University Physics — Year 1 · Bachelor Year 1

15Angular Momentum

A figure skater spinning with arms outstretched pulls them in and, with no push from anyone, whirls twice as fast. A door yields to a fingertip at the handle and resists a shoulder at the hinge. Planets sweep out equal areas in equal times — Kepler noticed it four centuries ago, long before anyone could say why. And the Moon, tugged by the tides it raises, drifts away from the Earth by a few centimeters a year while our days grow imperceptibly longer. Behind all four stands one vector quantity, the angular momentum, and one law, its rate of change equals the moment of the forces. This chapter defines both, proves the law for a particle and for a system, and puts it to work on pendulums, orbits and spinning things.

Arms pulled in, the skater spins faster: with no external torque about the vertical axis, angular momentum is conserved and a smaller moment of inertia means a larger angular velocity.

15.1 Moment of a force

Definition 15.1 (Moment about a point; about an axis)

The moment (or torque) about a point $O$ of a force $\vect F$ applied at $M$ is the vector

\vect{\mathcal M}_O(\vect F) = \vect{OM}\wedge\vect F \qquad (\mathrm{N}\,\mathrm{m}),

perpendicular to the plane of $\vect{OM}$ and $\vect F$ , of norm $OM\cdot F\sin\alpha = F\,d$ , where $d$ — the lever arm — is the distance from $O$ to the line of action of $\vect F$ . The moment about an oriented axis $\Delta$ (unit vector $\vect e_\Delta$ ) through $O$ is the scalar $\mathcal M_\Delta = \vect{\mathcal M}_O\cdot\vect e_\Delta$ , independent of the choice of $O$ on $\Delta$ ; it vanishes if $\vect F$ is parallel to $\Delta$ or meets it. Two opposite forces on different lines of action form a couple, whose moment $\vect F\wedge\vect{AB}$ is the same about every point.

Example 15.2 (The door)

A $20\,\mathrm{N}$ push perpendicular to a door at $0.80\,\mathrm{m}$ from the hinges has moment $16\,\mathrm{N}\,\mathrm{m}$ about the hinge axis; the same push at $0.10\,\mathrm{m}$ , $2\,\mathrm{N}\,\mathrm{m}$ ; along the door (line of action through the hinges), zero. The hinge’s own reaction, meeting the axis, has no moment about it: that is what makes the axis the natural place to take moments.

The moment of F about O: the cross product OM F, of norm F\,d with d the distance from O to the line of action (the lever arm), directed perpendicular to the figure — here toward the reader. — The moment of $\vect F$ about $O$ : the cross product $\vect{OM}\wedge\vect F$ , of norm $F\,d$ with $d$ the distance from $O$ to the line of action (the lever arm), directed perpendicular to the figure — here toward the reader.

15.2 Angular momentum of a particle

Definition 15.3 (Angular momentum)

The angular momentum about $O$ of a particle of mass $m$ at $M$ with velocity $\vect v$ is

\vect L_O = \vect{OM}\wedge m\vect v \qquad (\mathrm{kg}\,\mathrm{m}^{2}/\mathrm{s}),

and about an axis $\Delta$ through $O$ , $L_\Delta = \vect L_O\cdot\vect e_\Delta$ . For a motion in a plane perpendicular to $\Delta$ , in polar coordinates about the axis, $L_\Delta = mr^2\dot\theta$ ; for a circle of radius $R$ , $L_\Delta = mR^2\omega = mRv$ .

Proof. $\vect{OM} = r\vect e_r$ , $\vect v = \dot r\vect e_r + r\dot\theta\vect e_\theta$ : $\vect{OM}\wedge m\vect v = mr^2\dot\theta\,\vect e_r\wedge\vect e_\theta = mr^2\dot\theta\,\vect e_z$ . ∎

Theorem 15.4 (Angular momentum theorem)

In an inertial frame, for a point $O$ fixed in that frame,

\frac{\dd\vect L_O}{\dd t} = \sum\vect{\mathcal M}_O(\vect F_i), \qquad \frac{\dd L_\Delta}{\dd t} = \sum\mathcal M_\Delta(\vect F_i)

for a fixed axis $\Delta$ . If the total moment about $O$ vanishes, $\vect L_O$ is conserved.

Proof. $\dd(\vect{OM}\wedge m\vect v)/\dd t = \vect v\wedge m\vect v + \vect{OM}\wedge m\vect a = \vect 0 + \vect{OM}\wedge\sum\vect F_i$ , using Newton’s second law ( $O$ fixed so that $\dd\vect{OM}/\dd t = \vect v$ ). Project on $\vect e_\Delta$ . ∎

Corollary 15.5 (Central forces: planar motion and the law of areas)

A particle subject only to a central force (always directed along $\vect{OM}$ , toward or away from the fixed center $O$ ) keeps $\vect L_O$ constant. Its motion stays in the plane through $O$ perpendicular to $\vect L_O$ , and in that plane $r^2\dot\theta = C$ (constant): the radius vector $\vect{OM}$ sweeps out area at the constant areal velocity

\frac{\dd A}{\dd t} = \frac12 r^2\dot\theta = \frac{L_O}{2m} = \frac{C}{2} .

Proof. $\vect{OM}\wedge\vect F = \vect 0$ for $\vect F \parallel \vect{OM}$ : $\vect L_O$ is constant. $\vect{OM}\perp\vect L_O$ always, hence the plane; $L_O = mr^2\dot\theta$ is constant. The area swept during $\dd t$ is the triangle of sides $\vect{OM}$ and $\vect v\dd t$ : $\dd A = \tfrac12\abs{\vect{OM} \wedge\vect v}\dd t = \tfrac12 r^2\dot\theta\,\dd t$ . ∎

The law of areas for a central force: in equal times the radius vector from the center

O

sweeps equal areas, so the particle moves fast when close to

O

and slowly when far — Kepler’s second law, a consequence of the conservation of

\vect L_O

alone, whatever the law of force.

Example 15.6 (The pendulum by moments)

Simple pendulum, axis $\Delta$ through the pivot perpendicular to the plane of swing: $L_\Delta = m\ell^2\dot\theta$ ; the tension meets the axis (no moment), the weight’s moment is $-mg\ell\sin\theta$ (lever arm $\ell\sin\theta$ , restoring). The theorem gives $m\ell^2\ddot\theta = -mg\ell\sin\theta$ , the equation of Example 12.12 without ever writing the tension — the chief advantage of moments: forces through the axis disappear.

Example 15.7 (Pulling a puck inward)

A puck slides on frictionless ice in a circle, held by a string through a hole at the center; the string is pulled so that the radius halves. The tension is central: $L = mr^2\omega$ is conserved, so $\omega$ is multiplied by $4$ , the speed $r\omega$ by $2$ , the kinetic energy by $4$ — the extra energy is the work of the pull, which the puck resists all the way in. The skater’s arms are the same physics with the mass distributed (Exercise 15.4).

15.3 Systems of points and rotation about a fixed axis

Theorem 15.8 (Angular momentum of a system)

For a system of particles, with $\vect L_O = \sum_i\vect{OM_i}\wedge m_i\vect v_i$ and $O$ fixed in an inertial frame,

\frac{\dd\vect L_O}{\dd t} = \sum\vect{\mathcal M}_O(\vect F_{\mathrm{ext}}):

only the external forces’ moments count; the internal forces, which obey the action–reaction law along the line joining the particles, contribute nothing.

Proof. Sum the particle theorems. For an internal pair, $\vect{OM_i}\wedge\vect F_{j\to i} + \vect{OM_j}\wedge\vect F_{i\to j} = (\vect{OM_i} - \vect{OM_j})\wedge\vect F_{j\to i} = \vect{M_jM_i}\wedge\vect F_{j\to i} = \vect 0$ since the force is along $\vect{M_jM_i}$ . ∎

Proposition 15.9 (Rigid rotation about a fixed axis)

If the particles keep fixed distances $r_i$ from an axis $\Delta$ and all turn about it at the same angular velocity $\omega$ (a rigid rotation),

L_\Delta = J_\Delta\,\omega, \qquad J_\Delta = \sum_i m_ir_i^2, \qquad E_k = \tfrac12 J_\Delta\omega^2,

$J_\Delta$ being the moment of inertia about $\Delta$ , and the theorem reads $J_\Delta\dot\omega = \sum\mathcal M_\Delta(\vect F_{\mathrm{ext}})$ .

Proof. Each particle contributes $m_ir_i^2\omega$ to $L_\Delta$ and $\tfrac12 m_i(r_i\omega)^2$ to $E_k$ . Chapter 19 extends this to continuous solids. ∎

Example 15.10 (The skater)

Arms out, a skater’s body has $J \approx 4\,\mathrm{kg}\,\mathrm{m}^{2}$ plus two $3\,\mathrm{kg}$ arms at $0.8\,\mathrm{m}$ : $J_1 = 4 + 2 \times 3 \times 0.64 = 7.8\,\mathrm{kg}\,\mathrm{m}^{2}$ . Arms in ( $0.2\,\mathrm{m}$ ): $J_2 = 4.2\,\mathrm{kg}\,\mathrm{m}^{2}$ . The ice’s reaction and the weight have no moment about the vertical axis: $J\omega$ is conserved and the spin rate rises by $J_1/J_2 = 1.85$ — from $2$ to $3.7$ turns per second — while the kinetic energy rises by the same factor, paid by the muscles pulling the arms in.

Particles rigidly rotating about an axis: each turns on its own circle at the common , contributing m_ir_i2 to the angular momentum about the axis. Mass far from the axis counts as the square of its distance — the skater’s outstretched arms. — Particles rigidly rotating about an axis: each turns on its own circle at the common $\omega$ , contributing $m_ir_i^2\omega$ to the angular momentum about the axis. Mass far from the axis counts as the square of its distance — the skater’s outstretched arms.

Remark 15.11 (Statics)

A solid at rest has $\vect L_O = \vect 0$ constant: the external forces must have zero resultant and zero total moment about any point. The second condition is the lever law of the balance, the seesaw and the crowbar: a small force far from the pivot balances a large one close to it.

15.4 Exercises

Exercise 15.1 ★

A $50\,\mathrm{N}$ force on a wrench $0.30\,\mathrm{m}$ from the nut: moment when the force is perpendicular to the handle; at $60^\circ$ to it; along it.

Solution

Solution of Exercise 15.1.

$Fd = 50 \times 0.30 = 15\,\mathrm{N}\,\mathrm{m}$ ; $Fd\sin 60^\circ = 13\,\mathrm{N}\,\mathrm{m}$ ; along the handle the line of action passes through the nut: $0$ .

Exercise 15.2 ★

Angular momentum of the Earth about the Sun (circular orbit, $r = 1.50 \times 10^{11}\,\mathrm{m}$ , $v = 29.8\,\mathrm{km}/\mathrm{s}$ , $m = 5.97 \times 10^{24}\,\mathrm{kg}$ ), and its areal velocity.

Solution

Solution of Exercise 15.2.

$L = mrv = 5.97\times10^{24} \times 1.50\times10^{11} \times 2.98\times10^4 = 2.7 \times 10^{40}\,\mathrm{kg}\,\mathrm{m}^{2}/\mathrm{s}$ ; $\dd A/\dd t = L/2m = rv/2 = 2.2 \times 10^{15}\,\mathrm{m}^{2}/\mathrm{s}$ .

Exercise 15.3 ★

A conical pendulum (string $1.0\,\mathrm{m}$ , $30^\circ$ from the vertical, Exercise 12.9). Compute $L$ about the vertical axis through the pivot per unit mass, and explain why it is conserved although the string’s tension and the weight act.

Solution

Solution of Exercise 15.3.

$r = \ell\sin 30^\circ = 0.50\,\mathrm{m}$ , $\omega = 3.4\,\mathrm{rad}/\mathrm{s}$ : $L_z/m = r^2\omega = 0.84\,\mathrm{m}^{2}/\mathrm{s}$ . The tension’s line of action passes through the pivot (zero moment about any axis through it); the weight is parallel to the vertical axis (zero moment about it): $L_z$ is conserved.

Exercise 15.4 ★

A skater’s trunk has $J = 4.0\,\mathrm{kg}\,\mathrm{m}^{2}$ about the spin axis; each arm ( $3.0\,\mathrm{kg}$ ) is treated as a point at $0.80\,\mathrm{m}$ , then at $0.20\,\mathrm{m}$ . Spin rate after pulling the arms in, starting from $2.0\,$ turns per second; ratio of kinetic energies; where does the extra energy come from?

Solution

Solution of Exercise 15.4.

$J_1 = 4.0 + 2 \times 3.0 \times 0.64 = 7.8\,\mathrm{kg}\,\mathrm{m}^{2}$ , $J_2 = 4.0 + 2 \times 3.0 \times 0.04 = 4.2\,\mathrm{kg}\,\mathrm{m}^{2}$ ; $\omega_2 = \omega_1J_1/J_2 = 1.85\omega_1$ : $3.7\,$ turns per second. $E_k = L^2/2J$ : ratio $J_1/J_2 = 1.85$ ; the muscles, pulling the arms inward against the centrifugal tendency, do the work.

Exercise 15.5 ★★

Derive the pendulum equation $\ell\ddot\theta + g\sin\theta = 0$ from the angular momentum theorem about the pivot axis, and give the moment of each force.

Solution

Solution of Exercise 15.5.

$L_\Delta = m\ell^2\dot\theta$ ; tension: zero moment (meets the axis); weight: $-mg\ell\sin\theta$ (lever arm $\ell\sin\theta$ , restoring). $m\ell^2\ddot\theta = -mg\ell\sin\theta$ .

Exercise 15.6 ★★

The Earth’s distance to the Sun varies from $1.471 \times 10^{11}\,\mathrm{m}$ (perihelion) to $1.521 \times 10^{11}\,\mathrm{m}$ (aphelion). Using the law of areas, find the ratio of its speeds at these points; with a mean speed of $29.8\,\mathrm{km}/\mathrm{s}$ , give both speeds.

Solution

Solution of Exercise 15.6.

$r_pv_p = r_av_a$ : $v_p/v_a = r_a/r_p = 1.034$ ; with $(v_p + v_a)/2 \approx 29.8\,\mathrm{km}/\mathrm{s}$ : $v_p = 30.3\,\mathrm{km}/\mathrm{s}$ , $v_a = 29.3\,\mathrm{km}/\mathrm{s}$ .

Exercise 15.7 ★★

A $0.20\,\mathrm{kg}$ puck circles at $2.0\,\mathrm{m}/\mathrm{s}$ on a $0.50\,\mathrm{m}$ string through a hole; the string is pulled until the radius is $0.25\,\mathrm{m}$ . New speed and angular velocity; kinetic energies before and after; work done by the pull. Why is the tension able to do work here?

Solution

Solution of Exercise 15.7.

$L = mrv$ conserved: $v_2 = v_1r_1/r_2 = 4.0\,\mathrm{m}/\mathrm{s}$ , $\omega_2 = v_2/r_2 = 16\,\mathrm{rad}/\mathrm{s}$ (from $4\,\mathrm{rad}/\mathrm{s}$ ). $E_1 = 0.40\,\mathrm{J}$ , $E_2 = 1.6\,\mathrm{J}$ : the pull does $1.2\,\mathrm{J}$ . The tension is radial, but the puck’s displacement now has a radial component (it spirals in): the work $\vect T\cdot\dd\vect\ell$ is no longer zero.

Exercise 15.8 ★★

A uniform plank of mass $20\,\mathrm{kg}$ and length $4.0\,\mathrm{m}$ rests on a pivot $1.5\,\mathrm{m}$ from its left end; a child of $30\,\mathrm{kg}$ sits at the left end. Where must a $45\,\mathrm{kg}$ child sit for balance? What does the pivot support?

Solution

Solution of Exercise 15.8.

Moments about the pivot (lengths from the pivot): child $30 \times 1.5 = 45$ (left); plank’s weight at its center, $0.5$ to the right: $20 \times 0.5 = 10$ (right); second child at $x$ to the right: $45x$ . Balance: $45 = 10 + 45x$ , $x = 0.78\,\mathrm{m}$ . The pivot supports the total weight, $95 \times 9.81 = 932\,\mathrm{N}$ .

Exercise 15.9 ★★

An Atwood machine ( $m_1 = 2.0\,\mathrm{kg}$ , $m_2 = 1.0\,\mathrm{kg}$ ) hangs over a pulley of radius $0.10\,\mathrm{m}$ and moment of inertia $0.010\,\mathrm{kg}\,\mathrm{m}^{2}$ on which the string does not slip. Using the angular momentum theorem for the pulley and Newton’s law for the masses, find the acceleration and the two tensions. Compare with the massless pulley.

Solution

Solution of Exercise 15.9.

Masses: $m_1a = m_1g - T_1$ , $m_2a = T_2 - m_2g$ ; pulley: $J\dot\omega = (T_1 - T_2)R$ with $a = R\dot\omega$ . Adding: $a = (m_1 - m_2)g/(m_1 + m_2 + J/R^2) = 9.81/(3.0 + 1.0) = 2.5\,\mathrm{m}/\mathrm{s}^{2}$ (against $3.3\,\mathrm{m}/\mathrm{s}^{2}$ ); $T_1 = m_1(g - a) = 14.7\,\mathrm{N}$ , $T_2 = m_2(g + a) = 12.3\,\mathrm{N}$ — unequal, as the pulley needs a net moment.

Exercise 15.10 ★★★

A comet passes its perihelion at $0.50\,\mathrm{AU}$ from the Sun at $40\,\mathrm{km}/\mathrm{s}$ . When it is $10\,\mathrm{AU}$ away, what is the component of its velocity perpendicular to the radius vector? Can the law of areas give its full speed?

Solution

Solution of Exercise 15.10.

$r_1v_1 = r_2v_{\perp 2}$ : $v_{\perp 2} = 40 \times 0.5/10 = 2.0\,\mathrm{km}/\mathrm{s}$ . The law of areas gives only the perpendicular component; the radial one needs energy conservation (Chapter 16).

Exercise 15.11 ★★★

A star like the Sun ( $R = 7.0 \times 10^{8}\,\mathrm{m}$ , rotation period $25\,\mathrm{d}$ ) collapses into a neutron star of radius $10\,\mathrm{km}$ , keeping its mass and angular momentum ( $J \propto MR^2$ ). New period? Equatorial speed? What does the latter tell you about the assumption?

Solution

Solution of Exercise 15.11.

$J\omega$ constant with $J \propto R^2$ : $T' = T(R'/R)^2 = 25 \times 86400 \times (10^4/7\times10^8)^2 = 4.4 \times 10^{-4}\,\mathrm{s}$ . Equatorial speed $2\pi R'/T' = 1.4 \times 10^{8}\,\mathrm{m}/\mathrm{s}$ , half the speed of light: the collapse cannot keep all the angular momentum — much is shed (and relativity enters).

Exercise 15.12 ★★★

A bead slides without friction along a straight rod rotating in a horizontal plane at constant $\omega$ about a vertical axis through one end. Write $L_z$ of the bead; show that it is not conserved and compute the moment the rod must exert; deduce the transverse reaction of the rod on the bead.

Solution

Solution of Exercise 15.12.

$L_z = mr^2\omega$ , and $r$ changes: $\dd L_z/\dd t = 2mr\dot r\omega \neq 0$ . This must be the moment of the rod’s transverse reaction $N$ : $rN = 2mr\dot r\omega$ , so $N = 2m\dot r\omega$ — the rod pushes the bead sideways as it slides out (the Coriolis term of Chapter 18).

15.5 Problem: Why the Moon is leaving

Problem 15.1

Weekend problem — the tides the Moon raises on the Earth pull it forward and the Earth back: how angular momentum flows from our days into the Moon’s orbit, centimeter by centimeter, and where the energy goes

Data: $G = 6.67 \times 10^{-11}\,\mathrm{SI}$ ; Earth $M = 5.97 \times 10^{24}\,\mathrm{kg}$ , $R = 6.37 \times 10^{6}\,\mathrm{m}$ , spin $\Omega = 2\pi/(86\,164\,\mathrm{s})$ , moment of inertia $J = 0.33\,MR^2$ ; Moon $m = 7.35 \times 10^{22}\,\mathrm{kg}$ , orbit treated as circular, $r = 3.84 \times 10^{8}\,\mathrm{m}$ ; lunar laser ranging measures $\dd r/\dd t = 3.8\,\mathrm{cm}/\mathrm{yr}$ ; the day lengthens by $2.3\,\mathrm{ms}$ per century; $1$ century $= 3.16 \times 10^{9}\,\mathrm{s}$ .

Part I — The angular momenta.

Compute the orbital angular velocity $\omega$ of the Moon from Newton’s law (circular orbit), and check it against the $27.3\,\mathrm{d}$ sidereal month.
Compute the Moon’s orbital angular momentum $L_{\text{orb}} = mr^2\omega$ .
Show that for a circular orbit $L_{\text{orb}} = m\sqrt{GMr}$ , and deduce $\dd L_{\text{orb}}/\dd r = L_{\text{orb}}/2r$ .
Compute the Earth’s spin angular momentum $L_{\text{spin}} = J\Omega$ and compare with $L_{\text{orb}}$ .
The Moon spins once per month: estimate its spin angular momentum ( $J_m \approx 0.4\,mR_m^2$ , $R_m = 1.74 \times 10^{6}\,\mathrm{m}$ ) and justify neglecting it.
What external moments act on the Earth–Moon system about its center of mass? Argue that, to a good approximation, the total $L = L_{\text{spin}} + L_{\text{orb}}$ is conserved.

Part II — The tidal torque. The Moon raises two bulges of water on the Earth; because the Earth spins faster than the Moon orbits, friction drags the bulges ahead of the Earth–Moon line by a small angle.

Sketch the Earth, the leading bulge and the Moon. In which direction does the Moon’s attraction on the nearer bulge act, relative to the Earth’s spin?
Deduce the sign of the moment on the Earth’s spin: does the day lengthen or shorten?
By action and reaction, what moment acts on the Moon’s orbital motion? Does $L_{\text{orb}}$ grow or shrink?
From question 3, does $r$ grow or shrink? And the Moon’s orbital speed $v = \sqrt{GM/r}$ ?
Resolve the apparent paradox: the Moon is pushed forward, yet ends up moving more slowly.
Toward what final state does the exchange tend?

Part III — The numbers.

From the lengthening of the day, compute the relative change $\Delta\Omega/\Omega$ per century.
Deduce the change of $L_{\text{spin}}$ per century.
Deduce the change of $L_{\text{orb}}$ per century.
Using question 3, convert it into a change of $r$ per century and per year. Compare with the laser-ranging value.
By what fraction, and by how many seconds, does the sidereal month lengthen per century ( $T \propto r^{3/2}$ )?
Compute the change of the Earth’s rotational kinetic energy per century ( $\Delta E = L_{\text{spin}}\Delta\Omega$ to first order).
Compute the change of the Moon’s orbital mechanical energy, $E_{\text{orb}} = -GMm/2r$ , per century.

Part IV — The energy.

Show that the Earth loses more energy than the Moon gains, and compute the difference per century and the corresponding power. Where does it go?
Compare that power with humanity’s consumption, about $18\,\mathrm{TW}$ , and with the Earth’s internal heat flow, about $47\,\mathrm{TW}$ .
In the final state of question 12, day and month are equal: $\Omega = \omega = \sqrt{GM/r^3}$ . Writing the conservation of $L_{\text{spin}} + L_{\text{orb}}$ , check that $r \approx 5.5 \times 10^{8}\,\mathrm{m}$ satisfies it, and give the common period in days.
Why will that state in fact never be reached? (Think of the Sun’s tides.)
The Moon already shows the Earth one face. Explain with the same mechanism, and say why it happened to the Moon first.
Summarize in three lines: what is conserved, what is exchanged, what is dissipated.

Solution

Solution of Problem 15.1.

1. $m\omega^2r = GMm/r^2$ : $\omega = \sqrt{GM/r^3} = \sqrt{3.98\times 10^{14}/5.66\times10^{25}} = 2.65 \times 10^{-6}\,\mathrm{rad}/\mathrm{s}$ , period $2\pi/\omega = 2.37 \times 10^{6}\,\mathrm{s} = 27.4\,\mathrm{d}$ .

2. $L_{\text{orb}} = 7.35\times10^{22} \times 1.475\times10^{17} \times 2.65\times10^{-6} = 2.9 \times 10^{34}\,\mathrm{kg}\,\mathrm{m}^{2}/\mathrm{s}$ .

3. $v = \sqrt{GM/r}$ , $L = mrv = m\sqrt{GMr}$ ; $\dd L/\dd r = \tfrac12 m\sqrt{GM/r} = L/2r$ .

4. $J = 0.33 \times 5.97\times10^{24} \times 4.06\times10^{13} = 8.0 \times 10^{37}\,\mathrm{kg}\,\mathrm{m}^{2}$ ; $\Omega = 7.29 \times 10^{-5}\,\mathrm{rad}/\mathrm{s}$ ; $L_{\text{spin}} = 5.8 \times 10^{33}\,\mathrm{kg}\,\mathrm{m}^{2}/\mathrm{s}$ , a fifth of $L_{\text{orb}}$ .

5. $J_m \approx 0.4 \times 7.35\times10^{22} \times 3.0\times10^{12} = 8.9 \times 10^{34}\,\mathrm{kg}\,\mathrm{m}^{2}$ , times $\omega$ : $2.4 \times 10^{29}\,\mathrm{kg}\,\mathrm{m}^{2}/\mathrm{s}$ , $10^{-5}$ of the orbit’s: negligible.

6. The Sun’s attraction acts on the system’s center of mass (no moment about it to first order); its tidal effect on the pair is small: the total $L$ is conserved to the accuracy needed here.

7. The nearer bulge, ahead of the Earth–Moon line, is pulled by the Moon with a force whose tangential component opposes the Earth’s spin.

8. Negative moment on the spin: the Earth slows, the day lengthens.

9. The bulge pulls the Moon forward: positive moment on the orbit, $L_{\text{orb}}$ grows.

10. $L_{\text{orb}} \propto \sqrt r$ grows, so $r$ grows; $v \propto 1/\sqrt r$ decreases.

11. The forward pull does positive work, raising the Moon’s energy; a higher orbit is a slower orbit: the energy goes into height (potential), more than the kinetic energy lost.

12. Tidal locking of the Earth: day equal to month, no leading bulge, no torque.

13. $\Delta\Omega/\Omega = -2.3 \times 10^{-3}\,\mathrm{s}/86\,164\,\mathrm{s} = -2.7 \times 10^{-8}$ per century.

14. $\Delta L_{\text{spin}} = L_{\text{spin}}\Delta\Omega/\Omega = -1.55 \times 10^{26}\,\mathrm{kg}\,\mathrm{m}^{2}/\mathrm{s}$ per century.

15. $\Delta L_{\text{orb}} = +1.55 \times 10^{26}\,\mathrm{kg}\,\mathrm{m}^{2}/\mathrm{s}$ per century.

16. $\Delta r = 2r\,\Delta L/L = 2 \times 3.84\times10^8 \times 1.55 \times10^{26}/2.9\times10^{34} = 4.1\,\mathrm{m}$ per century, $4.1\,\mathrm{cm}/\mathrm{yr}$ : close to the measured $3.8\,\mathrm{cm}/\mathrm{yr}$ (the day’s lengthening also has a non-tidal part).

17. $\Delta T/T = \tfrac32\Delta r/r = 1.6\times10^{-8}$ : $2.37\times 10^6 \times 1.6\times10^{-8} = 0.04\,\mathrm{s}$ per century.

18. $\Delta E_{\text{rot}} = L_{\text{spin}}\Delta\Omega = 5.8\times10^{33} \times 7.29\times10^{-5} \times (-2.7\times10^{-8}) = -1.1 \times 10^{22}\,\mathrm{J}$ per century.

19. $\Delta E_{\text{orb}} = GMm\Delta r/2r^2 = 2.93\times10^{37} \times 4.1/(2 \times 1.475\times10^{17}) = +4.1 \times 10^{20}\,\mathrm{J}$ per century.

20. Net $-1.1\times10^{22} + 4\times10^{20} \approx -1.06 \times 10^{22}\,\mathrm{J}$ per century: $3.4\times10^{12}\ \mathrm{W}$ , dissipated as heat by tidal friction in the oceans and the solid Earth.

21. A fifth of humanity’s power, a fourteenth of the geothermal flow — a small but permanent heater.

22. $L_{\text{tot}} = 3.5\times10^{34}$ . At $r = 5.5\times10^8$ : $\omega = \sqrt{GM/r^3} = 1.55 \times 10^{-6}\,\mathrm{rad}/\mathrm{s}$ , $J\omega = 1.2\times10^{32}$ , $m\sqrt{GMr} = 7.35\times10^{22} \times 4.68\times10^{11} = 3.44\times10^{34}$ ; sum $3.45\times10^{34}$ : consistent. Period $2\pi/\omega = 4.1 \times 10^{6}\,\mathrm{s} = 47\,\mathrm{d}$ .

23. The Sun’s tides keep slowing the Earth below the Moon’s orbital rate, so the Earth–Moon torque would then reverse; besides, the Sun will have evolved long before (the time scale is tens of billions of years).

24. The Earth raised tides on the Moon, which braked its spin until one face stayed toward us; being lighter and closer to a heavier partner, the Moon’s locking was far faster.

25. Conserved: the total angular momentum. Exchanged: angular momentum from the Earth’s spin to the Moon’s orbit, about $1.6 \times 10^{26}\,\mathrm{SI}$ per century. Dissipated: mechanical energy, a few terawatts, as tidal heat.

Terms defined in this chapter

See all 393 terms in the glossary