Order-of-Magnitude Physics – Solution Set 5

Andrew Myers, Eugene Chiang, Ryan Olf, and Alwin Mao

Fall 2014

1 Planet Formation

Consider two identical spheres having radius s, elastic modulus M , and density ρ. Theycollide at relative velocity v.

When they collide, they deform: see Figure 1, which shows the moment of maximumdeformation.

(a) Derive an order-of-magnitude formula for r, the radius of the “smooshed cap”, interms of the given variables. Note that r 6= ∆s.

Hint: The (dimensionless) strain is the gradient of the displacement, not the displace-ment itself. There is more than one strain in this problem. Select the maximum gradientas that will dominate the stress.

I can identify two strains. The first is ε1 ∼ ∆s/s (radial strain). The second is ε2 ∼ ∆s/r(strain along the chord). The second dominates because r ∼

√s∆s (from geometry 1) and so

ε2 ∼ ∆s/r ∼√

∆s/s > ε1.I’ll do this using force balance. The stress is ∼Mε2. The force is ∼Mε2r

2 ∼ M√

∆s/s× s∆s.This force brings the two spheres to rest. Each sphere carries ∆p ∼ mv ∼ ρs3v momentum, whichgets zero’d out over ∆t ∼ ∆s/v time (the time to smoosh). So we have:

M

√∆s

ss∆s ∼ ∆p

∆t∼ ρs3v

∆s/v. (1)

Solve for ∆s ∼ (ρv2/M)2/5s whence r ∼ (ρv2/M)1/5s .

This theory of colliding elastic spheres is called “Hertz’s law of contact”.

(b) Solids have surface tension, too, because they have unsaturated bonds on the surfacelooking for partners to bind with. Laboratory measurements give a surface tension γ forsilicate similar to that of water.

Derive an order-of-magnitude formula for the critical speed vcrit below which the spheresstick upon impact, and above which they bounce. Express your formula in terms of allthe variables given.

1r =√r2, r2 = s2 − (s − ∆s)2 = s2(1 − (1 − ∆s

s)2).∆s

sis smaller than 1 so we can Taylor expand this. Thus,

r2 ∼ s2(1− (1− 2 ∆ss

)) = s2(2 ∆ss

) ∼ s∆s so r ∼√s∆s

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Evaluate vcrit in cm/s for s = 1µm and s = 1 m and thereby develop a sense for howfar particle-particle sticking by chemical bonds can take you down the road to planetformation.

If you wish, you can check your formula against what Chokshi et al. (1993, AstrophysicalJournal, v407, 806) found. Your answer should match in the scalings but is not expectedto match in the numerical pre-factor.

Surface tension has units of energy per area, so we can calculate a binding energy associatedwith the cap: BE ∼ γr2 ∼ γ(ρv2/M)2/5s2. If this surface energy exceeds the initial kinetic energyof the spheres, KE ∼ mv2 ∼ ρs3v2, then I figure they’ll stick. Setting BE ∼ KE we solve for

vcrit ∼ γ5/6/(s5/6M1/3ρ1/2) . For s = 1µm, γ ∼ 102 erg/cm2, M ∼ 1012 erg/cm3, and ρ ∼ 2 g/cm3,

we get vcrit ∼ 7 cm/s. For s = 1 m, vcrit ∼ 7 × 10−5 cm/s — which is ludicrously slow. So themoral is that small particles stick much more easily than big ones.

Equation (28) of Chokshi et al. confirms our answer, with a claimed numerical prefactor of3.86.

(Historical note: I had derived this scaling on the bus once and was all excited to write it upinto a short note. After I got into office and fired up ADS, I was crestfallen to find Chokshi hadbeat me to it by 20 years.)

2 “If Frogs Can Fly...”

“... then there is no reason why John Major cannot be Prime Minister.” — Reactionof the British press after researchers at the Nijmegen High Field Magnet Laboratory inthe Netherlands announced they had successfully levitated a 1-cm diameter baby frog intheir Bitter magnet.

A Bitter magnet (after the physicist Francis Bitter) is essentially a solenoid where theusual coil of wire is replaced by vertically thin, radially wide plates stacked helically—like a continuous, tightly wound, spiral staircase. You can see pictures of Bitter platesin slideshows on http://www.magnet.fsu.edu/mediacenter/slideshows/index.html, cre-ated by the National High Magnetic Field Lab at Florida State University.

Bitter plates are punched with holes so that water can cool them when current flowsthrough the plates. The bore hole (axis) of the solenoid is left empty for researchers’test subjects. In the case of Andre Geim’s experiments, subjects have included waterdroplets, hazelnuts, mice, and pizza.2

(a) To acquire intuition about magnetic fields, consider the analogous and hopefullymore familiar case of electric fields.

Consider two charges, e and −e, fixed to a rigid rod of length d. This electric dipole isplaced in an electric field ~E(~x) that varies with position ~x. Assume that the field variesover a lengthscale � d.

2In 2000, Andre Geim (of graphene and Scotch tape fame) was awarded the Ig Nobel Prize for his levitationexperiments, sharing it with Sir Michael Berry (of Berry phase fame) who provided the underlying theory.

2

�s

s

2r

Figure 1: Two smooshed spheres. Notwithstanding the author’s ineptitude with Keynote, bothspheres are smooshed inward by ∆s. The radius of the smooshed, flattened cap is r.

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Write down the net force ~F felt by the electric dipole. For simplicity you may orientthe dipole so that it is parallel with the field. Your expression need only be good toorder-of-magnitude.

Hint: The grad operator is your friend.

If the electric field were strictly uniform, then there would be no net force.But the electric field is not uniform. Say the force on the negative charge is −eE(x). The force

on the positive charge is eE(x+ d) ∼ e[E(x) + (∂E(x)/∂x)d] using a first-order Taylor expansion.

Adding up these forces gives ed(∂E/∂x) or ~F ∼ ed∇E . The combination ed ∼ µ is called theelectric dipole moment. Full credit will be awarded regardless of the sign of your answer (“signsare for the birds — unless you’re doing stability analysis, in which case signs are everything”).

(b) By analogy, write down an expression for the net force ~F felt by a magnetic dipoleof moment ~µ immersed in a spatially varying field ~B(~x).

Working by analogy, ~F ∼ µ∇B , where µ is the magnetic dipole moment.

(c) Model the Nijmegen magnet as follows. The plates are composed of copper, havean inner radius of R1 ≈ 2 cm, an outer radius R2 ≈ 20 cm, and are stacked such thatthere are n ≈ 10 turns per cm. The length of the solenoid is L ≈ 30 cm. The magnet isoriented so that its axis points parallel to Earth’s gravity.

Decide where the frog can levitate (top, middle, or bottom of solenoid). Estimate themagnetic field required to levitate the frog. Give both a symbolic answer and a numericalestimate in Teslas.

If you want to float a test magnet at the top of the solenoid, and the solenoid field is oriented

so that its North pole points vertically up (opposite to gravity), then you have to orient your testmagnet so that its North pole points down. Then the two North poles are aimed at each other,and the repulsive magnetic force can balance gravity. Diamagnetic materials — like the frog —have induced fields that automatically point in the opposite direction as the applied external field,so you automatically induce a S-N magnet when you place it in a N-S external field.

You can’t get levitation at the bottom of the solenoid, because the induced field will just forcethe frog out: the repulsive magnetic force points in the same direction as gravity. And you can’tget levitation in the middle of the solenoid, because there’s no gradient in the external B in themiddle, and our formula for magnetic force F = µ∇B tells us that it’s the gradient that matters.

To estimate µ for the frog, recall from lecture that the magnetic field on the surface of aspherical/cubical dipole is B ∼ µ/r3, where r is the size of the sphere/cube. So we have µ ∼Br3 ∼ B × Volume (for a spherical/cubical frog). Now for a diamagnetic material, B ∼ χBapplied

where χ ∼ −1 × 10−5 is the magnetic susceptibility. So putting it all together, the force F ∼χBapplied× Volume×∇B ∼ χBapplied× Volume×Bapplied/L where L is the length scale over whichthe external field varies. Let’s take L ∼ 10 cm as this is a characteristic size for the solenoid’scircular plates, i.e., the characteristic lengthscale of the “fringing fields” at the top of the solenoid.

The frog will levitate if the magnetic force is greater than its weight W = ρg × Volume. So

when we set F ∼ W , the Volume cancels out, and we get Bapplied ∼√ρgL/χ which for ρ ∼ 2

g/cm3 and L ∼ 10 cm gives 40000 G = 4 T . The actual field was closer to 10 T.

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(d) How much power is dissipated in the electromagnet? State in Watts.

In cgs units, the magnetic field inside a solenoid is given by B = 4πnI, where n is the numberof turns per length and I is the current. Plugging in, a B = 40000G field with n ∼ 10/cm requiresI ∼ 300 statamps or 3000 Amps.

Driving that much current dissipates a lot of power. To figure out how much, we need toestimate the resistance of the Bitter coil. Purcell’s cheat sheet says that the resistivity of copper is2 · 10−8 Ohms per meter. I’ll eat my pride and do the rest of the calculation in SI units. From thedimensions of the coil given in the problem, we can estimate that the cross sectional area is

A ∼ height of each turn× radial thickness = 10−1 cm× (20− 2) cm ≈ 1.8 cm2 ≈ 1.8 · 10−4 m2.

The total length of wire is approximately

L ∼ number of turns× circumference of 1 turn ∼ 300× 2π · 20 cm ≈ 40000 cm = 400 m

and the power is dissipated is then

P = I2R = I2ρL

A= (3000 amps)2 · (2 · 10−8 Ohm m) · 400 m

1.8 · 10−4 m2

or 4 · 105 W or about the power consumption of 200 Americans or 800 Chinese3 —just to lift onemeasly frog!

(e) Imagine now a Bitter magnet large and powerful enough to levitate you. Repeat (c)and (d). Is human levitation practical?

Note that the mass of our frog does not appear in answer for part (c), because both the massand the number of dipoles scale with Volume. At first you might think this means it would alsotake about 4 Tesla to lift a human, but remember that to get B we needed to use the length scaleover which B changes appreciably, which we said was comparable to the inner/outer radii of themagnet. But that distance will have to be much larger if we want to levitate a human being—maybe 2 meters instead of 2 cm. In other words, our magnet needs an inner hole radius that is atminimum 100 times bigger. This increases L by a factor of 100, and because B ∝ L1/2, we wouldneed a magnetic field about 10 times stronger or about 40 T .

To estimate the new power, let’s say the inner cylinder radius is now 200 cm = 2 m and theouter cylinder radius is 400 cm = 4 m — notice we’re changing the outer radius by only a factorof 20 (if we changed everything by a factor of 100, the device would be gargantuan). Let’s alsoadjust the height of the magnet so that it’s 300 cm = 3 m and can fit a person (maybe curledup into a ball). We’ll also assume that the number of turns per length n is the same. Then Agoes up by a factor of 10, and L goes up by a factor of 10 (from the height of the magnet) × 20

(from the increased outer circumference) = 200, and so the resistance is about 20 × bigger . Thecurrent goes as the magnetic field, and the power goes as the current squared, so at the end of the

day the power goes up by 102 × 20 or to 8× 108 W or nearly 1 GW. We’d need big nuclear or

coal power plant just to lift one person! So no, this isn’t practical.

3http://en.wikipedia.org/wiki/List of countries by electricity consumption

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These are the same numbers given in Andrey Geim’s (very interesting) Physics Today article.4

For the lowdown on a more practical form of levitation, check out this site5 about supercon-ducting magnetic trains. The problem with these is that the magnets must be cooled with liquidhelium or nitrogen, which greatly adds to the cost.

3 Watering Houseplants on Vacation

From Sterl Phinney:

A open-ended circular pipe of radius r is tilted at an angle of 45◦ to the vertical. Wateris supplied at the top to replace what gravity pushes out the bottom.

For all parts, provide symbolic expressions in addition to numerical estimates.

(a) What is the maximum pipe radius rmax for which the flow will be laminar, notturbulent, assuming the flow is driven purely by gravity and not pressure? You can watchG.I. Taylor’s “Low Reynolds Flow” video for guidance — assume a critical Reynoldsnumber of 2000 for the transition between laminar and turbulent flow.

For gravity-driven flow in a pipe, we have g sin 45◦ ∼ ν∇2v (gravity balances viscous accelera-tion), or v ∼ 0.7gr2/ν after approximating ∇ ∼ 1/r2. The Reynolds number is vr/ν ∼ 0.7gr3/ν2.

Set this equal to the critical Reynolds numberRecrit ∼ 2000 and solve for rmax ∼ (Recritν2/(0.7g))1/3

or 0.07 cm or 0.7 mm .

(b) To keep the pipe supplied with water, the top end is stuck into a hole in the bottomof a cup. Suppose the pipe has radius rmax and is L = 5 cm long.

There is a depth z∗ of water in the cup below which the rate of outflow from the pipedoes not depend on the depth in the cup, and above which the rate begins to increasesignificantly. Estimate z∗.

The greater the depth of water z in the cup, the greater the hydrostatic pressure ρgz at thebottom of the cup, and the greater the pressure gradient across the pipe of length L. The pressuregradient acceleration in the Navier-Stokes equation is ∇P/ρ ∼ ρgz/L/ρ ∼ gz/L, which we want tocompare to the gravitational acceleration g (dropping the annoying sin 45◦ which is really crampingmy style). Then the transition between pressure-driven flow and gravity-driven flow occurs at

gz∗/L ∼ g or z∗ ∼ L ∼ 5 cm .

(c) How long would it take to empty the cup from depth z∗?

The mass flow rate through the pipe is M ∼ ρvr2 ∼ ρ(gr2/ν)r2 ∼ ρRe4/3critν

5/3/g1/3. The massof water in the cup — which we will assume to be a standard coffee-sized mug — is ρπR2z∗ whereR ∼ 2 cm is the assumed radius of the mug and z∗ ∼ L ∼ 5 cm. Then the timescale to empty the

cup is M/M ∼ πR2z∗Re−4/3crit ν−5/3g1/3 ∼ 54 s or 1 minute .

4http://www.ru.nl/hfml/research/levitation/diamagnetic/5http://www.fzu.cz/ jirsa/htm/principles.htm

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(d) What changes would you make to this device to water your houseplant while awayon vacation?

We need to make sure we give the plant enough water but not too much water: say a coffeemug of water a day for a 7-day vacation.

We will have maximum flexibility by moving to a pressure-driven system rather than a gravity-driven system. So ∇P/ρ ∼ ρgz/L/ρ ∼ gz/L needs to dominate g sin θ. We tilt the pipe so that θ issmall enough that this is always true — heck we can just make θ = 0 and kill off gravity altogether.6

Then the Poiseuille flow relation we derived in class holds: ∇P/ρ ∼ gz/L ∼ νv/r2 which impliesv ∼ r2gz/(Lν). We want M ∼ ρvr2 ∼ ρgzr4/(Lν) to be a coffee mug a day. For z ∼ z∗ wewere getting a coffee mug a minute, so to slow this down by a factor of 1 day / 1 minute ∼ 1000,

we can decrease the radius of the pipe by 103/4 ∼ 5.6, down to 0.1 mm . Or we can get away with

slightly less of a reduction in r by also increasing L — but increasing L alone is impractical , since

we would have to increase it by a factor of 1000 (from 5 cm to 5000 cm = 50 m).Now we’ve emptied the coffee mug in 1 day — but we have 6 more days of vacation. So unless

we want to run back home and re-fill the mug, we need to supply more water: not by increasingthe depth of the water above z∗ (because that would increase the flow rate and we already went tothe trouble of reducing it) but by increasing R—the radius of the reservoir of water. We replace

the coffee mug with a wider container , one whose R is say 6 cm instead of 2 cm, so that M is

increased by 32 ∼ 9, which means we can even extend our vacation by a day or two. Who saysorder-of-magnitude physics can’t be used for both profit and pleasure?

6Some systems make θ negative and utilize capillary action against gravity, without pressure. Although that’soutside the scope of this problem, this problem illustrates why capillary action can sometimes be easier to use forlonger vacations.

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