Phys 239
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Quantitative Physics
Problem Set 1 Solutions
Laser pointer to moon
Some roadside billboard convinces you to participate in a program to shine your laser pointer at the center
of the Moon at a particular time, the goal being to illuminate the Moon enough to be seen by the naked eye
from the Earth. How many people/laser pointers would this take? Put the number in context/perspective. Is
there an optical way to make it easier? What are the minimum requirements?
A typical laser pointer has an optical power of 5 mW, and a beam diameter of something like D =1 mm.
The beam will therefore diverge due to diffraction with a characteristic angle of λ/D ≈ 5 × 10−4 radians. At
the distance to the moon, r ∼ 4 × 108 m, the beam will be 2 × 105 m across, covering an area of 3 × 1010 m2 .
What does it take to be noticed? We can construct several criteria. The first might be that we need to
illuminate with an intensity similar to that of the sun bathing the moon in light—but preferably at a place
unilluminated by the sun. The sun puts something like 1000 W/m2 of light intensity onto the surface. So
over the illuminated area, we need 3 × 1013 W, which will require about 1016 laser pointers. That’s about a
million laser pointers per person on the planet—assuming you can get them all onto the same hemisphere
to see the moon at the same time. It’s also about three times the entire power budget of human society
(12 TW). Not about to happen. It would also require everyone to point to within the 0.5 mR divergence of
the laser beam. Forgive my lack of confidence...
We could improve things substantially if we increase D, perhaps by expanding the lasers through small
telescopes (we’re going to need a lot of scopes). Once you get above D ∼ 0.1 m, turbulence in the atmosphere
begins to dominate the divergence, limiting one in practice to one arcsecond (5 µR). If we could achieve this
for all the laser pointers, and point them to similar tolerance (hard), then we gain a factor of 104 in area,
thus need only 1012 laser pointers, or about 100 per person on the planet (or one 0.5 W laser each).
But do we really need solar-level illumination? If we go for a small spot from expanded beams, we may get
away with less light. What can the eye detect? Our eye works in exposure times of about one tenth of a
second. How many photons can we detect in this period? Clearly must be greater than one. And we would
not expect our eyes to be 100% quantum-efficient. In fact, both eyes and camera film are about 1% efficient,
so that we can detect light above 1000 photons per second. [We can get this by realizing that the dimmest
stars we can see are 6.5 magnitude, interpreted through conversion/calibration values.] This corresponds to
a flux of ∼ 10−11 W/m2 , or about 4 × 10−16 W entering our pupil. (Another estimate: a laser pointer on a
wall 300 m away is easily visible at night, which gets spread into 2π sr, putting 5 mW times π(0.003)2 /2πr2
into our 3 mm radius pupil. This works out to 2.5 × 10−13 W.) The laser light that hits the lunar surface will
be largely absorbed (the moon is dark), with 10% reflected (albedo=0.1). The reflected light is spread into
2π sr. Our 6 mm pupil picks out π(0.003)2 /2πr2 of the reflected light, or 3 × 10−5 /1018 = 3 × 10−23 of the
emerging photons, and we need to receive 1000 photons per second. So we need to hit the moon with 3×1026
photons per second, accounting for the albedo. A visible light photon of 2 eV energy is 3 × 10−19 J, so we’re
hitting the spot on the moon with 108 W, or 2 × 1010 5 mW laser pointers (expanded through telescopes).
Now we’re down to a few per person on the planet. The contrast may be tough: to see something at the
edge of detection right next to the illuminated portion of the moon. And it’s still a grand-scale effort not to
be approached lightly.
As an aside, one might ask whether it works out better to point the laser pointers at one of the Apollo
retroreflectors: it would be like a bike reflector shining brightly under illumination. But the physical size is
< 1 m2 , so only 3 × 10−11 of the photons hit the reflector. On the plus side, the beam returns in a smaller
bundle—not spread over 2π sr. Diffraction of the individual 38 mm corner cubes produces an effective return
diameter of 10 arcsec, or 5 × 10−5 radians (not just λ/D due to funky total internal reflection polarization
foolery). So the solid angle is this spread squared, or 25 × 10−10 sr. Thus the gain over 2π is 2.5 × 109 . If
we credit an additional factor of ten because now we are not fighting the 0.1 albedo of surface reflection, we
come out about even: the problem is no easier or harder than hitting the surface itself. But hold-on... The
return spot, at 5 × 10−5 rad is 20 km across at the earth. So you’d have to pack all of the 1016 laser pointers
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into a small patch of the earth or the returns will not overlap properly and you’ll lose the gain. So using the
reflectors is even harder than just using the surface, in practice!
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Terraform Mars
Let’s terraform Mars, making its atmosphere breathable! We would certainly need to use raw materials from
Mars itself. Lots of oxygen is locked up in its sands and rusty rocks. Our goal is to create the same density
(partial pressure) of oxygen on Mars as on Earth, so that each lungfull contains the same amount of O2 .
Let’s skip the N2 , and go with less total pressure of nearly pure oxygen (plus a greenhouse gas cocktail that
brings the temperature up to something comfortable). Don’t forget that Mars—which is half the radius of
Earth—has weaker gravity than Earth.
a. If using the entire surface for materials, how deep must one process the soils to extract enough oxygen
for the job?
b. How much energy will it take to break the oxygen bonds, and how long if sunlight is used as the energy
source?
Part a: On Earth, about 20% of the atmosphere, whose total pressure is about 105 Pa (N/m2 ) is O2 . The
ideal gas law has P = nkT , where n is the number density of particles. If T is made to be the same on Mars,
then the number density, and thus mass density of oxygen will be the same for both. 20% of the 1.3 kg/m3
air density on Earth is 0.26 kg/m3 , so this is what we seek on Mars for the almost pure-oxygen atmosphere.
But what is the effective (scale) height of the atmosphere for Mars? Assuming a similar density of Earth
and Mars, the surface gravity will scale like R, since mass scales like R3 and gravity scales as M/R2 . Being
half the diameter, the surface gravity is half that on Earth, or 5 m/s2 .We seek a total surface pressure of
2 × 104 Pa, so the column of oxygen over a 1 m2 patch of surface has a weight of mg = 2 × 104 N, or
m = 4 × 103 kg. At a density of 0.26 kg/m3 , this occupies 16 × 103 m3 , extending 16 km above our 1 m2
patch. This is the scale height—the profile will actually be exponential with 16 km as the characteristic
scale.
Okay, so we need to extract 4 tonnes of O2 for each square meter of surface. If we assume a sand-like
composition (sandstone, quartz), we’ll get all our oxygen from SiO2 , with a molecular weight of 64 g/mol,
exactly half of which is oxygen. So we need 8 tonnes of sand per square meter. Water is 1 tonne per cubic
meter, and we can guess that sand will be about twice this. So we need to dig 4 m deep across the entire
surface of Mars to put enough oxygen in the atmosphere. It’s a mind-boggling excavation project on a
planetary scale. Don’t expect your plants to like the pure-oxygen atmoephere, though.
Part b: How about the energy required? If we assume 4 eV per bond broken, for each square meter we need
to break 8 × 106 g/64 g/mol = 1.25 × 105 mol, coming to 1.25 × 105 · 6 × 1023 · 6.4 × 10−19 J, or 5 × 1010 J of
energy input (not counting excavation energy).
Mars is about 50% further from the sun than Earth. So our solar constant of ∼ 1000 W/m2 becomes
500 W/m2 for Mars. An average square meter on Mars receives one fourth of this due to geometry (surface
vs. projected area), and some further loss is associated with the substantial lofted dust in the Martian
atmosphere, so we’ll use the round number of 100 W/m2 for the available input. Let’s optimistically assume
we can use this at 50% efficiency, so we have 50 W of usable energy on our square meter patch.
It will then take 5 × 1010 J/50 W = 109 s, or 30 years to harvest the necessary energy. But wait! Where did
the solar panels come from? From the leftover silicon, of course! But hold-on: which came first?
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Photosynthesis
a. Pick a familiar plant or tree and estimate its photosynthetic efficiency (by doing more than guessing
the final answer).
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b. Ethanol from corn is optimistically 20% over break-even, meaning that 1.2 J of energy is derived from
1.0 J of input energy (mechanized agriculture, harvest, processing to ethanol). If corn were used to
completely supplant oil—including the energy input needed to run the enterprise—how much land area
would this take for the U.S.? It is instructive to draw this on a map of the U.S. for scale.
Part a: Let’s pick a tree as our example. After 50 years, a fast-growing tree (plenty of water, nutrients)
might be 40 m tall, have a trunk 0.4 m in diameter, and have a foliage area of 300 m2 (10 m across by 30 m
high). The trunk splits into branches higher up, but let’s simplify by saying all the wood is equivalent to
a cylinder, with a volume of 40 m × 0.1 m2 , or 4 m3 . Add some more for the root system, or 5 m3 . Dry
wood might have a density of 700 kg/m3 , and burns at 4 Cal/g = 1.6 × 107 J/kg. This comprises an energy
content of about 5 × 1010 J, or 109 J per year, on average.
The solar exposure of the tree in full sun is 1000 W/m2 × 300 m2 = 3 × 105 W. The tree cleverly avoids
significant projection effects by being a 3-D shape (unlike a solar panel), so the exposure is approximately the
same at sunrise as it is at high noon–though geometry changes can effect maybe a factor-of-two difference.
But let’s say this is a deciduous tree, operating half the year, and half of the 24 hour day. So the tree receives
one fourth of 3 × 105 W × 3 × 107 s, coming to 2.5 × 1012 J of input. The tree held on to 109 J, translating
to 0.04% efficiency.
The leaves shed each year should also be counted. A blanket of leaves perhaps 5 thick is laid out under the
tree, in an area of about 100 m2 . The 5 number is essentially the same as the number of leaves one would
pierce if shooting a bullet straight up through the tree, and related to the probability (fractional area) of
clear shots through the tree, where one would see the sky (e−5 ≈ 0.6%). Dry leaves are perhaps twice as
thick as paper, so the compressed thickness would be the equivalent of 10 sheets of paper, about 1 mm thick
(a ream of 500 sheets is about 5 cm thick). So the compressed leaf volume is 100 m2 ×0.001 m = 0.1 m3 , with
a mass around 150 kg (leaves sink). This seems a bit heavy, but in any case note that we add 150 kg/year
to the bulk mass of the tree, which was 5 m3 after 50 years, or also 0.1 m3 per year (70 kg). So crudely
speaking, the lost leaves may effectively double the photosynthetic efficiency to the neighborhood of 0.1%.
How about an energy performer like a potato? With a leafy footprint of maybe 0.5 m2 and production of half
a dozen 1-pound potatoes (total about 3 kg) at 4 Cal/g in a growing season, we would get about 400 W/m2
during the 1.5 × 107 s season, so 3 × 109 J, making 5 × 106 J of spuds. This is about 0.2%. Add the leafy
mass, and maybe push this to 0.4%.
Part b: An Iowa cornfield—tended and fertilized as it is—operates at 1.5% photosynthetic efficiency. In the
growing season of 107 s, each square km of corn converts a dose of 0.015×200 W/m2 × 106 m2 × 107 s = 3 × 1013 J
into chemical energy, the 200 W/m2 accounting for day/night, geometry (our square km is flat and 2-D),
and weather. If we need to generate 6 units of energy on the field for every 1 we end up using as ethanol
fuel, then 5 × 1012 J will be available in the form of ethanol each year for each square kilometer of land.
We get 40% of our 10,000 W personal power budget from oil. Thus 4,000 W per person. The U.S., with
3 × 108 ppl, demands 4 × 1019 J of oil energy per year. So we need about 107 km2 , or 3,000 km on a side.
Bear in mind we used optimistic numbers: 1.5% may be high, and only the corn kernels contribute: not the
stalks/leaves/cobs/husks. The 20% gain is also on the optimistic side.
Plotted on a map of the U.S., the 3000 km square is alarming. The corn belt is much smaller than this.
Do we have enough water to convert the rest of the land to corn productivity? What will we eat? Sugar
cane does a much better job: some say 6:1 rather than 1.2:1. So we would only need one fifth the land as
we would for corn ethanol: still 1400 km square. And only Florida and Hawaii have the climate (for now!).
Step back and appreciate what you’ve done: with very little input, you’ve order-of-magnituded your way
into casting doubt on the corn industry aspirations. Might be handy before Congress.
4
Energy Storage
Many of our options for renewable energy are intermittent in nature, yet our society wants to pump along
without interruption. So in order to implement renewables on a large scale, we have to implement storage
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schemes. Imagine you want to store energy locally at your own home (will give you a sense of individual
scale), and you want to explore alternatives to batteries. For scale, a rechargeable NiMH AA battery is
typically rated at 2000 mA-hr at 1.5 V, and a golf-cart battery holds 150 A-h at 12 V. You contemplate three
storage media:
a. 1. A flywheel: given reasonable constraints on your own personal flywheel, how many batteries (of each
type) can you imagine supplanting?
b. Pumped storage: you erect a tank to hold water at some height, storing gravitational potential energy.
How much (in battery units) can you practically achieve?
c. Compressed air: high pressure tanks into which you can pump air. Careful to realize that some heating
in the compression means lost energy. How many batteries can you practically replace?
Now put this in context: how long will each of these storage devices run your house’s electricity demand, not
counting your natural gas and gasoline energy requirements?
For all three forms, I will dedicate a bedroom-sized volume to the endeavor, roughy 3 m across, and 2 m
high. A lead acid golf-cart battery has 12 V times 150 Ah of energy, or 1.8 kWh, with a hefty mass of 40 kg,
occupying about 0.015 m3 . 1 kWh is 1000 W for 3600 s, or 3.6 MJ.
4.1
Flywheel
The flywheel can be made of steel, with a diameter of 3 m and a height of 2 m. The mass will be about
8000 kg/m3 times πR2 h = 15 m3 for a total of 120 tons. Are we sure we want to do this? The moment of
inertia for a disk is 21 M R2 , or ∼ 105 kg·m2 . The fastest engines run at several thousand rotations per minute,
so let’s set an ambitious 3600 RPM, or 60 cycles per second, for an angular frequency just shy of 400 radians
per second. But this puts the outer radius whipping around at more than 500 m/s, which is supersonic!
Let’s scale back to 200 rad/sec to stay subsonic. The energy stored in the rotation is 12 Iω 2 = 2 GJ ≈
500 kWh. Pretty big haul! This would be equivalent to just over 300 lead-acid batteries with a mass of 10
tons and a volume of 5 m3 . So compared to batteries, the flywheel has slightly larger volume, more mass,
cheaper material, longer lifetime, but oh so dangerous in the house (especially when it comes to earthquakes).
Materials strengths also become relevant at these centripetal accelerations, which would need to be evaluated.
The flywheel might have to be in a vacuum and on some pretty amazing bearings. If the relevant parameters
in estimating the drag force in the presence of air are kinematic viscosity, ν, air density, speed at the
edge, area of the edge, and thickness of the air layer around the flywheel, then dimensional play (while
respecting how things ought to scale) would have Fd ∼ ρνvA/t. I have no idea if this is right. Throwing in
relevant parameters, with t = 5 cm yields a drag force of about 2.5 N. Acting on the 300 m/s surface, this
constitutes 750 W of power (1 hp) to keep the flywheel spinning, setting a timescale for rotation decay at
500 kWh/0.75 kW, or 700 h, or a month.
4.2
Pumped Storage
Using a cylindrical tank 3 m in diameter and 2 m high, we have our trusty 15 m3 , or 15 tons of water.
We build a derrick on which to place the tank, 10 m above the ground. We can therefore store 1.5 MJ of
gravitational potential energy, or about 0.5 kWh. Disappointing. A golf-cart battery stores three times more
energy. We coud get the same with a gross of AA batteries. Look for deals at Home Depot.
4.3
Compressed Air
Now we turn our bedroom space into a high-pressure tank. Most such tanks hold 3000 p.s.i., or about
200 atm of pressure. We’ll go with a roughly cylindrical shape for structural integrity, this time with domed
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bottom and top, but we’ll keep our now-familiar 15 m3 volume. Though the system undoubtedly will not
work this way, we can assess the energy required to compress the gas by taking a piston in a long cylinder
and compressing the volume by a factor of 200, ending up with 15 m3 at 200 atm. We will use our friend
P V = N kT . For amusement, what is N when P = 1.013 × 105 Pa, T = 273 K, and V = 0.0224 m3
(k = 1.38 × 10−23 J/K)? When the piston of area A pushes a distance dx against pressure P , the work done
is force times distance, or W = P Adx = P dV = N kT dV /V . We can integrate this in V from V0 to Vf ,
realizing that dV is negative, arriving at ∆E = N kT ln VV0f = P0 V0 ln VV0f if we hold the temperature constant.
In this case, V0 /Vf = 200, P0 = 105 Pa, and V0 = 3000 m3 . We get an energy of 1.6 GJ, or about 450 kWh,
or about 250 golf-cart batteries.
But when we compress the gas adiabatically, it will heat up. The rule for adiabatic processes is that no heat
flows in or out, and P V γ = constant, where γ = 57 for air. So our P dV integrand becomes P0 V0γ V −γ dV ,
integrating to P0 V0γ (1 − γ)−1 V 1−γ , becoming (γ − 1)−1 P0 (V0γ Vf1−γ − V0 ), realizing again that dV is negative.
This works out to 5.6 GJ of input. The first calculation would be valid if we let the heat escape (nonadiabatic), keeping T constant. In the adiabatic case, the gas heats up to 2300 K, and the pressure goes up
to more than 1500 atm (boom). This aside, we put in 5.6 GJ to keep 1.6 GJ, so the adiabatic system is less
than 30% efficient, when done quickly.
Wikipedia claims that if done slowly enough, the resulting isothermal compression can approach 100%
efficiency. I’m dubious: where does the thermal energy go, if not out to the environment? If we compress
slowly enough, the gas can cool and remain at its initial temperature. We can express the efficiency of the
adiabatic compression as the energy we keep divided by the energy put in, defining r ≡ V0 /Vf in the process.
The previous two paragraphs have the relevant pieces, and we find
ε=
(γ − 1)r1−γ ln r
(γ − 1)P0 V0 ln r
1
=
≈ 1 − (γ − 1)δ,
γ 1−γ
1−γ
1−γ
1
−
r
2
P0 V0 Vf
(1 − r
)
where the last step is what happens for a small step in volume, where r → 1 + δ (careful: need second order
epansion in δto get this right). As δ → 0, the efficiency approaches unity. For air, with γ = 1.4, ε ≈ 1 − 0.2δ.
So this looks favorable, but lots of little steps pile up, and the total efficiency for N steps is εN . Trying
numerically for r = 2 via 69 steps of 0.01, I get the same answer: ε69 (r = 1.01) ≈ ε(r = 2) ≈ 0.87. So I
don’t buy the Wikipedia claim. It’s inefficient.
It is also interesting to point out that pressure has the same units as energy density. An overpressure of
200 atm is 2 × 107 Pa, or 20 MJ/m3 . At 15 m3 , this is 300 MJ. Not immediately useful.
4.4
Context
Electrcity generation accounts for about a third of the energy use in the country (the other contributions are
transportation, industry, and residential direct energy (heating). Since the average American runs through
energy at a rate of 10,000 W, we can say that 3000 W of electricity is produced for each individual. If a third
of this is in the home (the rest in retail, industry, government), then the average American uses 24 kWh
in a day. This turns out to be about the right number per household. The scary flywheel would last 80
days. The water tower would last 30 minutes. The compressed air would last a little less than 20 days.
Practically speaking, we probably need a week’s worth of storage with no new input, so we could scale down
the flywheel substantially, forget about water tanks, deal with a 5 m3 compressed storage, or use about 90
golf-cart batteries. One note about the flywheel: 750 W to keep it up to speed is 18 kWh per day, and
comparable to the household electricity use without the flywheel.
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