Physics · Book 2 · Grades 10–12

High School Physics

High School Physics · Grades 10–12

13The Fundamental Interactions

Rub a balloon on your hair and it picks up scraps of paper — beating, with a few square centimetres of rubber, the gravitational pull of the entire planet. That easy victory is why atoms hold together and why heavy nuclei eventually shatter (Chapter 19). The universe runs on exactly four interactions; this chapter meets them, weighs them against each other, and assigns each its floor.

13.1 Matter from quarks to galaxies

Definition 13.1 (The structure ladder)

Matter is built in floors, each assembled from the one below: three quarks (pointlike, smaller than 10−18 m10^{-18}\,\mathrm{m}) bind into a nucleon — proton or neutron, about 10−15 m10^{-15}\,\mathrm{m} across; nucleons pack into a nucleus (10−15 m10^{-15}\,\mathrm{m} to 10−14 m10^{-14}\,\mathrm{m}); a nucleus plus its electron cloud is an atom (10−10 m10^{-10}\,\mathrm{m}); atoms bind into molecules and crystals (10−9 m10^{-9}\,\mathrm{m} up), which assemble into everyday objects (1 m1\,\mathrm{m}), planets (107 m10^{7}\,\mathrm{m}), planetary systems (1011 m10^{11}\,\mathrm{m}), galaxies (1021 m10^{21}\,\mathrm{m}) and the observable universe (1026 m10^{26}\,\mathrm{m}).

Definition 13.2 (The four interactions)

A fundamental interaction is a force not reducible to other forces; there are exactly four: gravitation, the electromagnetic interaction, the strong interaction and the weak interaction.

Remark 13.3

The floors are separated by voids: an atom is 10510^{5} times wider than its nucleus — scale the nucleus up to a 1 cm1\,\mathrm{cm} marble and the electron cloud is a kilometre across. Something must hold these sparse floors together, and gravity is almost never it.

The ladder of structure on a powers-of-ten axis: forty-four decades separate the quark from the observable universe. Sizes in metres.
The ladder of structure on a powers-of-ten axis: forty-four decades separate the quark from the observable universe. Sizes in metres.

13.2 Electric charge

Definition 13.4 (Electric charge)

Electric charge is the property of matter on which the electromagnetic interaction acts, as mass is the property gravitation acts on. It is measured in coulombs (C\mathrm{C}) and comes in two signs, positive and negative, which cancel when added; like charges repel, unlike charges attract.

Definition 13.5 (Elementary charge)

Charge is grained: every observed charge is a whole multiple of the elementary charge e=1.60×10−19 Ce = 1.60 \times 10^{-19}\,\mathrm{C}. The proton carries +e+e, the electron −e-e, the neutron 00; a charge qq holds N=∣q∣/eN = \abs{q}/e elementary charges.

Proposition 13.6 (Conservation of charge)

The total charge of an isolated system never changes: charge is neither created nor destroyed, only transferred — in practice by electrons, the lightest carriers.

Proof. Admitted at this level. ∎

Remark 13.7

No violation has ever been observed; the deep reason, a symmetry of electromagnetism, is given in the university volumes.

Definition 13.8 (Charging by friction and by contact)

Rubbing two insulators tears electrons off one and onto the other, leaving opposite charges of equal magnitude (charging by friction); touching a charged object to a neutral one shares the charge (charging by contact). Only electrons move; the total charge is conserved.

Example 13.9 (Counting electrons)

A balloon rubbed on hair acquires q=−1.6×10−8 Cq = -1.6 \times 10^{-8}\,\mathrm{C}: it has gained N=1.6×10−8/1.6×10−19=1.0×1011N = 1.6 \times 10^{-8} / 1.6 \times 10^{-19} = 1.0 \times 10^{11} electrons, and the hair is left with exactly +1.6×10−8 C+1.6 \times 10^{-8}\,\mathrm{C}.

13.3 Coulomb’s law

Theorem 13.10 (Coulomb’s law)

Two point charges q1q_1 and q2q_2 a distance dd apart exert on each other forces along the line joining them, of equal magnitudes and opposite directions — repulsive for like signs, attractive for unlike signs — with

F=k ∣q1q2∣d2,k=8.99×109 N m2/C2.F = k\,\frac{\abs{q_1 q_2}}{d^{2}}, \qquad k = 8.99 \times 10^{9}\,\mathrm{N}\,\mathrm{m}^{2}/\mathrm{C}^{2}.

Proof. Admitted at this level. ∎

Remark 13.11

Coulomb established the law in 1785 by measuring the twist of a thin torsion fibre; why the exponent is exactly 22 is answered in the university volumes.

Like charges repel (left), unlike charges attract (right); the two forces always have equal magnitudes and opposite directions. Like charges repel (left), unlike charges attract (right); the two forces always have equal magnitudes and opposite directions.
Like charges repel (left), unlike charges attract (right); the two forces always have equal magnitudes and opposite directions.

Remark 13.12 (A formal twin of gravitation)

Coulomb’s law is, symbol for symbol, the law of universal gravitation (Chapter 4) with charges in place of masses:

sourcemass mmcharge qq
lawF=G m1m2d2F = G\,\dfrac{m_1 m_2}{d^2}F=k ∣q1q2∣d2F = k\,\dfrac{\abs{q_1 q_2}}{d^2}
[1.5ex] constantG=6.67×10−11 N m2/kg2G = 6.67 \times 10^{-11}\,\mathrm{N}\,\mathrm{m}^{2}/\mathrm{kg}^{2}k=8.99×109 N m2/C2k = 8.99 \times 10^{9}\,\mathrm{N}\,\mathrm{m}^{2}/\mathrm{C}^{2}
signalways attractiveattractive or repulsive

The two differences drive everything below: electricity is overwhelmingly stronger, and it alone can cancel itself.

Example 13.13 (Two protons, both forces at once)

Two protons (mp=1.67×10−27 kgm_p = 1.67 \times 10^{-27}\,\mathrm{kg}) sit d=1.0×10−15 md = 1.0 \times 10^{-15}\,\mathrm{m} apart — nuclear neighbours. Electric repulsion:

FE=8.99×109×(1.60×10−19)2(1.0×10−15)2≈2.3×102 NF_E = 8.99 \times 10^{9} \times \frac{(1.60 \times 10^{-19})^{2}}{(1.0 \times 10^{-15})^{2}} \approx 2.3 \times 10^{2}\,\mathrm{N}

— the weight of a 2323-kg suitcase, carried by a particle of 10−2710^{-27} kilograms. Gravitational attraction: FG=6.67×10−11×(1.67×10−27)2/(1.0×10−15)2≈1.9×10−34 NF_G = 6.67 \times 10^{-11} \times (1.67 \times 10^{-27})^{2} / (1.0 \times 10^{-15})^{2} \approx 1.9 \times 10^{-34}\,\mathrm{N}. The ratio FE/FG=ke2/(Gmp2)≈1.2×1036F_E / F_G = k e^{2} / (G m_p^{2}) \approx 1.2 \times 10^{36} is distance-independent (d2d^{2} cancels): between elementary particles, gravity is 103610^{36} times too weak to matter — at any distance.

13.4 The strong and weak interactions

The example leaves a scandal: a helium nucleus holds two protons that repel with tens of newtons — and yet helium exists.

Definition 13.14 (The strong interaction)

The strong interaction binds quarks into nucleons and nucleons into nuclei. At 10−15 m10^{-15}\,\mathrm{m} it is attractive and far stronger than the electric repulsion (thousands of newtons between nucleons); beyond a few 10−1510^{-15} metres it vanishes: its range is about 10−15 m10^{-15}\,\mathrm{m}. It grips protons and neutrons alike and ignores charge.

The tug-of-war inside a nucleus: at 10-15 m the strong attraction (thousands of newtons) beats the electric repulsion.
The tug-of-war inside a nucleus: at 10−1510^{-15} m the strong attraction (thousands of newtons) beats the electric repulsion.

Remark 13.15 (Why nuclei exist — and why heavy ones give way)

The two forces in the nucleus scale differently. The strong glue is short-ranged: a nucleon binds only its few neighbours within 10−15 m10^{-15}\,\mathrm{m}. The electric repulsion is long-ranged: every proton pushes on every other proton. As nuclei grow, repulsion accumulates faster than glue: heavy nuclei need extra neutrons (glue without charge), and beyond about 8080 protons even that fails — the heaviest nuclei live on borrowed time (Chapter 19).

Definition 13.16 (The weak interaction)

The weak interaction, of range shorter still, lets a neutron turn into a proton (and back): it drives the β\beta radioactivity of Chapter 19.

13.5 Who rules which floor

Method 13.17 (Auditing a scale)

To find which interaction rules a structure of size dd, check range and cancellation:

  1. d≤10−14 md \leq 10^{-14}\,\mathrm{m}: the strong interaction rules — it beats the electric repulsion, and gravity is 103610^{36} out.
  2. atoms to mountains (10−10 m10^{-10}\,\mathrm{m} to 104 m10^{4}\,\mathrm{m}): the strong force is out of range; the electromagnetic interaction rules — bonds, cohesion, friction, contact.
  3. planets and beyond (d≥106 md \geq 10^{6}\,\mathrm{m}): matter is neutral to fantastic precision, so electricity cancels itself; mass has one sign and always adds — gravitation rules the sky.

The weak interaction holds no floor: it binds nothing, but arbitrates transformations (β\beta decay).

Who rules where: strong for nuclei, electromagnetic from atoms to mountains, gravitation for planets and beyond. Sizes in metres.
Who rules where: strong for nuclei, electromagnetic from atoms to mountains, gravitation for planets and beyond. Sizes in metres.

13.6 Exercises

Exercise 13.1 ★

Give the order of magnitude of each size and its floor of the ladder: proton 1.7×10−15 m1.7 \times 10^{-15}\,\mathrm{m}; water molecule 2.8×10−10 m2.8 \times 10^{-10}\,\mathrm{m}; grain of sand 2×10−4 m2 \times 10^{-4}\,\mathrm{m}; Earth 1.3×107 m1.3 \times 10^{7}\,\mathrm{m}; Milky Way 9.5×1020 m9.5 \times 10^{20}\,\mathrm{m}.

Solution

Solution of Exercise 13.1.

Proton: 10−1510^{-15} (nucleon floor); water molecule: 10−1010^{-10} (molecule floor — single molecules sit at atomic sizes); sand: 10−410^{-4} (everyday matter); Earth: 10710^{7} (planet); Milky Way: 102110^{21} (galaxy). All in metres.

Exercise 13.2 ★

A balloon rubbed on hair carries q=−4.8×10−9 Cq = -4.8 \times 10^{-9}\,\mathrm{C}. Has it gained or lost electrons, and how many? What is the hair’s charge afterwards, and which law says so?

Solution

Solution of Exercise 13.2.

Negative charge: it gained N=4.8×10−9/1.6×10−19=3.0×1010N = 4.8 \times 10^{-9}/1.6 \times 10^{-19} = 3.0 \times 10^{10} electrons. The hair carries +4.8×10−9 C+4.8 \times 10^{-9}\,\mathrm{C}, by conservation of charge (Proposition 13.6).

Exercise 13.3 ★

Charges q1=2.0×10−6 Cq_1 = 2.0 \times 10^{-6}\,\mathrm{C} and q2=−3.0×10−6 Cq_2 = -3.0 \times 10^{-6}\,\mathrm{C} sit 30 cm30\,\mathrm{cm} apart. Compute the force; attractive or repulsive? Compare the forces on q1q_1 and on q2q_2.

Solution

Solution of Exercise 13.3.

F=8.99×109×2.0×10−6×3.0×10−60.302=5.39×10−20.090≈0.60 NF = 8.99 \times 10^{9} \times \dfrac{2.0 \times 10^{-6} \times 3.0 \times 10^{-6}}{0.30^2} = \dfrac{5.39 \times 10^{-2}}{0.090} \approx 0.60\,\mathrm{N}, attractive (unlike signs). The two forces have the same magnitude, opposite directions.

Exercise 13.4 ★

Two charges attract with a force F0F_0. What does it become if: the distance doubles; the distance is divided by 33; one charge doubles; both charges and the distance double?

Solution

Solution of Exercise 13.4.

F0/4F_0/4; 9F09F_0; 2F02F_0; F0F_0 (numerator ×4\times 4, denominator ×4\times 4).

Exercise 13.5 ★

Name the interaction responsible for: (a) the Moon orbiting the Earth; (b) the cohesion of a salt crystal; (c) the cohesion of a helium nucleus; (d) the β\beta decay of carbon-14; (e) the balloon of Exercise 13.2 sticking to a wall.

Solution

Solution of Exercise 13.5.

(a) gravitation; (b) electromagnetic; (c) strong; (d) weak; (e) electromagnetic.

Exercise 13.6 ★★

The two protons of a helium nucleus sit d=2.0×10−15 md = 2.0 \times 10^{-15}\,\mathrm{m} apart (mp=1.67×10−27 kgm_p = 1.67 \times 10^{-27}\,\mathrm{kg}). Compute their electric repulsion, their gravitational attraction, and the ratio. What keeps the nucleus whole?

Solution

Solution of Exercise 13.6.

FE=8.99×109×(1.60×10−19)2/(2.0×10−15)2=2.30×10−28/4.0×10−30≈58 NF_E = 8.99 \times 10^{9} \times (1.60 \times 10^{-19})^2/(2.0 \times 10^{-15})^2 = 2.30 \times 10^{-28}/4.0 \times 10^{-30} \approx 58\,\mathrm{N}; FG=6.67×10−11×(1.67×10−27)2/4.0×10−30≈4.7×10−35 NF_G = 6.67 \times 10^{-11} \times (1.67 \times 10^{-27})^2/4.0 \times 10^{-30} \approx 4.7 \times 10^{-35}\,\mathrm{N}. Ratio FE/FG≈1.2×1036F_E/F_G \approx 1.2 \times 10^{36}. The strong interaction holds the nucleus, gravity is irrelevant.

Exercise 13.7 ★★

In a hydrogen atom, the electron (me=9.11×10−31 kgm_e = 9.11 \times 10^{-31}\,\mathrm{kg}) sits d=5.3×10−11 md = 5.3 \times 10^{-11}\,\mathrm{m} from the proton (mp=1.67×10−27 kgm_p = 1.67 \times 10^{-27}\,\mathrm{kg}). Compute the electric and gravitational forces between them and their ratio. What holds atoms together?

Solution

Solution of Exercise 13.7.

FE=2.30×10−28/(5.3×10−11)2=2.30×10−28/2.81×10−21≈8.2×10−8 NF_E = 2.30 \times 10^{-28}/(5.3 \times 10^{-11})^2 = 2.30 \times 10^{-28}/2.81 \times 10^{-21} \approx 8.2 \times 10^{-8}\,\mathrm{N}; FG=6.67×10−11×1.67×10−27×9.11×10−31/2.81×10−21≈3.6×10−47 NF_G = 6.67 \times 10^{-11} \times 1.67 \times 10^{-27} \times 9.11 \times 10^{-31} / 2.81 \times 10^{-21} \approx 3.6 \times 10^{-47}\,\mathrm{N}. Ratio ≈2.3×1039\approx 2.3 \times 10^{39}: the electromagnetic interaction holds atoms together.

Exercise 13.8 ★★

Sphere A carries qA=8.0×10−9 Cq_A = 8.0 \times 10^{-9}\,\mathrm{C}; an identical sphere B is neutral. They are touched together, then separated to 10 cm10\,\mathrm{cm}. What charge does each carry (check the conservation of charge)? Compute the force between them — attractive or repulsive?

Solution

Solution of Exercise 13.8.

By contact the charge is shared: qA=qB=4.0×10−9 Cq_A = q_B = 4.0 \times 10^{-9}\,\mathrm{C} (total still 8.0×10−9 C8.0 \times 10^{-9}\,\mathrm{C}). Then F=8.99×109×(4.0×10−9)2/0.102=1.4×10−5 NF = 8.99 \times 10^{9} \times (4.0 \times 10^{-9})^2/0.10^2 = 1.4 \times 10^{-5}\,\mathrm{N}, repulsive (like signs).

Exercise 13.9 ★★

Two identical charges q=1.0×10−6 Cq = 1.0 \times 10^{-6}\,\mathrm{C} repel with 0.90 N0.90\,\mathrm{N}: how far apart are they? Where does the force drop to 0.10 N0.10\,\mathrm{N}?

Solution

Solution of Exercise 13.9.

d=kq2/F=8.99×109×1.0×10−12/0.90≈0.10 md = \sqrt{k q^2/F} = \sqrt{8.99 \times 10^{9} \times 1.0 \times 10^{-12}/0.90} \approx 0.10\,\mathrm{m}. A force 99 times smaller needs a distance 33 times larger: d=0.30 md = 0.30\,\mathrm{m}.

Exercise 13.10 ★★

Two protons in neighbouring molecules sit d=1.0×10−10 md = 1.0 \times 10^{-10}\,\mathrm{m} apart. Compute their electric repulsion. What does the strong interaction contribute at this distance? Which interaction binds molecules?

Solution

Solution of Exercise 13.10.

F=2.30×10−28/(1.0×10−10)2=2.3×10−8 NF = 2.30 \times 10^{-28}/(1.0 \times 10^{-10})^2 = 2.3 \times 10^{-8}\,\mathrm{N}. The strong interaction contributes nothing: 10−1010^{-10} m is 10510^{5} times its range. Molecules are bound by the electromagnetic interaction.

Exercise 13.11 ★★

Each of two coins holds about 102310^{23} electrons. Move one electron in a million between them and set the coins 1.0 m1.0\,\mathrm{m} apart. Compute each coin’s charge, then the force. The weight of what mass equals it? Conclude on the neutrality of everyday matter.

Solution

Solution of Exercise 13.11.

q=1023/106×1.6×10−19=1.6×10−2 Cq = 10^{23}/10^{6} \times 1.6 \times 10^{-19} = 1.6 \times 10^{-2}\,\mathrm{C} on each coin. F=8.99×109×(1.6×10−2)2/1.02≈2.3×106 NF = 8.99 \times 10^{9} \times (1.6 \times 10^{-2})^2/1.0^2 \approx 2.3 \times 10^{6}\,\mathrm{N} — the weight of 2.3×106/9.81≈2.3×105 kg2.3 \times 10^{6}/9.81 \approx 2.3 \times 10^{5}\,\mathrm{kg}, a loaded freight train, from a millionth of the electrons. Everyday matter must be (and is) neutral to far better than one part in a million.

Exercise 13.12 ★★★

A uranium nucleus has diameter 1.4×10−14 m1.4 \times 10^{-14}\,\mathrm{m} and 9292 protons. Compute the repulsion between two protons at opposite ends. Can the strong interaction bind this pair — why not? How many repelling proton pairs are there? Why do heavy nuclei need extra neutrons, and why beyond some size can even neutrons not save them?

Solution

Solution of Exercise 13.12.

F=2.30×10−28/(1.4×10−14)2=2.30×10−28/1.96×10−28≈1.2 NF = 2.30 \times 10^{-28}/(1.4 \times 10^{-14})^2 = 2.30 \times 10^{-28}/1.96 \times 10^{-28} \approx 1.2\,\mathrm{N} — still enormous at this scale. No: they are 1414 times the strong range apart; each is glued only to its nearest neighbours. Pairs: 92×912=4186\frac{92 \times 91}{2} = 4186, all repelling at any distance, while the glue does not grow with size. Neutrons add attraction without repulsion and dilute the protons; but repulsion grows with the square of the proton number and glue only with the number of neighbours, so beyond about 8080 protons no neutron budget balances the books — such nuclei decay (Chapter 19).

Exercise 13.13 ★★★

Two balls of mass 1.0 g1.0\,\mathrm{g} hang from 50 cm50\,\mathrm{cm} threads tied to the same point. Given the same charge qq, they settle 6.0 cm6.0\,\mathrm{cm} apart. Each ball is in equilibrium under its weight, the thread’s tension and the electric force: show that F=mgtan⁡θF = mg\tan\theta (θ\theta: angle of thread to vertical) and that here tan⁡θ≈0.060\tan\theta \approx 0.060. Compute FF, then qq, then the number of elementary charges moved.

Solution

Solution of Exercise 13.13.

Equilibrium: vertically Tcos⁡θ=mgT\cos\theta = mg, horizontally Tsin⁡θ=FT\sin\theta = F, so F=mgtan⁡θF = mg\tan\theta. Here sin⁡θ=3.0/50=0.060\sin\theta = 3.0/50 = 0.060, and for small angles tan⁡θ≈sin⁡θ=0.060\tan\theta \approx \sin\theta = 0.060. Then F=1.0×10−3×9.81×0.060≈5.9×10−4 NF = 1.0 \times 10^{-3} \times 9.81 \times 0.060 \approx 5.9 \times 10^{-4}\,\mathrm{N}; q=dF/k=0.060×5.9×10−4/8.99×109≈1.54×10−8 Cq = d\sqrt{F/k} = 0.060 \times \sqrt{5.9 \times 10^{-4}/8.99 \times 10^{9}} \approx 1.54 \times 10^{-8}\,\mathrm{C}; N=1.54×10−8/1.6×10−19≈9.6×1010N = 1.54 \times 10^{-8}/1.6 \times 10^{-19} \approx 9.6 \times 10^{10} elementary charges.

Exercise 13.14 ★★★

Data: ME=5.97×1024 kgM_E = 5.97 \times 10^{24}\,\mathrm{kg}, MM=7.35×1022 kgM_M = 7.35 \times 10^{22}\,\mathrm{kg}, Earth–Moon distance d=3.84×108 md = 3.84 \times 10^{8}\,\mathrm{m}. Compute the Earth–Moon gravitational force. Gravity is switched off: what charges +Q+Q on the Earth and −Q-Q on the Moon keep the same attraction? How many electrons is QQ, and what do they weigh (me=9.11×10−31 kgm_e = 9.11 \times 10^{-31}\,\mathrm{kg})? Comment.

Solution

Solution of Exercise 13.14.

F=6.67×10−11×5.97×1024×7.35×1022/(3.84×108)2=2.93×1037/1.47×1017≈2.0×1020 NF = 6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 7.35 \times 10^{22} / (3.84 \times 10^{8})^2 = 2.93 \times 10^{37}/1.47 \times 10^{17} \approx 2.0 \times 10^{20}\,\mathrm{N}. Setting kQ2/d2=FkQ^2/d^2 = F: Q=dF/k=3.84×108×2.0×1020/8.99×109≈5.7×1013 CQ = d\sqrt{F/k} = 3.84 \times 10^{8} \times \sqrt{2.0 \times 10^{20}/8.99 \times 10^{9}} \approx 5.7 \times 10^{13}\,\mathrm{C}. That is N=5.7×1013/1.6×10−19≈3.6×1032N = 5.7 \times 10^{13}/1.6 \times 10^{-19} \approx 3.6 \times 10^{32} electrons, of mass 3.6×1032×9.11×10−31≈3.2×102 kg3.6 \times 10^{32} \times 9.11 \times 10^{-31} \approx 3.2 \times 10^{2}\,\mathrm{kg}: a few hundred kilograms of electrons could do the work of 7.35×1022 kg7.35 \times 10^{22}\,\mathrm{kg} of Moon — electricity’s strength and gravity’s weakness in one number.

Exercise 13.15 ★★★

In a salt crystal, Na+^{+} and Cl−^{-} ions (charges ±e\pm e) sit d=2.8×10−10 md = 2.8 \times 10^{-10}\,\mathrm{m} apart; a Cl−^{-} ion has mass 5.9×10−26 kg5.9 \times 10^{-26}\,\mathrm{kg}. Compute the force between neighbouring ions and the weight of a Cl−^{-} ion; give the ratio. Then explain how gravity wins at large scales — no mountain tops 104 m10^{4}\,\mathrm{m} — though each bond beats it by fifteen orders of magnitude.

Solution

Solution of Exercise 13.15.

F=2.30×10−28/(2.8×10−10)2=2.30×10−28/7.84×10−20≈2.9×10−9 NF = 2.30 \times 10^{-28}/(2.8 \times 10^{-10})^2 = 2.30 \times 10^{-28}/7.84 \times 10^{-20} \approx 2.9 \times 10^{-9}\,\mathrm{N}; weight P=5.9×10−26×9.81≈5.8×10−25 NP = 5.9 \times 10^{-26} \times 9.81 \approx 5.8 \times 10^{-25}\,\mathrm{N}; ratio F/P≈5×1015F/P \approx 5 \times 10^{15}. Each bond has a fixed strength, but weight grows with the whole mass above: pile up rock and the load on the bottom layer grows without limit while its bonds do not. Around 104 m10^{4}\,\mathrm{m} of rock the load crushes the electric bonds — gravity wins not by strength per particle but by adding up when charge cancels.

13.7 Problem: The audit of the four forces

Problem 13.1

Weekend problem — why the world neither collapses nor flies apart: the four interactions audited floor by floor, down to matter’s neutrality to two parts in 101810^{18}

Atoms do not implode, nuclei mostly hold, the Moon neither crashes nor escapes; this problem audits who deserves the credit. Data: e=1.60×10−19 Ce = 1.60 \times 10^{-19}\,\mathrm{C}, k=8.99×109 N m2/C2k = 8.99 \times 10^{9}\,\mathrm{N}\,\mathrm{m}^{2}/\mathrm{C}^{2}, G=6.67×10−11 N m2/kg2G = 6.67 \times 10^{-11}\,\mathrm{N}\,\mathrm{m}^{2}/\mathrm{kg}^{2}, mp=1.67×10−27 kgm_p = 1.67 \times 10^{-27}\,\mathrm{kg}, me=9.11×10−31 kgm_e = 9.11 \times 10^{-31}\,\mathrm{kg}, ME=5.97×1024 kgM_E = 5.97 \times 10^{24}\,\mathrm{kg}, MM=7.35×1022 kgM_M = 7.35 \times 10^{22}\,\mathrm{kg}, Earth–Moon distance 3.84×108 m3.84 \times 10^{8}\,\mathrm{m}.

Part I — The ladder.

  1. List the floors of the ladder from quark to galaxy, one order of magnitude of size each.
  2. Scale model: the nucleus becomes a 1 cm1\,\mathrm{cm} marble; how wide is the atom?
  3. What fraction of an atom’s volume does its nucleus occupy?
  4. A hair is 70 µm70\,\text{µ}\mathrm{m} thick. How many atoms span it?
  5. How many powers of ten from the quark ceiling (10−18 m10^{-18}\,\mathrm{m}) to a galaxy (1021 m10^{21}\,\mathrm{m})?

Part II — The electric floor.

  1. In a hydrogen atom the electron sits d=5.3×10−11 md = 5.3 \times 10^{-11}\,\mathrm{m} from the proton: compute the electric force between them.
  2. Compute the gravitational force between them, and the ratio of the two. Which one holds the atom together?
  3. What changes if gravity is switched off inside atoms? And if electricity is?
  4. Why are the “contact” forces of everyday life — the floor pushing on your feet, friction — electromagnetic in disguise?
  5. Each of two coins holds about 102310^{23} electrons. Transfer one electron in a billion between them, set them 1.0 m1.0\,\mathrm{m} apart: compute the charges and the force. Conclusion?
  6. Which interaction rules the floors from 10−10 m10^{-10}\,\mathrm{m} to 104 m10^{4}\,\mathrm{m}, and why does the strong interaction not compete?

Part III — The nuclear floor.

  1. Two protons sit 1.0×10−15 m1.0 \times 10^{-15}\,\mathrm{m} apart in a nucleus: compute their electric repulsion.
  2. Unrestrained, what acceleration would this force give a proton? Compare with gg.
  3. List the three properties the strong interaction needs to explain why nuclei exist, why neutrons are gripped too, and why nuclei are tiny.
  4. In a uranium nucleus (diameter 1.4×10−14 m1.4 \times 10^{-14}\,\mathrm{m}, 9292 protons), compute the repulsion between two protons at opposite ends. Can the strong interaction bind that pair directly? Explain why repulsion wins as nuclei grow, and what the neutrons are for.
  5. In one sentence, as promised: what does the weak interaction do?

Part IV — The astronomical floor, and the verdict.

  1. Compute the gravitational force between the Earth and the Moon.
  2. The Earth contains about 1.8×10511.8 \times 10^{51} protons (and as many electrons), the Moon about 2.2×10492.2 \times 10^{49}. What fraction ff of each body’s proton charge, left uncancelled, would make the electric force match the gravitational one?
  3. Why does gravitation, the weakling of question 7, rule astronomy? Two reasons.
  4. Verdict — one sentence per floor (nucleus, atom to mountain, planet and up): what holds it together? Then answer the title: why does the world neither collapse nor fly apart?
Solution

Solution of Problem 13.1.

1. Quark <10−18 m< 10^{-18}\,\mathrm{m}; nucleon 10−15 m10^{-15}\,\mathrm{m}; nucleus 10−1510^{-15}–10−14 m10^{-14}\,\mathrm{m}; atom 10−10 m10^{-10}\,\mathrm{m}; molecule 10−9 m10^{-9}\,\mathrm{m}; everyday object 1 m1\,\mathrm{m}; planet 107 m10^{7}\,\mathrm{m}; planetary system 1011 m10^{11}\,\mathrm{m}; galaxy 1021 m10^{21}\,\mathrm{m}.

2. Scale factor 10510^{5}: the atom is 1 cm×105=1 km1\,\mathrm{cm} \times 10^{5} = 1\,\mathrm{km} wide.

3. (10−15/10−10)3=10−15(10^{-15}/10^{-10})^3 = 10^{-15}: matter is 99.999 999 999 999 9 %99.999\,999\,999\,999\,9\,\% empty.

4. 7×10−5/10−10=7×1057 \times 10^{-5}/10^{-10} = 7 \times 10^{5} atoms — about a million across one hair.

5. From 10−1810^{-18} to 102110^{21}: 3939 powers of ten.

6. FE=8.99×109×(1.60×10−19)2/(5.3×10−11)2≈8.2×10−8 NF_E = 8.99 \times 10^{9} \times (1.60 \times 10^{-19})^2 / (5.3 \times 10^{-11})^2 \approx 8.2 \times 10^{-8}\,\mathrm{N}.

7. FG=6.67×10−11×1.67×10−27×9.11×10−31/(5.3×10−11)2≈3.6×10−47 NF_G = 6.67 \times 10^{-11} \times 1.67 \times 10^{-27} \times 9.11 \times 10^{-31}/(5.3 \times 10^{-11})^2 \approx 3.6 \times 10^{-47}\,\mathrm{N}; ratio FE/FG≈2.3×1039F_E/F_G \approx 2.3 \times 10^{39}. Electricity holds the atom; gravity is a spectator.

8. Without gravity: nothing measurable changes (it is 103910^{39} below). Without electricity: no atoms, no molecules, no matter at all.

9. “Contact” is the electric repulsion between the electron clouds of atoms brought to about 10−10 m10^{-10}\,\mathrm{m}: nothing ever touches. Floor reaction and friction are Coulomb’s law in bulk.

10. q=1023×10−9×1.6×10−19=1.6×10−5 Cq = 10^{23} \times 10^{-9} \times 1.6 \times 10^{-19} = 1.6 \times 10^{-5}\,\mathrm{C} each; F=8.99×109×(1.6×10−5)2≈2.3 NF = 8.99 \times 10^{9} \times (1.6 \times 10^{-5})^2 \approx 2.3\,\mathrm{N} — a plainly visible force from one electron in a billion. Everyday matter is neutral to far better than 10−910^{-9}.

11. The electromagnetic interaction. The strong force is out of range beyond about 10−14 m10^{-14}\,\mathrm{m} — 10410^{4} times smaller than a single atom.

12. F=8.99×109×(1.60×10−19)2/(1.0×10−15)2≈2.3×102 NF = 8.99 \times 10^{9} \times (1.60 \times 10^{-19})^2 / (1.0 \times 10^{-15})^2 \approx 2.3 \times 10^{2}\,\mathrm{N}.

13. a=F/mp=230/1.67×10−27≈1.4×1029 m/s2a = F/m_p = 230/1.67 \times 10^{-27} \approx 1.4 \times 10^{29}\,\mathrm{m}/\mathrm{s}^{2}, about 1.4×10281.4 \times 10^{28} times gg: no everyday force comes close.

14. Stronger than 230 N230\,\mathrm{N} at 10−15 m10^{-15}\,\mathrm{m} (nuclei exist); acting on protons and neutrons alike, blind to charge (neutrons are gripped); range about 10−15 m10^{-15}\,\mathrm{m} (nuclei stay tiny — the glue cannot reach further).

15. F=2.30×10−28/(1.4×10−14)2≈1.2 NF = 2.30 \times 10^{-28}/(1.4 \times 10^{-14})^2 \approx 1.2\,\mathrm{N}. No: 1414 times the range. Repulsion acts across all 92×912=4186\frac{92\times91}{2} = 4186 proton pairs; the glue binds only nearest neighbours, so growth favours repulsion. Neutrons add glue and spacing without adding repulsion — until, beyond about 8080 protons, no mixture balances and the nucleus decays.

16. The weak interaction turns a neutron into a proton (and back), causing β\beta decay — it transforms; it never binds.

17. F=6.67×10−11×5.97×1024×7.35×1022/(3.84×108)2≈2.0×1020 NF = 6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 7.35 \times 10^{22}/(3.84 \times 10^{8})^2 \approx 2.0 \times 10^{20}\,\mathrm{N}.

18. QE=1.8×1051×1.6×10−19=2.9×1032 CQ_E = 1.8 \times 10^{51} \times 1.6 \times 10^{-19} = 2.9 \times 10^{32}\,\mathrm{C}, QM=3.5×1030 CQ_M = 3.5 \times 10^{30}\,\mathrm{C}. Matching forces means f2 k QEQM=GMEMMf^2\,k\,Q_E Q_M = G M_E M_M:

f=2.93×10378.99×109×2.9×1032×3.5×1030=3.2×10−36≈1.8×10−18:f = \sqrt{\frac{2.93 \times 10^{37}}{8.99 \times 10^{9} \times 2.9 \times 10^{32} \times 3.5 \times 10^{30}}} = \sqrt{3.2 \times 10^{-36}} \approx 1.8 \times 10^{-18} :

an imbalance of two parts in 101810^{18} would cancel gravity.

19. Charge comes in two signs and cancels (question 18 shows how perfectly), so the stronger force switches itself off at large scales; mass has one sign, adds forever, and nothing screens it.

20. Nucleus: the strong interaction outmuscles the protons’ repulsion at 10−15 m10^{-15}\,\mathrm{m}. Atom to mountain: the electromagnetic interaction binds electrons to nuclei and atoms to each other. Planet and up: gravitation, unopposed once charge cancels, holds planets, orbits and galaxies. The world does not collapse because each floor is propped by a binder stiffer than the load, and does not fly apart because matter is neutral to about 10−1810^{-18} — so every scale has exactly one force in charge, and it is attractive where it must be.