1 Newton mechanics

1. A World Before Newton — From Aristotle to Galileo

Before Newton, motion was not mathematics but metaphysics. Aristotle’s physics had dominated Europe for nearly two millennia:

  • Aristotelian worldview: Objects had a natural place. A stone fell because it sought the Earth; fire rose because it sought the heavens. Motion required a continuous mover. If you stopped pushing, motion ceased.

This model explained everyday intuition but failed under scrutiny.

The Aristotelian Dilemma

  • Why does an arrow keep flying after it leaves the bowstring?
  • Why do planets not fall to Earth?

Medieval scholars answered with “impetus theory” — the idea that a force imparted continues for a while before dissipating. It was a step toward inertia but still tied motion to metaphysical “qualities.”


Enter Galileo (1564–1642)

Galileo Galilei shattered Aristotelian orthodoxy with experiments:

  1. Inclined Plane Experiment
    Galileo rolled spheres down inclined planes to slow gravity’s effect.
    He discovered uniform acceleration:

    s=12gt2s = \tfrac{1}{2} g t^2

    where ss is displacement, tt is time, and gg is the gravitational acceleration.

    → A universal constant gg governs free fall, independent of mass.

    Exercise 1.1: Drop a hammer and a feather in a vacuum. Which hits first?

    Hammer & Feather in vacuum

    Answer: They land simultaneously, as acceleration gg is independent of mass.

  2. Law of Inertia (proto-form)
    Galileo noted that a ball rolling on a frictionless plane would not stop by itself.
    If no force acted, its velocity remained constant.

    dvdt=0if F=0\frac{d \boldsymbol v }{dt} = 0 \quad \text{if } \boldsymbol F = 0

    This was the embryo of Newton’s First Law.

    Exercise 1.2: A puck slides across smooth ice and slows only gradually. Which force acts to slow it?
    Answer: Friction with the ice and air drag. In an ideal frictionless surface, it would never stop.


Kepler and the Celestial Puzzle

Johannes Kepler (1571–1630) provided another piece. Using Tycho Brahe (1546-1601) ’s data, he extracted three laws of planetary motion:

  1. Planets move in ellipses with the Sun at one focus.
  2. They sweep out equal areas in equal times.
  3. The square of the orbital period is proportional to the cube of the semi-major axis (T2a3T^2 \propto a^3).

But Kepler had no cause. Why ellipses? Why T2a3T^2 \propto a^3? He described the heavens beautifully but lacked a mechanism.


Descartes and the Mechanical Dream

René Descartes (1596–1650) imagined a clockwork universe filled with vortices of subtle matter pushing the planets along. It was poetic but untestable. Still, the dream of mechanization of the cosmos lingered.


A World Waiting for Newton

Thus by the mid-17th century:

  • Galileo gave Earth-bound laws of motion.
  • Kepler gave planetary laws.
  • Descartes gave philosophical mechanism.

But no one unified them.
Into this gap, in 1642, the year Galileo died, Isaac Newton was born.


2. The Genesis of Laws

Newton did not invent physics, but he wove it into a system. He absorbed Galileo’s inertia, Kepler’s harmonies, and Descartes’ mechanics. In 1687, he published Philosophiæ Naturalis Principia Mathematica — the Principia.

Here, three Laws of Motion and the Law of Universal Gravitation were laid out.

For the first time, terrestrial and celestial motions were governed by the same principles. Apples and moons followed the same law.

We stand today on that unification.

3. The First Law – Inertia and Isolation

Newton’s First Law of Motion deepened Galileo’s idea of inertia:

“Every body perseveres in its state of rest, or of uniform motion in a straight line, unless it is compelled to change that state by forces impressed upon it.”

Mathematical Formulation

If no net force acts on a body:

dvdt=0v=constant\frac{d \boldsymbol v }{dt} = 0 \quad \Rightarrow \quad \boldsymbol v = \text{constant}

Equivalently, in terms of position:

r(t)=r0+v0t\boldsymbol r(t) = \boldsymbol r_{0} + \boldsymbol v_{0} t

Thus, an inertial frame is defined as one in which Newton’s laws hold and objects maintain their velocity in the absence of forces.

Proof via Galileo’s Inclined Plane

Consider a ball rolling down one slope, across a flat surface, and up another slope:

  • With friction, it rises less high.
  • With reduced friction, it rises higher.
  • In the limit of zero friction, it would rise to the same height — or if perfectly flat, continue forever.

This experiment demonstrates that uniform motion needs no continuous cause.


Example Problem 3.1 – Sliding on Ice

Problem: A hockey puck is struck across frictionless ice with speed v0v_0. How far does it travel in 10 s?
Solution: Since a=0a = 0, velocity is constant:
x=v0tx = v_0 t
If v0=5m/sv_0 = 5 \,\text{m/s}, after t=10st=10\,\text{s} it has traveled 50m50 \,\text{m}.


4. The Second Law – Force and Genius

Newton’s Second Law provides the causal link:

“The change of motion is proportional to the motive force impressed; and is made in the direction of the straight line in which that force is impressed.”

Mathematically:

F=ma=md2rdt2\boldsymbol F = m \boldsymbol a = m \frac{d^2 \boldsymbol r}{dt^2} a=dvdt=d2rdt2\boldsymbol a = \frac{d \boldsymbol v}{dt} = \frac{d^2 \boldsymbol r}{dt^2}

This single relation makes the universe calculable.

Proof from the Lagrangian

Take the Lagrangian L=TV=12mr˙2V(r)L = T - V = \tfrac{1}{2} m \dot{\boldsymbol r}^2 - V(\boldsymbol r).

The Lagrangian is defined as the difference between the kinetic energy TT and the potential energy VV.

Euler–Lagrange equation:

ddt(Lr˙)Lr=0\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\boldsymbol r}}\right) - \frac{\partial L}{\partial \boldsymbol r} = 0

gives

mr¨+V=0mr¨=Fm \ddot{\boldsymbol r} + \nabla V = 0 \quad \Rightarrow \quad m \ddot{\boldsymbol r} = \boldsymbol F

Thus, Newton’s F=maF = ma emerges naturally from variational principles.


Time-Translation Symmetry → Energy Conservation

If LL has no explicit time dependence, then Energy EE is:

E=r˙Lr˙LE = \dot{\boldsymbol r} \cdot \frac{\partial L}{\partial \dot{\boldsymbol r}} - L

is conserved.

For a particle:

E=12mv2+V(r)=constantE = \tfrac{1}{2} m v^2 + V(\boldsymbol r) = \text{constant}

Thus time symmetry = energy conservation.

Proof

From the definition,

E=q˙iLq˙iL,E = \dot{q}_i \frac{\partial L}{\partial \dot{q}_i} - L,

its time derivative is

dEdt=q¨iLq˙i+q˙iddt(Lq˙i)dLdt.\frac{dE}{dt} = \ddot{q}_i \frac{\partial L}{\partial \dot{q}_i} + \dot{q}_i \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right) - \frac{dL}{dt}.

Using the Euler–Lagrange equation

ddt(Lq˙i)=Lqi,\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right) = \frac{\partial L}{\partial q_i},

we obtain

dEdt=Lt.\frac{dE}{dt} = -\frac{\partial L}{\partial t}.

Therefore, if LL has no explicit time dependence,

Lt=0dEdt=0,\frac{\partial L}{\partial t} = 0 \quad \Rightarrow \quad \frac{dE}{dt} = 0,

so EE is conserved.

This is a direct example of Noether’s theorem:
time-translation symmetry corresponds to conservation of energy.


Example Problem 4.1 – Block on an Inclined Plane

Problem: A block of mass mm slides down a frictionless incline at angle θ\theta. Find its acceleration aa.

Block on an Inclined Plane Problem

Solution:
Force along incline: F=mgsinθF = mg \sin\theta
Newton’s Second Law: F=maF = ma
So a=gsinθa = g \sin\theta.

Block on an Inclined Plane Solution

If θ=30\theta = 30^\circ, a=4.9m/s2a = 4.9 \,\text{m/s}^2.


Example Problem 4.2 – Simple Pendulum (small angle)

Simple Pendulum

We consider a simple pendulum consisting of a point mass mm attached to a rigid, massless rod (or string) of length \ell. The pivot is frictionless, and the angle from the vertical is denoted by θ(t)\theta(t).

The bob moves on a circular arc of radius \ell. Its tangential acceleration is

atan=θ¨,a_{\text{tan}} = \ell \ddot{\theta},

and its radial (centripetal) acceleration is

arad=θ˙2.a_{\text{rad}} = \ell \dot{\theta}^{2}.

Two forces act on the bob:

  • The tension TT along the rod.
  • The gravitational force mgmg vertically downward.

The tangential component of gravity is

Ftan=mgsinθ,F_{\text{tan}} = -mg \sin\theta,

where the minus sign indicates that gravity restores the pendulum toward θ=0\theta=0.

Applying Newton’s second law in the tangential direction,

matan=Ftan,m a_{\text{tan}} = \sum F_{\text{tan}},

we find

m(θ¨)=mgsinθ.m(\ell \ddot{\theta}) = -mg \sin\theta.

Simplifying,

θ¨+gsinθ=0.\ddot{\theta} + \frac{g}{\ell} \sin\theta = 0.

This is the exact nonlinear equation of motion of a simple pendulum.

For small angular displacements θ1|\theta|\ll 1 (in radians), we can use

sinθθ.\sin\theta \approx \theta.

The equation then becomes

θ¨+gθ=0.\boxed{ \ddot{\theta} + \frac{g}{\ell}\,\theta = 0 }.

This is the familiar simple harmonic oscillator equation, with angular frequency

ω0=g,T=2πω0=2πg.\omega_0 = \sqrt{\tfrac{g}{\ell}}, \quad T = \frac{2\pi}{\omega_0} = 2\pi\sqrt{\tfrac{\ell}{g}}.
  • Torque Method:
    About the pivot, the torque due to gravity is

    τ=mgsinθ,\tau = -mg\ell \sin\theta,

    and with moment of inertia I=m2I=m\ell^2,

    Iθ¨=τθ¨+gsinθ=0.I\ddot{\theta} = \tau \quad\Rightarrow\quad \ddot{\theta} + \frac{g}{\ell}\sin\theta=0.
  • Lagrangian Method:

    T=12m2θ˙2,V=mg(1cosθ),T = \tfrac{1}{2}m\ell^2\dot{\theta}^2, \quad V = mg\ell(1-\cos\theta), L=TV.L = T - V.

    Applying the Euler–Lagrange equation yields the same result.

Since

sinθ=θθ36+,\sin\theta = \theta - \tfrac{\theta^3}{6} + \cdots,

the relative error is θ2/6\sim \theta^2/6:

  • At θ=100.17\theta = 10^\circ \approx 0.17 rad → error 0.5%\approx 0.5\%.
  • At θ=200.35\theta = 20^\circ \approx 0.35 rad → error 2%\approx 2\%.

Thus, the small-angle equation is accurate for modest oscillations.

Equation of motion:

θ¨+glθ=0\ddot{\theta} + \frac{g}{l}\theta = 0

Solution:

θ(t)=θ0cos ⁣(glt)\theta(t) = \theta_0 \cos\!\left(\sqrt{\tfrac{g}{l}}\,t\right)

Period:

T=2πlgT = 2\pi \sqrt{\tfrac{l}{g}}

5. The Third Law – Symmetry and Struggle

Newton’s Third Law introduces mutuality:

“To every action there is always opposed an equal reaction.”

Mathematically:

F12=F21\boldsymbol F_{12} = - \boldsymbol F_{21}

Proof → Momentum Conservation

For a two-particle system:

m1r¨1=F12,m2r¨2=F21m_1 \ddot{\boldsymbol r}_1 = \boldsymbol F_{12}, \quad m_2 \ddot{\boldsymbol r}_2 = \boldsymbol F_{21}

Adding:

m1r¨1+m2r¨2=F12+F21=0m_1 \ddot{\boldsymbol r}_1 + m_2 \ddot{\boldsymbol r}_2 = \boldsymbol F_{12} + \boldsymbol F_{21} = 0

So:

ddt(m1r˙1+m2r˙2)=0\frac{d}{dt}(m_1 \dot{\boldsymbol r}_1 + m_2 \dot{\boldsymbol r}_2) = 0

Thus total momentum P\boldsymbol P = (m1v1+m2v2)(m_1 \boldsymbol v_{1} + m_2 \boldsymbol v_{2}) is conserved.
This is the mathematical heart of collisions and recoil.


Space-Translation Symmetry → Momentum Conservation

If the Lagrangian is invariant under rr+ϵ\mathbf r \to \boldsymbol r + \epsilon, then:

P=Lr˙=mv\boldsymbol P = \frac{\partial L}{\partial \dot{\boldsymbol r}} = m \boldsymbol v

is conserved.

Example Problem 5.1 – Elastic Collision

Problem: Two masses m1,m2m_1, m_2 collide elastically in 1D. Initial velocities v1,v2v_1, v_2. Find v1,v2v_1', v_2'.

Solution:
Momentum:

m1v1+m2v2=m1v1+m2v2m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'

Energy:

12m1v12+12m2v22=12m1(v1)2+12m2(v2)2\tfrac{1}{2} m_1 v_1^2 + \tfrac{1}{2} m_2 v_2^2 = \tfrac{1}{2} m_1 (v_1')^2 + \tfrac{1}{2} m_2 (v_2')^2

Solving:

v1=(m1m2)v1+2m2v2m1+m2,v2=(m2m1)v2+2m1v1m1+m2v_1' = \frac{(m_1 - m_2)v_1 + 2 m_2 v_2}{m_1 + m_2}, \quad v_2' = \frac{(m_2 - m_1)v_2 + 2 m_1 v_1}{m_1 + m_2}

Example Problem 5.2 – The Rocket Equation (Tsiolkovsky)

Consider a rocket of instantaneous mass M(t)M(t) moving with velocity v(t)v(t) in free space (no external forces).
During a small interval dtdt, the rocket expels mass dM|dM| backward with exhaust velocity uu relative to the rocket.

1 Momentum Conservation

The change in momentum of the rocket is

dProcket=Mdv.dP_{\text{rocket}} = M\,dv.

The expelled fuel has velocity (vu)(v - u) in the inertial frame, so its momentum is

dPexhaust=(dM)(vu).dP_{\text{exhaust}} = (dM)(v - u).

Here dM<0dM<0 because the rocket loses mass.
Conservation of total momentum gives

Mdv+(dM)(vu)=vdM,M\,dv + (dM)(v - u) = v\,dM,

which simplifies to

Mdv=udM.M\,dv = -u\,dM.

2 Differential Form

Dividing by dtdt,

Mdvdt=udMdt.M \frac{dv}{dt} = -u \frac{dM}{dt}.

3 Integration

Rewriting,

dv=udMM.dv = -u \frac{dM}{M}.

Integrating from (M0,v0)(M_0,v_0) to (M,v)(M,v),

v0vdv=uM0MdMM,\int_{v_0}^{v} dv' = -u \int_{M_0}^{M} \frac{dM'}{M'},

which yields

vv0=uln ⁣(M0M).v - v_0 = u \ln\!\left(\frac{M_0}{M}\right).

Thus the general rocket equation is

v(t)=v0+uln ⁣(M0M(t)).\boxed{\,v(t) = v_0 + u \ln\!\left(\tfrac{M_0}{M(t)}\right)\,}.

4 Notes on Interpretation

  • The simplified form

    v=uln ⁣(M0M)v = u \ln\!\left(\tfrac{M_0}{M}\right)

    assumes v0=0v_0=0 (rocket starts at rest).

  • uu is the exhaust velocity relative to the rocket. Some texts define uu with a minus sign (backward direction), which can flip signs in the formula.

  • This derivation assumes no external forces (no gravity, no drag). Including them modifies the equation significantly.

5 Physical Meaning

The equation shows that rocket velocity grows logarithmically with the ratio of initial to final mass:

M0M=“mass ratio”.\frac{M_0}{M} = \text{“mass ratio”}.

A higher exhaust velocity uu or larger mass ratio leads to greater final speed.

The rocket equation comes from momentum conservation.
For rocket mass MM expelling exhaust mass dmdm at velocity uu:

Mdvdt=udMdtM \frac{dv}{dt} = -u \frac{dM}{dt}

Integrating:

v=uln ⁣(M0M)v = u \ln\!\left(\frac{M_0}{M}\right)

This is the Tsiolkovsky rocket equation.


6. Why These Laws Still Matter

Newton’s three laws form the foundation of every mechanics course, every engineering calculation, every orbit planned by space agencies.
But they also encode deep symmetries:

  • Space translation symmetry → momentum conservation
  • Rotation symmetry → angular momentum conservation
  • Time translation symmetry → energy conservation

We now see how these emerge directly from Newton’s framework.


7. A Question from the Heavens

In the 1670s, Newton pondered: What keeps the Moon in its orbit?
Kepler had given the empirical laws, but not the cause. Newton realized the same force that pulls an apple also governs the Moon.

If the Moon, distance r60Rr \approx 60 R_\oplus, falls toward Earth with acceleration a=g/602a = g/60^2, then it is the same inverse-square law that acts near the surface.


Example Problem 7.1 – Moon’s Acceleration

Problem: Earth’s g=9.8m/s2g = 9.8 \,\text{m/s}^2, radius R=6.4×106mR_\oplus = 6.4 \times 10^6 \,\text{m}. What is the Moon’s centripetal acceleration at r=60Rr = 60 R_\oplus?

Solution:

a=g6029.836000.0027m/s2a = \frac{g}{60^2} \approx \frac{9.8}{3600} \approx 0.0027 \,\text{m/s}^2

This matches the observed centripetal acceleration of the Moon. Newton saw the connection: apples and moons obey the same law.


8. The Law of Universal Gravitation

Newton proposed:

F=Gm1m2r2r^\boldsymbol F = - \frac{G m_1 m_2}{r^2} \hat r

Every mass attracts every other. This is a central force, always directed along the line joining the bodies.


Proof of the Shell Theorem

Newton proved that:

  • A uniform spherical shell attracts an external particle as if all mass were concentrated at its center.
  • Inside the shell, the net force is zero.

This requires calculus and geometry, but the result is profound: planets can be treated as point masses for orbital motion.


Example Problem 8.1 – Weight on Mt. Everest

Problem: Earth’s radius R=6.4×106mR = 6.4 \times 10^6 \,\text{m}, height of Everest h=8.8×103mh = 8.8 \times 10^3 \,\text{m}. By how much is gravity reduced at the summit?

Solution:

g=g(RR+h)2g' = g \left(\frac{R}{R+h}\right)^2 gg(6.4×1066.4×106+8.8×103)20.999997\frac{g'}{g} \approx \left(\frac{6.4 \times 10^6}{6.4 \times 10^6 + 8.8 \times 10^3}\right)^2 \approx 0.999997

Gravity is only 0.0003%0.0003\% weaker — imperceptible.


9. Central Forces and Orbital Motion

For central forces:

L=r×mv\boldsymbol L = \boldsymbol r \times m \boldsymbol v

Torque:

dLdt=r×F\frac{d\boldsymbol L}{dt} = \boldsymbol r \times \boldsymbol F

But if Fr\boldsymbol F \parallel \boldsymbol r, then r×F=0\boldsymbol r \times \boldsymbol F = 0, so:

dLdt=0L=const.\frac{d\boldsymbol L}{dt} = 0 \quad \Rightarrow \quad \boldsymbol L = \text{const.}

Thus rotation symmetry → angular momentum conservation.


Example Problem 9.1 – Earth–Sun System

Problem: If Earth is twice as far from the Sun, how must its velocity change to conserve angular momentum?

Solution:
L=mvrL = mvr constant. If r2rr \to 2r, then vv/2v \to v/2.


10. The Kepler Problem Revisited

Under an inverse-square force, the orbit equation is:

r(θ)=p1+ecosθr(\theta) = \frac{p}{1 + e \cos \theta}

where ee is eccentricity.

  • 0e<10 \leq e < 1: ellipse
  • e=1e = 1: parabola
  • e>1e > 1: hyperbola

Proof of Kepler’s 2nd Law

From angular momentum conservation:

dAdt=12r2θ˙=L2m=constant\frac{dA}{dt} = \frac{1}{2} r^2 \dot{\theta} = \frac{L}{2m} = \text{constant}

Thus equal areas are swept in equal times.


Example Problem 10.1 – Satellite Period

Problem: A satellite orbits Earth at altitude h=Rh = R_\oplus. Find its period.

Solution:
Orbit radius a=2R=1.28×107ma = 2R_\oplus = 1.28 \times 10^7 \,\text{m}.

Kepler’s 3rd Law:

T2=4π2GMa3T^2 = \frac{4\pi^2}{GM} a^3

With GM=3.986×1014GM = 3.986 \times 10^{14},

T2.0×104s5.6hoursT \approx 2.0 \times 10^4 \,\text{s} \approx 5.6 \,\text{hours}

11. Kepler’s Three Laws through Newton’s Eyes

Newton derived Kepler’s empirical rules:

  1. Elliptical Orbits: From inverse-square law.
  2. Equal Areas in Equal Times: From angular momentum conservation.
  3. Harmonic Law:
T2=4π2GMa3T^2 = \frac{4\pi^2}{GM} a^3

Example Problem 11.1 – Comet’s Orbit

Problem: A comet on a parabolic orbit (e=1e=1) approaches the Sun. What is its speed at perihelion distance rpr_p?

Solution:
For parabolic orbit, specific energy E=0E=0:

12v2GMrp=0v=2GMrp\tfrac{1}{2} v^2 - \frac{GM}{r_p} = 0 \quad \Rightarrow \quad v = \sqrt{\frac{2GM}{r_p}}

Thus parabolic comets move at escape velocity at perihelion.


12. The Halley Encounter

In 1684, Edmond Halley visited Cambridge. Astronomers puzzled over: if gravity obeys an inverse-square law, what is the shape of planetary orbits?
Halley asked Newton. Newton replied: “It would be an ellipse.”

This casual remark became the Principia, financed by Halley. Without his push, the unification of terrestrial and celestial mechanics might have remained unpublished.

Example Problem 12.1 – Halley’s Comet Period

Problem: Halley’s Comet has semi-major axis a=17.8AUa = 17.8\,\text{AU}. Find its orbital period.

Solution:
Kepler’s 3rd law (in AU and years): T2=a3T^2 = a^3
So T=(17.8)375.3yearsT = \sqrt{(17.8)^3} \approx 75.3 \,\text{years}
Exactly the observed value.


13. The Clockwork Universe

Newton’s mechanics revealed a cosmos like a machine: predictable tides, eclipses, and comets. The Enlightenment embraced this deterministic worldview.

But behind it lay deep conservation laws — the operational code of the universe.

Example Problem 13.1 – Tides

Problem: Why do tides occur twice daily?
Answer: The Moon’s gravity creates bulges on both the near and far side of Earth. As Earth rotates, each location passes through two bulges per day.


14. Legacy of Gravitation

Newton’s gravity guided ships, calendars, and later spacecraft.
Even Einstein’s relativity reduces to Newton’s law at human scales.

Example Problem 14.1 – Escape Velocity

Problem: Find escape velocity from Earth’s surface.

Solution:
Energy conservation:

12mv2GMmR=0\tfrac{1}{2} m v^2 - \frac{GMm}{R} = 0 v=2GMR11.2km/sv = \sqrt{\frac{2GM}{R}} \approx 11.2 \,\text{km/s}

15. Rigid Body Mechanics

Newtonian mechanics extends from particles to extended bodies.

15.1 Rotation About a Fixed Axis

Moment of inertia:

I=imiri2I = \sum_i m_i r_i^2

Kinetic energy:

T=12Iω2T = \tfrac{1}{2} I \omega^2

Angular momentum:

L=IωL = I \omega

Example Problem 15.1 – Uniform Rod

Problem: A uniform rod of length LL rotates about its center. Find II.

Solution:

I=L/2L/2x2MLdx=ML212I = \int_{-L/2}^{L/2} x^2 \, \frac{M}{L} dx = \frac{M L^2}{12}

15.2 Euler’s Equations

For general rigid-body rotation:

ddt(Iω)+ω×(Iω)=τ\frac{d}{dt}(I \mathbf \omega) + \mathbf \omega \times (I \mathbf \omega) = \mathbf \tau

These govern spinning tops, gyroscopes, and spacecraft.

Example Problem 15.2 – Figure Skater Spin

Problem: A skater pulls arms inward, reducing II by half. What happens to ω\omega?

Solution:
Angular momentum conserved: Iω=const.I \omega = \text{const.}
If II/2I \to I/2, then ω2ω\omega \to 2\omega.
She spins twice as fast.


16. Non-Inertial Frames

Newton’s laws hold in inertial frames. But on Earth, which rotates, we must add fictitious forces.

16.1 Coriolis and Centrifugal Forces

In a rotating frame with angular velocity Ω\vec \Omega:

a=a2Ω×vΩ×(Ω×r)\boldsymbol a' = \boldsymbol a - 2 \boldsymbol \Omega \times \boldsymbol v' - \boldsymbol \Omega \times (\boldsymbol \Omega \times \boldsymbol r)
  • 2Ω×v-2 \boldsymbol \Omega \times \boldsymbol v' = Coriolis force
  • Ω×(Ω×r)-\boldsymbol \Omega \times (\boldsymbol \Omega \times \boldsymbol r) = Centrifugal force

Example Problem 16.1 – Deflection of Falling Body

Problem: A body falls from height hh at latitude ϕ\phi. How far is it deflected east by Coriolis force?

Solution (approx):
Time of fall t=2h/gt = \sqrt{2h/g}.
Eastward acceleration 2Ωsinϕvz\approx 2 \Omega \sin\phi \cdot v_z.
Displacement:

x13Ωsinϕgt3x \approx \tfrac{1}{3} \Omega \sin\phi \, g t^3

For h=100m,ϕ=45h=100\,\text{m}, \phi=45^\circ: x2cmx \approx 2\,\text{cm}.

Example Problem 16.2 – Cyclones

Problem: Why do cyclones rotate counterclockwise in the Northern Hemisphere?

Answer: The Coriolis force deflects moving air to the right of its motion. In low-pressure systems, inflowing air curves counterclockwise in the north, clockwise in the south.


17. The Human Newton

Newton the scientist was brilliant, but Newton the man was complex. He poured decades into not only physics but also alchemy, theology, and political battles.
As Master of the Mint, he reformed English coinage, hunting counterfeiters with forensic precision. As President of the Royal Society, he promoted English science — and sometimes crushed rivals.

Yet, his mechanics transcended personal flaws: a framework both eternal and universal.

Example Problem 17.1 – The Mint’s Gold Coin

Problem: If a counterfeit coin weighs 1% less than a true coin, how would Newton detect it with a balance?

Solution: Conservation of momentum in balances implies equal torques.
A 1% mass difference causes measurable imbalance if compared against a standard reference. Newton’s obsession with precision ensured no counterfeiter escaped long.


18. Legacy and Transition

By 1727, when Newton died, his mechanics had become the intellectual foundation of the Enlightenment.
For 200 years, engineers and astronomers relied solely on Newton.

But cracks emerged:

  • Mercury’s precession defied Newton’s predictions.
  • Light, treated as particles by Newton, showed wave phenomena.
  • Atoms and electrons demanded quantum rules.

Example Problem 18.1 – Mercury’s Orbit

Problem: Mercury’s perihelion precesses an extra 4343'' per century beyond Newton’s gravity. Why?

Answer: General relativity explains the correction: spacetime curvature modifies Newton’s law at high precision.

Example Problem 18.2 – When Newton Fails

  • At near-light speeds → Special Relativity replaces F=maF=ma with F=dp/dtF = dp/dt where p=γmvp=\gamma mv.
  • At atomic scales → Quantum Mechanics replaces trajectories with wavefunctions.

Yet Newton remains the first approximation:

  • For vcv \ll c and large scales, Newton is still exact enough to guide rockets to Mars.

19. Why We Still Study Newtonian Mechanics

Despite relativity and quantum physics, Newton is still taught first — because:

  • It is accurate at human scales.
  • It provides the grammar of physics: translating words → equations → predictions.
  • It reveals the link between symmetry and conservation laws.

Recap of Symmetries and Conservation

  • Space translation invariance → momentum conservation
    P=mivi=const.\boldsymbol P = \sum m_i \boldsymbol v_i = \text{const.}

  • Rotation invariance → angular momentum conservation
    L=r×mv=const.\boldsymbol L = \boldsymbol r \times m \boldsymbol v = \text{const.}

  • Time translation invariance → energy conservation
    E=T+V=const.E = T + V = \text{const.}

These are the seeds of Noether’s theorem, fully blossomed in the 20th century.

Example Problem 19.1 – Bungee Jumper

Problem: A 70 kg jumper leaps with cord modeled as spring k=50N/mk=50\,\text{N/m}, unstretched length L=30mL=30\,\text{m}. Find maximum stretch.

Solution:
Energy conservation:

mg(L+x)=12kx2mg(L+x) = \tfrac{1}{2}kx^2

Solve quadratic: x27.7mx \approx 27.7\,\text{m}.
So maximum fall = L+x57.7mL+x \approx 57.7\,\text{m}.


20. Summary

  • Three Laws of Motion: inertia, force, reaction.
  • Universal Gravitation: unites Earth and sky.
  • Conservation Laws: momentum, angular momentum, energy emerge from symmetry.
  • Rigid Body & Non-Inertial Frames: extend laws to real-world complexity.
  • Newton the Man: brilliant, flawed, but foundational.

Newton’s vision was not final truth, but the first map of the universe in mathematics.
Every student who learns mechanics steps into the same tradition.


Epilogue

From Woolsthorpe’s apple orchard to spacecraft leaving the Solar System, Newton’s equations endure.

  • Engineers still use F=maF=ma.
  • Astronomers still predict orbits with T2a3T^2 \propto a^3.
  • Architects still compute loads with Newtonian statics.

Even when surpassed by Einstein or Schrödinger, Newton is never discarded.

He once wrote:

“I do not know what I may appear to the world, but to myself I seem to have been only like a boy playing on the seashore, finding a smoother pebble or a prettier shell than ordinary, whilst the great ocean of truth lay all undiscovered before me.”

Four centuries later, we still explore that ocean with the compass Newton gave us.