete"Estimate how much snow falls over the Earth each year."

My attempt:

I suppose snow climates begin from 50 latitude to north.(I will take into account Northern Hemisphere only, at first.)

I estimate annual average precipitation in this area about h=2 mm/day.

Now i suppose that the water equivalent of falling snow is roughly h/2 = 1mm/day.

What follows is pure geometry:

Surface area of the considered area is:

s=S/2*(1-sin(f))

where

S is Earth's surface area

f is latitude in radians

So the total water equivalent of snow falling in each year in the area is 365*s*h/2

If we take into account the Southern Hemisphere also then for rough estimation we multiply previous result simply by 2.

Final result would be 365*s*h

Now numerical answer:

S=5.1*10^8 km^2

f=0.873 rad

h=2*10^(-6) km

s=S/2*(1-sin(f))=6*10^7 km^2

365*s*h=4.4*10^4 km^3

This is a cube with edge length about 35 km.

This problem is interesting, but a bit vague.

ReplyDeleteAre you counting only the percentage of snow on land, or that falls anywhere?

@ Martin

How did you come up with a 2mm per day average?

I will tackle another part of the problem: The one where it asks to calculate the energy to throw all those snow balls.

Assuming I had n snow balls (I will try and work Martin's solution), I would use the balistic equation to determine the range to throw one snow ball at 45 degrees, figure out the velocity needed, and use KE = 1/2 m v^2 to figure out the energy. Multiply this by n snowballs and I have the energy!

@Martin

ReplyDeleteWoah! A cube of water 35 km on a side! A bit later I will post my take on this. =)

@Brad,

I agree I was a bit vague, but if there is a snowball war I will assume that it is the snow falling on land, and assume that the snowball military machine can access it on any terrain! ;)

As for you solution, perhaps you can consider this from a purely energy standpoint?

Consider how much energy (neglecting friction etc) it takes to throw a snowball straight up. That energy is converted to potential energy of PE = mgh.

That is the energy of a snowball tossed to that height.

Now, the human has to not only expend that energy, but the energy required to move his arm fast enough to launch the snowball! Here is where we can estimate. Figure for how much his arm weighs, and how much energy it takes to get it to a speed needed to launch the snowball a given height.

Then, for each single snow ball, the energy to throw it is:

Etot = mgh + (1/2 Ma v^2)c

Where c is a scaling factor for the arm.

You can calculate a ratio of energy to move arm to total energy to throw the snowball.

You can do this for n snowballs, but for the whole mass of snow that martin calculated, just calculate Mgh for the whole 35 km on a side cube! Perhaps the height is 15 m.

Then add the scaled arm energy and there you have it! =)

your stupid. A snowball that big would melt all the snow. it is like the dinasaur killing comet. Everyone would be dead!

ReplyDeleteThat is much simpler! And I should have figured that out since we are talking about conservation of energy in my physics class. :-(

ReplyDeleteHow about the energy needed to pick up the snowball in the first place? ;)

@anonymous

ReplyDeleteA moments thought will show you that throwing a snowball that big will not melt all the snow. What happens if you throw a snowball? Does it melt? What makes you think throwing the equivalent of a 35 km on a side snowball will melt it? Conservation of energy after all. ;)

@ Brad

Humm... I guess that is what the scaling constant is for! ;)

Let's just say mghN where N is the number of snowballs, let h~ 1m. Add that to the end total.

@Brad

ReplyDeleteI looked at various meteorology charts.

Now, i suspect that h=2mm/day is maybe slightly overrated. More appropriate is h=1.5mm/day, i think. In this case the edge length of the cube reduces to 32 km.

Here is a cool site:

ReplyDeletehttp://modis-snow-ice.gsfc.nasa.gov/intro.html

Its doesnt in the tropical regions (lets say from 23 S to 23 N lattitude), so that leaves with (90-23 = 66 ) lattitudes on either hemisphere. Which means we can probably say that about 2/3rd of Earth's total area gets some kind of snowfall.

ReplyDeleteNow assuming it snows twice as much inside the acrtic circles than it does between 23-66 lattitude, we can call the total snowfall 3x , x being the amount of snowfall between 23 and 66. Of course, the amount of snowfall in this region vary a lot with altitude as well as other factors like distance from the sea, we can probably take Boulder, which falls in region, as a standard. So we can probably multiply the annual snowfall of Boulder with 6 (for both hemispheres ) and get a good estimate of annual snowfall in Earth . I beleive the average snowfall is abt 5 inches, and it must about 4 months (120 days of the year).

so the total volume of snow comes out to be

120 * 5" * 2/3 * (4* Pi * r^2) where r is about 6400 km .

To get the no of protons and stuff, we have to multiply this by the density of ice , which is lower than but can be taken as the density of water (1000 kg /m^3).

to be continued .....

so the no of protons would be [(V* rho)/18]*N*10 where V is the aforementioned volume, rho the density of water, and N the Avogadro no. (remember the molecular wt of water is 18 and each molecule contains 10 protons, 2 from Hyd atoms and 8 from the O atoms).

ReplyDeleteAs for energy of melting, that would be mass*L where mass = V/rho as above, and Latent heat of melting for ice is 80 cal/g

to vaporize this , first it has to be brought to 100 C water for which the heat required is

m * C * dT where c is the specific heat of water and dT= 100 K . Add to this the energy required to vaporize this water which would be (mass * Latent heat of vaporization) .