Larry Weinstein, Column Editor
Old Dominion University, Norfolk, VA 23529
weinstein@odu.edu
Fermi Questions
Solutions for Fermi Questions, March 2016
w Question 1: Wind
Globally, how much kinetic energy is carried by the
wind?
so that the mass of the atmosphere is
m = PA/g
= (105 N/m2)(4 3 1014 m2 )(10 N/kg)
= 4 3 1018 kg.
Answer: Since KE = (1/2)mv2, we need to estimate
mass and velocity. The mass equals the surface density
times the area. If we remember the surface air pressure
(15 psi or 105 Pa), then we just need to divide by g to get
the density
s = P /g
= (105 N/m2)/(10 N/kg)
= 104 kg/m2.
If we don’t remember the air pressure, then we can
estimate it from the volume density of air and the height
of the atmosphere. The height of the atmosphere is not
well defined. The volume density decreases steadily as we
go up from the surface. We want the “constant density
height,” the height the atmosphere would be if the density
was the same as at the surface. So how can we estimate
this? Let’s estimate the height at which the density decreased by a factor of two.
Mountain climbers can breathe almost normally on a
33103 m (104 ft) peak. However, they can barely survive
without supplemental oxygen at the top of Mt. Everest,
which is at 104 m. Airplanes also fly at an altitude of
104 m to significantly reduce air drag. Therefore, the constant density height of the atmosphere is more than 3 3
103 m and less than 33104 m, so we will estimate 104 m.
Air, as a gas, has about 10-3 times the density of water,
or 1 kg/m3. Alternatively, we should know that air is
composed primarily of N2 with a molecular weight of
about 30 g/mole and that one mole of gas at STP occupies
22.4 liters, giving a density of about 1 g/L or 1 kg/m3. This
gives a density of
s = rh
= (1 kg/m3)/(104 m)
= 104 kg/m2,
which agrees (as it should) with the density derived from
the pressure.
The area of the Earth is
A = 4pR2
= 10(63106m)2 = 4 3 1014 m2
THE PHYSICS TEACHER ◆ Vol. 54, 2016
Now we need to estimate the velocity. We know that
surface winds typically blow at speeds between 1 and
10 m/s (20 mph), but can occasionally reach 100 m/s
(200 mph) in tornados. We also know that the jet stream
blows significantly faster than surface winds. Let’s sweep
all the details under the (very lumpy) rug by estimating
the average global wind speed must be more than 1 m/s
and less than 100 m/s and therefore about 10 m/s. If you
prefer to place your upper bound at 300 m/s (the speed
of sound), then your estimate for the average wind speed
is 20 m/s.
Thus, the total global kinetic energy of the wind is
or about 105 megatons of TNT. That’s a LOT of energy.
Humans use about 2 TW of electrical power, so at 100%
efficiency (and without replenishment) the atmosphere
could provide electrical power for 108 s = 3 years. The
total insolation of the Earth is
P = (103 W/m2)p(6 3106m)2
= 1018 W,
so that even if only 10-3 of solar energy gets converted to
wind energy, the wind energy would be replenished in
105 s = 1 day.
Copyright 2016, Lawrence Weinstein.
w Question 2: Pacific wters
Bottled water from Fiji is very popular now. How
much does it cost to ship one liter of water from Fiji
to the U.S.?
Answer: Fiji Water ® is bottled in Fiji and shipped to
the U.S. by, surprisingly, a ship. Let’s use a large (but not
enormous) ship to transport our water. Enormous ships
like aircraft carriers and supertankers displace about 105
tons, so our ship will displace a more modest
104 tons. Since water has a density of 1 ton/m3, our ship
will displace 104 m3. As ships move, they push water
aside. We will estimate the cost of moving a ship from
Fiji to the U.S. by estimating the kinetic energy imparted to the water en route.
No. We won’t. I started solving the problem that way
and got a completely wrong answer. Let’s just estimate
the relative amount of fuel carried by the ship. In order
to travel the large (but unspecified and unestimated)
distance from Fiji to the U.S., the ship will burn more
than 1% and less than 100% of its displacement as fuel.
Thus, a 104 ton ship will burn 103 tons of fuel. This
means that the ship will burn 0.1 L of fuel oil for every
liter of Fiji Water (or other cargo). At today’s price of
$0.5 per liter, the ship will burn $0.05 (five cents) of fuel
for each liter of water. Even if we multiply by a few to
account for other costs (ship amortization, crew wages,
fuzzy dice for the rear view mirror, etc), shipping only
adds $0.2 to the cost of a liter of Fiji Water.
Shipping is cheap!
Now here’s the long, complicated, and wrong answer:
We need to estimate the ship’s cross sectional area and
its speed. It will be longer than 10 m and shorter than
103 m, so we’ll estimate a length of 102 m. This means
that it will have an average cross sectional area (below
the water line) of
At a typical efficiency  = 0.25 and fuel energy density
e = 4 3107 J/L, the ship will burn
We can estimate the travel time from Fiji to the U.S. in
several ways. The easiest is to do it directly. The travel
time is more than 1 day and less than 1 year (400
days), so we will estimate t = 20 days or 23106 s. On
the other hand, we know the distance to Fiji is more
than 103 km and less than 105 km, so we will estimate
104 km. At 10 m/s, the ship will travel 104 km in 106 s.
On the gripping hand, we can estimate the distance
to Fiji as ¼ of the globe’s circumference, also giving a
distance of 104 km and a time of 106 s. (Fiji is actually 8853 km from Los Angeles, so our estimates are
remarkably good.)
The total fuel used will then be
A = V/l
= 104 m3/102 m
= 102 m2.
Our ship will travel faster than 1 m/s (2 mph) and
slower than 100 m/s, giving an estimated speed of
10 m/s. Thus, it will push aside water at the rate
.
V = Av
= (102 m2)(10 m/s)
= 103m3/s
and
.
.
m= rV
= (103 kg/m3)(103m3/s)
= 106 kg/s.
The average speed of the water pushed aside will equal
the speed of the ship so that the power needed to push
water aside will be
470
THE PHYSICS TEACHER ◆ Vol. 54, 2016
Whoops. This much fuel will have a mass of 53106 kg
= 53103 tons. This is half the mass of the ship. We
must have missed something.
The sideways velocity of the water displaced by the
passage of the ship will be significantly less than the
speed of the ship because the ship is streamlined. This
should reduce the speed of the displaced water by a
factor of two to three, reducing the kinetic energy by a
factor of about six and therefore the fuel needed by a
factor of six.
That reduces the fuel needed from 53103 tons to
103 tons, in agreement with our previous (much simpler) estimate. Hooray for simplicity!
Copyright 2016, Lawrence Weinstein.