Experiment: Free Fall and Projectile Motion

Contents
  1. Historical context: the Aristotelian expectation and its collapse
  2. The observed phenomenon and its derivation
  3. Galileo's Inclined Planes (1604–1638)
  4. Absolute Ballistic Gravimetry (1995–present)
  5. Deviations and the domain of validity
  6. From Galileo's ratios to the equivalence principle

Every experimental chapter of this treatise follows the same arc: a phenomenon is stated as it is observed, the observation is confronted with a derivation from explicitly declared assumptions, and the residual disagreement — there is always one — marks the domain of validity of those assumptions. This chapter is the exemplar. Its subject, the free fall of heavy bodies and the flight of projectiles, is the oldest quantitative experiment in physics, and it remains one of the most precise: the same law of fall that Galileo extracted from a bronze ball and a water clock is today verified, in ballistic gravimeters, to nine significant figures. The derivation given in Section 20.2 is deliberately complete — every assumption numbered, every integration displayed — because it is the pattern to which all later derivations in this book, whether inline, in Long Proofs, or pending, are referred.

Historical context: the Aristotelian expectation and its collapse

For nearly two thousand years the educated expectation about falling bodies was Aristotle's: that a body falls with a speed proportional to its weight and inversely proportional to the resistance of the medium, so that a stone of ten pounds should reach the ground ten times faster than a stone of one pound. That doctrine, together with the companion difficulty of explaining why a projectile keeps moving after it leaves the hand, is rehearsed at length — and demolished — in the book that founded modern mechanics, Galileo's Discorsi e dimostrazioni matematiche intorno a due nuove scienze, composed under house arrest at Arcetri and printed by the Elzevirs at Leiden in 1638 [Galilei:1638]. There the Aristotelian position is put in the mouth of Simplicio, and Salviati answers it first with an argument requiring no apparatus at all: if a heavy body fell faster than a light one, then tying the two together should produce a compound body that falls both faster (being heavier than either) and slower (the light body retarding the heavy one) — a contradiction, from which Galileo concludes that all bodies must fall alike once the disturbance of the medium is removed [Galilei:1638].

The argument disposes of the Aristotelian ratio, but it does not by itself yield the law of fall. That required measurement, and here lies the historical significance of the experiments described in Section 20.3: free fall was the first phenomenon in history to be captured by a mathematical law that was extracted from, and tested against, quantitative data with an explicit appeal to reproducibility. The Third Day of the Discorsi defines uniformly accelerated motion, proves as a theorem that the distances traversed from rest are as the squares of the elapsed times, and derives as a corollary the odd-number rule — that the spaces covered in equal successive intervals of time stand in the ratios \(1:3:5:7:\dots\) — before reporting the inclined-plane measurements that confirm it; the Fourth Day then compounds uniform horizontal motion with naturally accelerated vertical motion and demonstrates that the resulting trajectory is a semi-parabola [Galilei:1638]. The conceptual order — definition, theorem, experiment — is precisely the one this treatise adopts.

What Galileo established kinematically, Newton grounded dynamically. In the Principia of 1687 the parabolic trajectory appears no longer as a brute geometrical fact but as the immediate consequence of the second law of motion applied to a uniform downward force, and Newton, in the scholium to the laws, credits Galileo with exactly this pair of discoveries: the squared-time law of descent and the parabola of projectiles [Newton:1687]. The present chapter retraces that inference in modern notation, and then holds it against four centuries of data.

The observed phenomenon and its derivation

Phenomenon 20.1.

Near the surface of the Earth, and wherever the resistance of the air may be neglected, projectiles follow arcs that are parabolic: the height depends quadratically on the horizontal distance, the axis of the arc is vertical, and the arc opens downward. Moreover a body released from rest falls through distances that grow as the square of the elapsed time, so that the spaces traversed in equal successive intervals stand in the ratios \(1:3:5:7:\dots\) (Galileo's odd-number rule). Both regularities are independent of the weight and composition of the body [Galilei:1638]. Rests on Equation (18.7) and Phenomenon 18.12.

This phenomenon admits a short, complete derivation from Newtonian dynamics, which we now give in full; it is the model derivation of the treatise.

Theorem 20.2 (Parabolic trajectory).

Consider a point projectile launched from the origin with speed \(v_{0} > 0\) at an angle \(\theta \in (0, 90^\circ)\) above the horizontal, in a uniform gravitational field and in the absence of air resistance, the ground being treated as flat and non-rotating. Then the motion remains in the vertical plane containing the initial velocity, and the trajectory is the arc of a parabola with vertical axis, opening downward,

\begin{equation}\tag{20.1} z = x\tan\theta - \frac{g\,x^{2}}{2 v_{0}^{2}\cos^{2}\theta}\ep \end{equation}

Rests on Postulate 19.8, Equation (18.7) and Phenomenon 20.1.

Derivation. Derives Theorem 20.2. We first state the assumptions explicitly.

  1. Point mass. The projectile is treated as a point particle of mass \(m\); its extension, internal structure and rotation are ignored.

  2. Uniform field. The gravitational field is \(\vect{g} = -g\,\hat{\vect{z}}\), constant in magnitude and direction over the region of flight; numerically \(g\) is close to the conventional standard value \(g_{0} = 9.80665\,\mathrm{m}/\mathrm{s}^{2}\) [BIPM:2019].

  3. No air resistance. The only force acting during flight is gravity.

  4. Flat, non-rotating Earth. The laboratory frame is treated as inertial: the curvature of the Earth and the centrifugal and Coriolis accelerations of its rotation are neglected.

Under assumptions 1–4, Newton's second law [Newton:1687] [Goldstein:2002] reads

\begin{equation}\tag{20.2} m\,\ddot{\vect{r}} = -m g\,\hat{\vect{z}}\ep \end{equation}

The mass cancels — an innocuous-looking step whose depth is the subject of Section 20.6 — leaving the kinematic statement \(\ddot{\vect{r}} = -g\,\hat{\vect{z}}\). Since the right-hand side is constant, two elementary integrations suffice. Integrating once from the initial condition \(\dot{\vect{r}}(0) = \vect{v}_{0}\),

\begin{equation}\tag{20.3} \dot{\vect{r}}(t) = \vect{v}_{0} - g t\,\hat{\vect{z}}\ec \end{equation}

and integrating again from \(\vect{r}(0) = \vect{0}\),

\begin{equation}\tag{20.4} \vect{r}(t) = \vect{v}_{0}\,t - \tfrac{1}{2} g t^{2}\,\hat{\vect{z}}\ep \end{equation}

Choose the \(x\)-axis along the horizontal component of \(\vect{v}_{0}\), so that \(\vect{v}_{0} = (v_{0}\cos\theta,\, 0,\, v_{0}\sin\theta)\). The \(y\) component of Equation (20.4) vanishes identically: the motion is confined to the vertical plane \(y = 0\), as claimed. In that plane the parametric solution is

\begin{equation}\tag{20.5} x(t) = v_{0}\cos\theta\;t\ec \qquad z(t) = v_{0}\sin\theta\;t - \tfrac{1}{2} g t^{2}\ep \end{equation}

Because \(\theta \neq 90^\circ\) we have \(\cos\theta \neq 0\), and the first equation can be solved for the time, \(t = x/(v_{0}\cos\theta)\). Substituting into the second eliminates \(t\) and yields Equation (20.1):

\begin{equation*} z = x\tan\theta - \frac{g\,x^{2}}{2 v_{0}^{2}\cos^{2}\theta}\ep \end{equation*}

This is a polynomial of degree exactly two in \(x\) with constant coefficients; its leading coefficient \(-g/(2 v_{0}^{2}\cos^{2}\theta)\) is strictly negative. A quadratic graph \(z = a x^{2} + b x\) with \(a < 0\) is a parabola whose axis is parallel to the \(z\)-axis and which opens downward — completing the square,

\begin{equation}\tag{20.6} z - z_{\mathrm{apex}} = -\frac{g}{2 v_{0}^{2}\cos^{2}\theta}\, \bigl(x - x_{\mathrm{apex}}\bigr)^{2}\ec \end{equation}

with the apex coordinates given in the corollary below. This proves the first claim of Phenomenon 20.1.

For the second claim, set \(\theta = 90^\circ\) and reverse the sign convention so that \(s(t)\) denotes the distance fallen from rest; then Equation (20.5) with \(v_{0} = 0\) gives

\begin{equation}\tag{20.7} s(t) = \tfrac{1}{2} g t^{2}\ep \end{equation}

Partition time into equal intervals of length \(\tau\). The distance covered during the \(n\)-th interval is

\begin{equation}\tag{20.8} \Delta_{n} = s(n\tau) - s\bigl((n-1)\tau\bigr) = \tfrac{1}{2} g \tau^{2}\bigl(n^{2} - (n-1)^{2}\bigr) = \tfrac{1}{2} g \tau^{2}\,(2n - 1)\ec \end{equation}

so that \(\Delta_{1} : \Delta_{2} : \Delta_{3} : \dots = 1 : 3 : 5 : \dots\) — Galileo's odd-number rule [Galilei:1638]. Finally, nothing in the derivation refers to \(m\) after the cancellation in Equation (20.2), which reproduces the observed independence of the motion from the weight and composition of the body.

Corollary 20.3 (Range, apex, time of flight).

With launch and landing at the same height, the time of flight, the horizontal range and the apex height are

\begin{equation}\tag{20.9} T = \frac{2 v_{0}\sin\theta}{g}\ec \qquad R = \frac{v_{0}^{2}\sin 2\theta}{g}\ec \qquad H = \frac{v_{0}^{2}\sin^{2}\theta}{2 g}\ec \end{equation}

and for fixed \(v_{0}\) the range is greatest at \(\theta = 45^\circ\), where \(R_{\max} = v_{0}^{2}/g\). Rests on Theorem 20.2 and Equation (20.5).

Proof.

Derives Corollary 20.3. From Equation (20.5), \(z(t) = t\,(v_{0}\sin\theta - \tfrac{1}{2} g t)\) vanishes at \(t = 0\) and at \(t = T = 2 v_{0}\sin\theta/g\). The range is \(R = x(T) = v_{0}\cos\theta \cdot T = 2 v_{0}^{2} \sin\theta\cos\theta/g = v_{0}^{2}\sin 2\theta/g\), using the double-angle identity. Since \(\sin 2\theta \leq 1\) with equality precisely when \(2\theta = 90^\circ\), the range is maximal at \(\theta = 45^\circ\). The apex occurs where \(\dot{z} = v_{0}\sin\theta - g t = 0\), i.e. at \(t = T/2\), by which symmetry of the parabola is recovered; substitution gives \(H = v_{0}^{2}\sin^{2}\theta/(2g)\).

Example 20.4.

A ball thrown at \(v_{0} = 10\,\mathrm{m}/\mathrm{s}\) and \(\theta = 45^\circ\) has, by Equation (20.9) with \(g = g_{0}\), a time of flight \(T \approx 1.44\,\mathrm{s}\), a range \(R = v_{0}^{2}/g_{0} \approx 10.2\,\mathrm{m}\) and an apex height \(H \approx 2.55\,\mathrm{m}\). These are the magnitudes of everyday experience, and the fact that everyday experience does not flagrantly contradict them is already a statement about air drag: for dense projectiles at such speeds the neglect of resistance in assumption 3 is a percent-level idealisation (Section 20.5). Rests on Corollary 20.3.

Remark 20.5.

The derivation used nothing but the second law and two integrations, yet its logical anatomy repays attention. Each of the four assumptions is independently falsifiable, and each does fail at some level of precision: assumption 1 for spinning or extended bodies, assumption 2 over heights where the field gradient matters (Section 20.4), assumption 3 for fast or light projectiles, and assumption 4 for long-range or long-duration motion (Section 20.5). The parabola of Equation (20.1) is therefore not “the” trajectory of a real projectile but the zeroth term of a systematic approximation — which is exactly what an idealised law of nature is. Rests on Theorem 20.2 and Equation (20.1).

Two experiments anchor the phenomenon, separated by almost four centuries and by eight orders of magnitude in timing precision: Galileo's inclined planes, which established the form of the law, and modern ballistic gravimetry, which turns the law around and uses free fall as a measuring instrument.

Galileo's Inclined Planes (1604–1638)

Tests Phenomenon 20.1 and Equation (20.8). Assuming Equation (18.7).

Apparatus

A hard, smooth, very round ball of bronze, and a straight wooden beam or moulding some twelve braccia long (roughly \(7\,\mathrm{m}\)), half a braccio wide and three finger-breadths thick, along whose upper face a channel a little more than one finger in breadth was cut, planed straight, and lined with polished parchment to make it as smooth as possible. One end of the beam could be raised one or two braccia above the horizontal, tilting the groove into an inclined plane of small, adjustable slope. Time was measured with a water clock: a large elevated vessel discharged through a thin pipe into a beaker during the motion, and the collected water was weighed after each run on a sensitive balance, the mass of the effluent serving as the measure of the elapsed time [Galilei:1638].

Procedure

The ball was released from rest at the top of the groove and timed over the full length of the channel; then over one quarter of the length, one half, two thirds, and other fractions; the tilt of the plane was varied; and, in Galileo's own report, each configuration was repeated “a full hundred times” to establish the reproducibility of the timings [Galilei:1638].

Observations and data

The distances traversed from rest were found proportional to the squares of the measured times, at every inclination of the plane: the full length was covered in exactly twice the time of the first quarter, and the spaces covered in equal successive time intervals followed the odd-number progression,

interval $n$space in intervalcumulative space
111
234
359
4716

(in units of the distance covered in the first interval), in agreement with Equation (20.8). Galileo states that repeated timings of the same descent never differed by more than one tenth of a pulse beat — of order \(0.1\,\mathrm{s}\) on a total descent time of several seconds — and that the result did not depend on the weight of the ball [Galilei:1638].

Interpretation

The measurements identify the motion as uniformly accelerated: equal increments of speed in equal times, the definition adopted in the Third Day of the Discorsi [Galilei:1638]. The inclined plane is itself the decisive methodological invention: rolling down a slope of inclination \(\alpha\) dilutes the effective acceleration by the factor \(\sin\alpha\), slowing the motion until a water clock of \(0.1\,\mathrm{s}\) resolution can resolve it, while leaving the form of the law — distance proportional to time squared — untouched. Extrapolating the diluted experiments to vertical inclination, Galileo obtained the law of free fall, Equation (20.7), without ever timing a free vertical drop. The observed independence from the ball's weight is the seed of the universality discussed in Section 20.6. (A modern reader will note what the water clock could not resolve: rolling, as opposed to sliding, reduces the acceleration by a further constant factor — \(5/7\) for a homogeneous sphere — which drops out of all of Galileo's ratios of distances and times, and so left his conclusions intact.)

Primary references

[Galilei:1638]; Newton's dynamical reading of these results is in the scholium to the laws of motion [Newton:1687].

Absolute Ballistic Gravimetry (1995–present)

Tests Equations (18.7) and (20.7). Assuming Phenomenon 20.1.

Apparatus

The FG5 class of absolute gravimeters realises Galileo's experiment with interferometric metrology [Niebauer:1995]. The falling body is a corner-cube retroreflector enclosed in a co-falling “drag-free” carriage inside an evacuated drop chamber, the carriage tracking the cube so as to shield it from residual gas drag and to carry the electrostatic environment with it. The falling cube forms one arm of a Michelson-type interferometer; the reference cube is suspended from an actively servo-controlled long-period isolation stage (a “superspring”) to decouple it from ground vibration. The length standard is an iodine-stabilised helium–neon laser at \(633\,\mathrm{nm}\); the time standard is a rubidium atomic clock [Niebauer:1995].

Procedure

The carriage releases the cube, which falls freely through about \(20\,\mathrm{cm}\) in about \(0.2\,\mathrm{s}\). Interference fringes — one per \(316.5\,\mathrm{nm}\) of fall, half the laser wavelength — are counted and time-stamped against the atomic clock, and the fringe record is scaled to roughly 200 position–time pairs per drop. Each drop is fitted by least squares to the free-fall trajectory of Equation (20.5) augmented by the vertical field gradient, and standard corrections are applied for solid-Earth tides, ocean loading, polar motion and local barometric pressure. A measurement campaign comprises hundreds of drops per hour, combined over one or more days; the statistical treatment of the drop-to-drop scatter follows the standard error analysis of repeated measurements [Niebauer:1995] [Taylor:1997].

Observations and data

quantityvalue
conventional standard $g_{0}$, exact [BIPM:2019]\(9.80665\,\mathrm{m}/\mathrm{s}^{2}\)
FG5 instrumental accuracy [Niebauer:1995]$\approx 2\times 10^{-8}\,\mathrm{m}/\mathrm{s}^{2}$ ($2\,\mu\mathrm{Gal}$)
drop-to-drop scatter [Niebauer:1995]$\approx 5\times 10^{-8}\,\mathrm{m}/\mathrm{s}^{2}$
geographic variation of $g$ (equator–pole, altitude)$\approx \pm0.03\,\mathrm{m}/\mathrm{s}^{2}$
free-air height gradient [Niebauer:1995]$\approx -3.1\times 10^{-6}\,\mathrm{m}/\mathrm{s}^{2}$ per metre

Here \(1\,\mathrm{Gal} = 10^{-2}\,\mathrm{m}/\mathrm{s}^{2}\) is the conventional unit of the gravimetric literature; the mean of a measurement set converges on the drop-to-drop scatter as \(1/\sqrt{N}\) in the number of drops [Taylor:1997]. The fitted trajectories reproduce \(s(t) = s_{0} + v_{0} t + \tfrac{1}{2} g t^{2}\) at the nanometre level over the whole drop, and the instrument accuracy of \(2\times 10^{-8}\,\mathrm{m}/\mathrm{s}^{2}\) against \(g \approx 9.8\,\mathrm{m}/\mathrm{s}^{2}\) means that the local acceleration of free fall is determined to about two parts in \(10^{9}\) — nine significant figures [Niebauer:1995].

Interpretation

Three centuries of refinement have inverted the logic of the experiment. Galileo needed the inclined plane to make the law of fall measurable; the FG5 assumes the law — Newtonian free fall in a locally uniform field with a known gradient — and measures its single parameter \(g\) so precisely that the instrument serves geodesy and metrology: monitoring ground-water and magma movements, realising the unit of force, and calibrating the watt-balance experiments that underpin the 2019 redefinition of the SI [BIPM:2019] [Tiesinga:2021]. Equally important is what the fits do not show: dropped test bodies of different construction yield the same \(g\) within the stated uncertainty, in line with the far sharper torsion-balance (Eötvös-type) tests of the universality of free fall discussed in The Equivalence Principle and Classical Tests. The residual systematic budget — laser stability, clock calibration, vertical alignment, residual drag — is itself a textbook exercise in the error analysis of Measurement, SI Units, and the Theory of Errors [Taylor:1997].

Primary references

[Niebauer:1995] [Tiesinga:2021]; conventions for \(g_{0}\) in [BIPM:2019].

Deviations and the domain of validity

The parabola of Equation (20.1) is exact only within assumptions 1–4 of the derivation. Each assumption fails observably, and the pattern of failure is itself lawful. We record here the two deviations that dominate at laboratory and ballistic scales, air resistance and the Earth's rotation, stating in each case what is observed; the corresponding derivations are pending and will follow the standards set by the exemplar above.

Phenomenon 20.6 (Air drag).

Real trajectories in air fall short of Equation (20.9) and are asymmetric: the descending branch is steeper than the ascending one, and the range at fixed muzzle speed is maximised at launch angles below \(45^\circ\). A body falling for long enough ceases to accelerate altogether and reaches a constant terminal speed — raindrops arrive at metres per second, not at the hundreds of metres per second that Equation (20.7) would assign to their fall height [Newton:1687] [Eiffel:1912]. Rests on Theorem 20.2 and Corollary 20.3.

The observed deviations follow from adding to Equation (20.2) a resistive force directed against the velocity,

\begin{equation}\tag{20.10} m\ddot{\vect{r}} = -m g\,\hat{\vect{z}} - f\!\left(\abs{\vect{v}}\right)\vect{v}\ec\qquad f>0\ec \end{equation}

in which \(f\) carries the SI unit \(\mathrm{kg}/\mathrm{s}\). Two regimes cover the practical cases [Landau:1976]. At low Reynolds number the resistance is linear in the velocity, \(f=b\) constant; at the Reynolds numbers of a macroscopic projectile in air it is quadratic, \(f=c\abs{\vect{v}}\) with \(c=\tfrac{1}{2}\rho\,C_{\mathrm{D}}A\) in \(\mathrm{kg}/\mathrm{m}\), built from the density \(\rho\) of the fluid, the frontal area \(A\) and a dimensionless drag coefficient \(C_{\mathrm{D}}\) which is itself a function of the Reynolds number and which drops abruptly at the drag crisis of a sphere [Eiffel:1912].

Proposition 20.7 (Terminal speed and the approach to it).

Let a body fall from rest along the vertical under Equation (20.10), and let \(v>0\) be its downward speed. Then \(v\) increases monotonically towards the terminal speed \(v_{\mathrm{t}}\) fixed by \(f(v_{\mathrm{t}})\,v_{\mathrm{t}}=mg\),

\begin{equation}\tag{20.11} v_{\mathrm{t}}=\frac{mg}{b}\ \ \text{(linear)}\ec\qquad v_{\mathrm{t}}=\sqrt{\frac{mg}{c}}\ \ \text{(quadratic)}\ec \end{equation}

and the approach to it is

\begin{equation}\tag{20.12} v(t)=v_{\mathrm{t}} \left(1-\ee^{-g t/v_{\mathrm{t}}}\right)\ec\qquad v(t)=v_{\mathrm{t}} \tanh\!\left(\frac{g t}{v_{\mathrm{t}}}\right)\ec \end{equation}

in the linear and quadratic cases respectively. In both the characteristic time is \(v_{\mathrm{t}}/g\). Rests on Equation (20.10) and Theorem 9.8.

Proof.

Derives Proposition 20.7. Along the vertical, Equation (20.10) reads \(m\,\dd v/\dd t = mg - f(v)\,v\). Dividing by \(m\) and using \(f(v_{\mathrm{t}})v_{\mathrm{t}}=mg\) to eliminate the constants,

\begin{equation}\tag{20.13} \dv{v}{t}=g\left(1-\frac{v}{v_{\mathrm{t}}}\right)\ec\qquad \dv{v}{t}=g\left(1-\frac{v^{2}}{v_{\mathrm{t}}^{2}}\right)\ec \end{equation}

for \(f=b\) and for \(f=c\abs{v}\) respectively, with Equation (20.11) for \(v_{\mathrm{t}}\) in each case. Each right-hand side is of class \(C^{1}\) in \(v\) and hence Lipschitz on any bounded interval, so by Theorem 9.8 the initial-value problem with \(v(0)=0\) has exactly one solution; it is therefore enough to check that Equation (20.12) solves it. In the linear case \(\dd v/\dd t=g\,\ee^{-gt/v_{\mathrm{t}}} =g\left(1-v/v_{\mathrm{t}}\right)\); in the quadratic case the identity \(\tanh'=1-\tanh^{2}\) gives \(\dd v/\dd t=g\left(1-\tanh^{2}(gt/v_{\mathrm{t}})\right) =g\left(1-v^{2}/v_{\mathrm{t}}^{2}\right)\); and both vanish at \(t=0\). Monotonicity and the limit follow from Equation (20.13), whose right-hand side is strictly positive for \(0\le v<v_{\mathrm{t}}\) and vanishes only at \(v=v_{\mathrm{t}}\), which is therefore approached but never attained in finite time.

Example 20.8 (A raindrop).

For a spherical water drop of radius \(1.0\,\mathrm{mm}\) — mass \(m=\tfrac{4}{3}\pi r^{3}\rho_{\mathrm{w}}=4.19\times 10^{-6}\,\mathrm{kg}\), frontal area \(A=\pi r^{2}=3.14\times 10^{-6}\,\mathrm{m}^{2}\) — falling in air of density \(\rho=1.2\,\mathrm{kg}/\mathrm{m}^{3}\) with \(C_{\mathrm{D}}=0.5\), Equation (20.11) gives

\begin{equation}\tag{20.14} v_{\mathrm{t}}=\sqrt{\frac{2mg}{\rho\,C_{\mathrm{D}}A}} \approx6.6\,\mathrm{m}/\mathrm{s}\ec \end{equation}

reached to within one per cent after about \(1.8\,\mathrm{s}\) of fall. Without resistance, Equation (20.7) would give a drop falling from \(1\,\mathrm{km}\) an arrival speed of \(\sqrt{2gh}\approx140\,\mathrm{m}/\mathrm{s}\) — a factor of twenty in speed and some four hundred in kinetic energy, which is the difference between rain and gunfire. Rests on Proposition 20.7.

Proposition 20.9 (Linear resistance: exact motion, and the lost range).

Let \(f=b\) be constant and write \(\tau=m/b=v_{\mathrm{t}}/g\). The solution of Equation (20.10) with \(\vect{r}(0)=\vect{0}\) and \(\dot{\vect{r}}(0)=\vect{v}_{0}\) is

\begin{align} \dot{\vect{r}}(t)&=\left(\vect{v}_{0} +v_{\mathrm{t}}\hat{\vect{z}}\right)\ee^{-t/\tau} -v_{\mathrm{t}}\hat{\vect{z}}\ec \tag{20.15}\\ \vect{r}(t)&=\tau\left(\vect{v}_{0} +v_{\mathrm{t}}\hat{\vect{z}}\right) \left(1-\ee^{-t/\tau}\right) -v_{\mathrm{t}}\,t\,\hat{\vect{z}}\ep \tag{20.16} \end{align}

For launch and landing at the same height, and to first order in the small parameter \(v_{0}/v_{\mathrm{t}}\), the range is

\begin{equation}\tag{20.17} R=R_{0}\left(1-\frac{4}{3}\, \frac{v_{0}\sin\theta}{v_{\mathrm{t}}}\right) +O\!\left(\frac{v_{0}^{2}}{v_{\mathrm{t}}^{2}}\right)\ec\qquad R_{0}=\frac{v_{0}^{2}\sin2\theta}{g}\ec \end{equation}

and \(\dd R/\dd\theta<0\) at \(\theta=45^\circ\), so that at fixed \(v_{0}\) the range is greatest at a launch angle strictly below \(45^\circ\). Rests on Equation (20.10), Corollary 20.3 and Corollary 9.9.

Proof.

Derives Proposition 20.9. The solution. Differentiating Equation (20.16) returns Equation (20.15), and differentiating once more,

\[ \ddot{\vect{r}} =-\frac{1}{\tau}\left(\vect{v}_{0} +v_{\mathrm{t}}\hat{\vect{z}}\right)\ee^{-t/\tau} =-\frac{1}{\tau}\left(\dot{\vect{r}} +v_{\mathrm{t}}\hat{\vect{z}}\right) =-\frac{b}{m}\dot{\vect{r}}-g\hat{\vect{z}}\ec \]

since \(v_{\mathrm{t}}/\tau=g\); the initial conditions are met by inspection. The system is linear with constant coefficients, so by Corollary 9.9 this is the only solution.

The range. Choose the \(x\)-axis along the horizontal component of \(\vect{v}_{0}\) and write \(v_{0x}=v_{0}\cos\theta\), \(v_{0z}=v_{0}\sin\theta\). Expanding \(1-\ee^{-u}=u-u^{2}/2+u^{3}/6+\dots\) in Equation (20.16) and keeping terms through first order in \(1/\tau\), with \(v_{\mathrm{t}}/\tau=g\),

\begin{equation}\tag{20.18} x(t)=v_{0x}\left(t-\frac{t^{2}}{2\tau}\right)\ec\qquad z(t)=v_{0z}t-\frac{1}{2}gt^{2} -\frac{v_{0z}t^{2}}{2\tau}+\frac{gt^{3}}{6\tau}\ep \end{equation}

The time of flight solves \(z(T)=0\); dividing by \(T\) and writing \(T=T_{0}+T_{1}\) about the drag-free value \(T_{0}=2v_{0z}/g\) of Equation (20.9), the residual at \(T_{0}\) is

\[ -\frac{v_{0z}T_{0}}{2\tau}+\frac{gT_{0}^{2}}{6\tau} =-\frac{v_{0z}^{2}}{g\tau}+\frac{2v_{0z}^{2}}{3g\tau} =-\frac{v_{0z}^{2}}{3g\tau}\ec \]

which against the slope \(-g/2\) gives \(T_{1}=-2v_{0z}^{2}/\left(3g^{2}\tau\right)\), that is \(T=T_{0}\left(1-v_{0z}/3v_{\mathrm{t}}\right)\). Substituting into the first of Equation (20.18),

\[ R=v_{0x}T-\frac{v_{0x}T^{2}}{2\tau} =R_{0}\left(1-\frac{v_{0z}}{3v_{\mathrm{t}}}\right) -R_{0}\frac{v_{0z}}{v_{\mathrm{t}}} =R_{0}\left(1-\frac{4}{3}\frac{v_{0z}}{v_{\mathrm{t}}}\right)\ec \]

using \(R_{0}=v_{0x}T_{0}=2v_{0x}v_{0z}/g\) in both terms, which is Equation (20.17).

The optimal angle. Write \(R(\theta)=\left(v_{0}^{2}/g\right)\sin2\theta \left(1-\varepsilon\sin\theta\right)\) with \(\varepsilon=4v_{0}/3v_{\mathrm{t}}\). Then

\[ \dv{R}{\theta}=\frac{v_{0}^{2}}{g} \left[2\cos2\theta\left(1-\varepsilon\sin\theta\right) -\varepsilon\sin2\theta\cos\theta\right]\ec \]

and at \(\theta=45^\circ\) the first bracket vanishes with \(\cos2\theta\), leaving \(\dd R/\dd\theta =-\varepsilon v_{0}^{2}/\left(g\sqrt{2}\right)<0\). The stationary point has therefore moved below \(45^\circ\).

Proposition 20.10 (The descending branch is the steeper, whatever the resistive law).

Let the motion obey Equation (20.10) with any continuous \(f>0\), launched from the origin with \(v_{0x}>0\) and \(v_{0z}>0\) and followed until it returns to \(z=0\). Then \(v_{x}\) is positive and strictly decreasing throughout the flight, and the slope of the trajectory at landing exceeds the launch slope in magnitude,

\begin{equation}\tag{20.19} \abs{\left.\dv{z}{x}\right|_{\text{landing}}}>\tan\theta\ep \end{equation}

With \(f\equiv0\) the two are equal, which is the symmetry of the parabola Equation (20.6). Rests on Equations (20.6) and (20.10).

Proof.

Derives Proposition 20.10. In components, Equation (20.10) reads \(m\dot{v}_{x}=-f v_{x}\) and \(m\dot{v}_{z}=-mg-f v_{z}\). The first is linear and homogeneous in \(v_{x}\) with continuous coefficient, so \(v_{x}(t)=v_{0x}\exp\left(-\int_{0}^{t}f/m\,\dd t'\right)>0\), and \(\dot{v}_{x}=-\left(f/m\right)v_{x}<0\): horizontal speed is positive and strictly decreasing. Hence \(x\) is strictly increasing and may be used as a parameter along the trajectory. For the slope \(p=\dd z/\dd x=v_{z}/v_{x}\),

\begin{equation}\tag{20.20} \dv{p}{t} =\frac{\dot{v}_{z}v_{x}-v_{z}\dot{v}_{x}}{v_{x}^{2}} =\frac{\left(-g-\frac{f}{m}v_{z}\right)v_{x} +\frac{f}{m}v_{z}v_{x}}{v_{x}^{2}} =-\frac{g}{v_{x}}\ec\qquad \dv{p}{x}=-\frac{g}{v_{x}^{2}}\ec \end{equation}

the resistance cancelling identically. So \(p\) decreases strictly and monotonically from \(p=\tan\theta\) at launch, through \(p=0\) at the apex \(x_{\mathrm{a}}\), to \(p=-q_{\mathrm{f}}<0\) at landing; each value of the slope is attained exactly once, and \(p\) may itself be used as parameter, with \(\dd x=-\left(v_{x}^{2}/g\right)\dd p\).

Let \(v_{\uparrow}(u)\) and \(v_{\downarrow}(u)\) denote the horizontal speed at the instants where the slope is \(+u\) on the way up and \(-u\) on the way down. The apex height is reached and then lost, so the two integrals

\begin{equation}\tag{20.21} H=\int_{0}^{x_{\mathrm{a}}}\!\!p\,\dd x =\frac{1}{g}\int_{0}^{\tan\theta}\!\!u\,v_{\uparrow}(u)^{2}\dd u\ec \qquad H=\int_{x_{\mathrm{a}}}^{x_{\mathrm{f}}}\!\!\left(-p\right)\dd x =\frac{1}{g}\int_{0}^{q_{\mathrm{f}}}\!\! u\,v_{\downarrow}(u)^{2}\dd u \end{equation}

are equal. Every instant of the descent is later than every instant of the ascent and \(v_{x}\) is strictly decreasing, so \(v_{\downarrow}(u)<v_{\uparrow}(u)\) wherever both are defined. Were \(q_{\mathrm{f}}\le\tan\theta\), the second integral would be strictly smaller than the first, contradicting their equality; hence \(q_{\mathrm{f}}>\tan\theta\), which is Equation (20.19). With \(f\equiv0\), \(v_{x}\) is constant, both integrands are \(u\,v_{0x}^{2}\), and the two limits coincide.

Derivation. Derives Phenomenon 20.6. Each clause of Phenomenon 20.6 is now accounted for. The shortfall against Equation (20.9) is Equation (20.17), first order in \(v_{0}/v_{\mathrm{t}}\) and negative. The asymmetry of the two branches is Proposition 20.10, which needs no assumption about the resistive law beyond its sign, and which is why the effect is seen for bodies of every kind. The optimum below \(45^\circ\) is the last claim of Proposition 20.9. The terminal speed is Proposition 20.7, and the numbers for a raindrop are Example 20.8: \(6.6\,\mathrm{m}/\mathrm{s}\), not the \(140\,\mathrm{m}/\mathrm{s}\) that Equation (20.7) would assign to a fall of \(1\,\mathrm{km}\).

Remark 20.11 (What has no closed form, and what to do about it).

Proposition 20.9 is exact only for the linear law. The two-dimensional quadratic problem — Equation (20.10) with \(f=c\abs{\vect{v}}\) — couples the two components through \(\abs{\vect{v}}=\sqrt{v_{x}^{2}+v_{z}^{2}}\) and has no solution in closed form; the vertical motion of Proposition 20.7 is integrable only because there the motion is one-dimensional and \(\abs{\vect{v}}=v\). What survives without integration is exactly what the three propositions above deliver, and the rest is obtained numerically. Whether a given experiment may ignore resistance is then an error-budget question, settled by comparing the estimated correction with the measurement uncertainty [Taylor:1997]: Galileo's dense bronze ball, at speeds of a few \(\mathrm{m}/\mathrm{s}\) on a plane whose water clock resolved \(0.1\,\mathrm{s}\), could; a musket ball could not. Rests on Propositions 20.9 and 20.10.

Phenomenon 20.12 (Coriolis deflection).

A body dropped from rest in a tall shaft does not land at the plumb point below its release: it lands displaced towards the east, in either hemisphere, by an amount that grows with the fall height — about \(1.5\,\mathrm{cm}\) for a \(100\,\mathrm{m}\) drop at latitude \(45^\circ\) [Reich:1832]. Long-range projectiles are likewise deflected sideways relative to the parabolic plane of Equation (20.1). Rests on Equation (20.1) and Phenomenon 29.53.

Assumption 4 is the one at fault, and the argument that replaces it can be made without ever leaving an inertial frame — which is worth doing once, because the fictitious-force bookkeeping of Section 29.5 is easy to perform and easy to mistrust.

Proposition 20.13 (Eastward deflection of a body dropped from rest).

Let a body be released from rest with respect to the ground, at a height \(h\ll R\) above the surface of the Earth at geographic latitude \(\lambda\), with \(R\) the Earth's radius and \(\omega=7.292\times 10^{-5}\,\mathrm{rad}/\mathrm{s}\) its sidereal rotation rate. To leading order in \(\omega\) and in \(h/R\) it strikes the ground east of the point vertically below its release by

\begin{equation}\tag{20.22} d=\frac{1}{3}\,\omega g\,t_{\mathrm{f}}^{3}\cos\lambda =\frac{\omega\cos\lambda}{3}\sqrt{\frac{8h^{3}}{g}}\ec\qquad t_{\mathrm{f}}=\sqrt{\frac{2h}{g}}\ec \end{equation}

in either hemisphere, since \(\cos\lambda>0\) at every latitude but the poles, where the deflection vanishes. Rests on Equation (20.7), Equation (19.14) and Postulate 19.19.

Proof.

Derives Proposition 20.13. Work in a frame that does not rotate with the Earth, with its origin at the Earth's centre, and take cylindrical coordinates \(\left(\varrho,\varphi,\zeta\right)\) about the polar axis with \(\varphi\) increasing eastward. Treat the Earth's attraction as central, \(\vect{F}=-F(r)\hat{\vect{r}}\). Then the torque about the centre vanishes identically, \(\vect{x}\times\vect{F}=\vect{0}\), so by Equation (19.14) the angular momentum is conserved; in particular its polar component

\begin{equation}\tag{20.23} L_{\zeta}=m\varrho^{2}\dot{\varphi}=\text{constant} \end{equation}

is constant throughout the fall.

At release the body shares the motion of the ground, so \(\dot{\varphi}(0)=\omega\), and it stands at radius \(r_{0}=R+h\), that is at \(\varrho_{0}=r_{0}\cos\lambda\) from the axis. To leading order in \(\omega\) the fall is radial and obeys Equation (20.7), so \(r(t)=r_{0}-\tfrac{1}{2}gt^{2}\) at unchanged latitude and \(\varrho(t)=r(t)\cos\lambda\). Then Equation (20.23) gives

\begin{equation}\tag{20.24} \dot{\varphi}(t)=\omega\left(\frac{r_{0}}{r(t)}\right)^{2}\ec \qquad\text{so}\qquad \dot{\varphi}-\omega \simeq\omega\,\frac{2\left(r_{0}-r\right)}{R} =\frac{\omega g t^{2}}{R}\ec \end{equation}

the body running ahead of the rotation because, as it approaches the axis, conservation of \(L_{\zeta}\) speeds its azimuthal rate up.

The ground point that lay directly below at release keeps the azimuthal rate \(\omega\) exactly. The body's lead in azimuth after the flight time \(t_{\mathrm{f}}\) of Equation (20.7) is therefore

\begin{equation}\tag{20.25} \Delta\varphi=\int_{0}^{t_{\mathrm{f}}} \left(\dot{\varphi}-\omega\right)\dd t =\frac{\omega g\,t_{\mathrm{f}}^{3}}{3R}\ec \end{equation}

and the arc this subtends at the ground, at distance \(R\cos\lambda\) from the axis, is \(d=R\cos\lambda\,\Delta\varphi\), which is Equation (20.22). Substituting \(t_{\mathrm{f}}=\sqrt{2h/g}\) gives the second form.

Three things were dropped and each is small, the numbers being quoted for a hundred-metre drop. Terms of relative order \(\omega t_{\mathrm{f}}\approx3\times 10^{-4}\) come from the departure of the fall from radial; terms of relative order \(h/R\approx2\times 10^{-5}\) come from replacing \(r_{0}\) by \(R\); and the plumb vertical is tilted from the radial direction by the centrifugal term of Equation (29.68), by about \(2\times 10^{-3}\) radians, but that tilt lies in the meridian plane and so displaces the plumb point north or south, not east.

Remark 20.14 (Why the elementary argument gives three halves of this).

The argument that suggests itself first — at release the body's eastward speed exceeds the ground's by \(\omega h\cos\lambda\), so in a time \(t_{\mathrm{f}}\) it should gain \(\omega h\cos\lambda\,t_{\mathrm{f}} =\tfrac{1}{2}\omega g t_{\mathrm{f}}^{3}\cos\lambda\) — returns three halves of Equation (20.22). It fails on two counts, which is why the discrepancy is not obviously either way: the excess is not held constant, since Equation (20.24) has the azimuthal rate rising during the fall; and what must be compared with the ground is the angle swept, not a linear speed measured at a different distance from the axis. The rotating-frame route reaches Equation (20.22) in fewer lines, through the Coriolis acceleration \(-2\vect{\omega}\times\vect{v}\) of Section 29.5, whose leading contribution \(-2\vect{\omega}\times\left(-gt\hat{\vect{z}}\right)\) has magnitude \(2\omega g t\cos\lambda\) and points due east, and integrates twice from rest to the same \(\tfrac{1}{3}\omega g t^{3}\cos\lambda\) [Landau:1976] [Coriolis:1835]. The two routes agree, as they must. Rests on Proposition 20.13.

Derivation. Derives Phenomenon 20.12. Proposition 20.13 is the statement, and the numbers follow from it. For a drop of \(h=100\,\mathrm{m}\) at \(\lambda=45^\circ\), Equation (20.22) gives \(t_{\mathrm{f}}=4.52\,\mathrm{s}\) and \(d=1.55\,\mathrm{cm}\), which is the figure quoted in Phenomenon 20.12. Reich's measurement is the test: dropping balls down a mine shaft at Freiberg, at latitude about \(50.9^\circ\) and through some \(158\,\mathrm{m}\), he reported a mean eastward deviation of about \(28\,\mathrm{mm}\) [Reich:1832], against the \(27.5\,\mathrm{mm}\) that Equation (20.22) gives for those parameters. The lateral deflection of a long-range projectile has the same origin, the Coriolis term acting on a horizontal velocity instead of a vertical one, and the whole effect is the free-fall counterpart of the Foucault pendulum [Foucault:1851] — a direct mechanical signature of assumption 4 failing, that is, of the laboratory frame not being inertial.

Beyond these two, the remaining assumptions set their own limits. The uniform-field assumption 2 fails at the level of the free-air gradient measured in Section 20.4 — already relevant inside a \(20\,\mathrm{cm}\) gravimeter drop at nine significant figures — and fails qualitatively for ranges comparable with the Earth's radius, where the trajectory closes into an ellipse about the Earth's centre: Newton's celebrated diagram of a cannonball fired from a mountaintop makes the parabola visibly a local approximation to an orbit [Newton:1687]. Assumption 1 excludes the spin of real projectiles, whose coupling to the air (the Magnus effect) bends every driven golf ball and curled football; these lie beyond the scope of this chapter.

From Galileo's ratios to the equivalence principle

One step of the derivation above deserves to be read twice. In Equation (20.2) the mass \(m\) multiplies both sides — once as the inertial mass, the coefficient of resistance to acceleration in Newton's second law, and once as the gravitational mass, the coupling of the body to the field. Their cancellation is what made the trajectory independent of the body, and that independence is not a theorem but an experimental fact: Galileo's inclined planes could not distinguish the bronze ball from a heavier one [Galilei:1638], Newton checked the equality with pendulums of many materials [Newton:1687], and its modern descendants — Eötvös-type torsion balances and satellite tests reviewed in The Equivalence Principle and Classical Tests — constrain any composition-dependence of free fall at the level of parts in \(10^{13}\) and beyond.

Einstein's insight of 1907, developed into the general theory of relativity [Einstein:1916], was to take this oldest of experimental facts as a founding principle: if all bodies fall alike, then free fall is not a forced motion at all but the natural, unforced state, and gravitation is a property of spacetime geometry rather than of the falling body. The uniform field \(\vect{g}\) of assumption 2 becomes, in that reading, the first-order description of a curved spacetime in a small laboratory. The bronze ball on the parchment-lined groove thus stands at the head of a straight line of inference that runs through this treatise to Part V — General Relativity and Cosmology; the reader will meet Phenomenon 20.1 again there, restated as the geodesic motion of a test body.