ARCHETYPE: The Tunnel Through the Earth

In 1679, Robert Hooke wrote to Isaac Newton with a puzzle designed to lure him back into natural philosophy: imagine dropping a ball into a hole bored straight through the Earth. What path does it take, and what motion does it perform? The correspondence that followed — testy, competitive, and ultimately productive — helped push Newton toward the Principia. The puzzle itself became physics' most beloved impossible experiment: a frictionless, airless tunnel through a uniform Earth, and a stone released at the lip. The answer is delicious. The stone does not plummet to the centre and stop; it oscillates — surface to centre to the far surface and back — with a round trip of about 84 minutes, the same 84 minutes as a satellite skimming the ground. Fall through Delhi and you emerge, momentarily at rest, off the coast of Chile; fail to grab the edge and you fall all the way home again, forever. Stranger still, a straight tunnel between any two cities — London to Paris, or pole to pole — takes the same 42 minutes one-way. For JEE, this is the classic bridge problem: it starts in gravitation (the shell theorem's $g \propto r$ inside the Earth — the cheque that module wrote, cashed here) and ends in oscillations, because the restoring force turns out proportional to displacement — the exact fingerprint that the next module will name simple harmonic motion. Solve the tunnel and you have already met SHM, one chapter early.

Key takeaways

Inside a uniform Earth, gravity is $g(r) = g_0\,r/R$ — linear in distance from the centre (shell theorem). A stone in a diametral tunnel feels $F = -\dfrac{mg_0}{R}\,x$ : a restoring force proportional to displacement.
That force law produces oscillation with $\omega^2 = \dfrac{g_0}{R} = \dfrac{GM}{R^3}$ and period $T = 2\pi\sqrt{\dfrac{R}{g_0}} \approx 5060\ \mathrm{s} \approx 84\ \mathrm{min}$ — one-way trip $42$ minutes.
Maximum speed at the centre: $v_{\max} = \sqrt{g_0R} \approx 7.9\ \mathrm{km/s}$ — exactly the ground-skimming orbital speed.
The chord theorem: a straight frictionless tunnel between any two surface points has the same period — every ride is 42 minutes, regardless of destination.
The period depends only on the planet's density: $T = \sqrt{3\pi/G\rho}$ — a denser planet is a faster oscillator, whatever its size.
Key idea: linear-in-displacement gravity inside the Earth turns the tunnel into a giant oscillator whose period equals the skimming orbit's — the bridge from gravitation to SHM.

The core insight

A body in a straight frictionless tunnel through a uniform Earth experiences a restoring force proportional to its displacement from the tunnel's midpoint — $F = -(mg_0/R)x$ — and therefore oscillates with period $T = 2\pi\sqrt{R/g_0} = 2\pi\sqrt{R^3/GM} \approx 84$ minutes, independent of the tunnel's endpoints and equal to the period of a surface-skimming orbit.

Where this lives today

Nobody will bore the tunnel — the route passes through rock at thousands of kelvin and pressures of millions of atmospheres — but the physics pays rent everywhere. The identity $\omega^2 = GM/R^3$ makes "tunnel period = orbital period" a working tool: asteroid scientists quote a rubble pile's critical spin (the rotation rate at which it sheds surface rocks) directly from its density via this article's $\sqrt{3\pi/G\rho}$ , and the same number sets the orbital period just above the surface of any body, from Ceres to a neutron star (for which $\rho \sim 5\times10^{17}\ \mathrm{kg/m^3}$ gives $T \sim$ a millisecond — matching the fastest pulsars). The 2015 Klotz calculation — 38 minutes with the real density profile — is a small classic of honest modelling, showing exactly how much the uniform-Earth idealization costs. And conceptually, this archetype is the door you are now standing in: the equation $a = -\omega^2x$ that ran the whole show is about to be adopted, next module, as the definition of simple harmonic motion — springs, pendulums, and vibrating molecules all obeying the law you just used on a planet.

The setup

Bore a straight tunnel through the Earth's centre, pole to pole (choosing the poles sidesteps the Earth's spin). Evacuate the air, ignore friction, pretend the rock is uniform and survivable — every one of these idealizations is fatal to engineering and harmless to physics. Release a stone at one lip. The whole problem is decided by one fact, proved in the shell-theorem module and merely cashed here: at interior radius $r$ , the shells of rock outside $r$ pull not at all, and the sphere inside $r$ acts as a point mass. For uniform density, the enclosed mass shrinks as $r^3$ while the inverse square concentrates as $1/r^2$ — net result, gravity falls off linearly on the way down:

g(r) = g_0\,\frac{r}{R} \qquad (1)

where: $g_0 = GM/R^2 = 9.8\ \mathrm{m/s^2}$ at the surface, dropping in a straight line to zero at the centre. Written as a force on the stone at displacement $x$ from the centre (taking outward as positive):

F = -\frac{mg_0}{R}\,x \qquad (2)

where: the minus sign is the physics — the force always points toward the centre, opposing the displacement. Pulled back when out, overshooting through the middle, pulled back again: the stone cannot rest anywhere but the centre, and it arrives there too fast to stay. This force law — restoring, and proportional to displacement — is the most important structural pattern in all of oscillation physics, and the next module adopts it as the very definition of simple harmonic motion.

Figure

The tunnel oscillator — linear interior gravity turns the Earth into a spring

Before deriving anything, read off the ride's biography from the force law. Acceleration is maximal ( $g_0$ ) at the lips, where the stone is momentarily at rest; it is zero at the centre, where the stone is fastest. The trip is perfectly symmetric — decelerating on the way up the far side exactly as it accelerated coming down — so the stone arrives at the antipode with zero speed, pauses for an instant, and falls back. Without friction this repeats forever: a planetary pendulum with an 84-minute tick.

The master pattern

Move 1 — certify the oscillator. Show the net force has the form $F = -kx$ with a positive constant $k$ ; here $k = mg_0/R$ from equation (2). Any motion governed by such a law repeats with angular frequency and period

\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{g_0}{R}}, \qquad T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{R}{g_0}} = 2\pi\sqrt{\frac{R^3}{GM}} \qquad (3)

where: the timetable $T = 2\pi\sqrt{m/k}$ is the standard result for linear restoring forces — the oscillations chapter, which opens two modules ahead, derives it in full; the tunnel borrows it with this signpost, exactly as mission planners borrow it daily. Numbers: $T = 2\pi\sqrt{6.37\times10^6/9.8} = 5060\ \mathrm{s} \approx 84\ \mathrm{min}$ ; one-way $\approx 42\ \mathrm{min}$ .

Move 2 — speed anywhere, by energy. The force–displacement graph is a straight line, so work is the area under it (the work-by-variable-force module's tool). From the surface to radius $r$ , per kilogram, the trapezoid gives $\tfrac{g_0}{2R}(R^2 - r^2)$ , so

v(r) = \sqrt{\frac{g_0}{R}\left(R^2 - r^2\right)}, \qquad v_{\max} = v(0) = \sqrt{g_0R} \approx 7.9\ \mathrm{km/s} \qquad (4)

where: the centre speed equals the ground-skimming orbital speed — no coincidence, as the second figure shows: the tunnel ride is a circular orbit watched edge-on.

Move 3 — the chord upgrade. For a straight tunnel that misses the centre, only the component of gravity along the tunnel accelerates the stone — and that component works out to $-(g_0/R)\times$ (displacement along the tunnel), the same constant (Derivation B). Same $\omega$ , same 42-minute ride, any chord.

variant asked	move	result
period / one-way time	eq. (3)	$84$ / $42$ min
speed at the centre	eq. (4) at $r = 0$	$\sqrt{g_0R} = 7.9\ \mathrm{km/s}$
speed at depth	eq. (4)	e.g. $r = R/2$ : $6.8\ \mathrm{km/s}$
chord tunnel (city to city)	Derivation B	same $T$ , gentler ride
another planet	$T = 2\pi\sqrt{R^3/GM} = \sqrt{3\pi/G\rho}$	density alone decides
relation to orbit	projection argument	tunnel = orbit's shadow

Derivation ACertifying the oscillator and its 84-minute period

We derive the tunnel's period, equation (3), from the interior force law. Strategy: establish $F = -kx$ honestly from the shell theorem's linear gravity, identify the constant $k$ , and apply the linear-oscillator timetable $T = 2\pi\sqrt{m/k}$ (stated here, derived in the oscillations chapter — the borrow is signposted, not smuggled). Watch the mass cancel: the ride's schedule is the same for a pebble and a train.

Step 1. Interior gravity from the shell theorem (cited, not re-proved): at radius $r$ inside a uniform Earth, only the enclosed sphere acts, giving $g(r) = g_0\,r/R$ — equation (1).

Step 2. Force on the stone at displacement $x$ from the centre, along the tunnel, outward positive:

F = -m\,g(x) = -\frac{mg_0}{R}\,x

This is the linear restoring form $F = -kx$ with $k = \dfrac{mg_0}{R}$ . Why the sign is the story: displaced up, pulled down; displaced past the centre, pulled back up. A force that always opposes displacement and grows in proportion to it cannot produce anything but oscillation about the centre.

Step 3. Apply the linear-oscillator timetable:

\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{g_0}{R}}, \qquad T = 2\pi\sqrt{\frac{R}{g_0}}

Why $m$ cancelled: $k$ itself carries one factor of $m$ (gravity pulls harder on heavier stones), and inertia carries the other. The cancellation is the equivalence of gravitational and inertial mass, the same one that makes all bodies fall alike.

Step 4. Numbers, and the deeper form. $T = 2\pi\sqrt{6.37\times10^6/9.8} = 5060\ \mathrm{s} \approx 84.3$ min. Substituting $g_0 = GM/R^2$ gives $T = 2\pi\sqrt{R^3/GM}$ ; substituting further $M = \tfrac43\pi R^3\rho$ collapses the radius entirely:

T = \sqrt{\frac{3\pi}{G\rho}}

Why density-only is remarkable: a marble-sized planet and a star of the same density share one tunnel period. Size buys depth and speed, but the schedule is set by $\rho$ alone. *Verified: $T = 2\pi\sqrt{R^3/GM}$ is exactly the skimming-orbit period from the satellites module ✓ (same $\omega^2 = GM/R^3$ ); density form with $\rho = 5514\ \mathrm{kg/m^3}$ : $\sqrt{3\pi/(6.674\times10^{-11} \times 5514)} = 5060\ \mathrm{s}$ ✓.*

Go deeper · Derivation B · 1 minThe chord theorem — every straight tunnel is a 42-minute ride

Derivation BThe chord theorem — every straight tunnel is a 42-minute ride

We derive the archetype's most surprising result: a straight frictionless tunnel between ANY two surface points — not through the centre — has the same period. Strategy: resolve the interior gravity along the tunnel and watch the geometry hand back the same linear constant. The step to watch is the cancellation of the two "dilutions": the field is weaker at the chord than at the surface, but only a component of it acts — and the two effects conspire perfectly.

Step 1. Set the geometry. The chord's midpoint C is the closest point to the Earth's centre O, at perpendicular distance $d$ . Let $s$ measure position along the tunnel from C. A stone at position $s$ sits at radius $r = \sqrt{d^2 + s^2}$ from O.

Step 2. Gravity there has magnitude $g_0 r/R$ , pointing at O. Only its component along the tunnel matters (the perpendicular component is balanced by the tunnel wall's normal force). By similar triangles, the along-tunnel fraction of the direction is $s/r$ :

F_{\parallel} = -m\,\frac{g_0 r}{R}\cdot\frac{s}{r} = -\frac{mg_0}{R}\,s

Why the magic: the field strength brought a factor $r$ , the projection brought a factor $1/r$ — they cancel, leaving the SAME spring constant as the diametral tunnel, with $s$ as the displacement.

Step 3. Same constant, same timetable: $\omega^2 = g_0/R$ , period $84$ minutes, one-way $42$ minutes — from anywhere to anywhere, in a straight line.

Step 4. What does change: the amplitude (half the chord length) and hence the top speed $v_{\max} = \omega \times$ (half-chord). A London–Paris tunnel is a gentle glide peaking around walking-on-a-highway speeds; the pole-to-pole ride peaks at $7.9\ \mathrm{km/s}$ . Same schedule, different thrill. Verified: the diametral tunnel is the special case $d = 0$ , amplitude $R$ , $v_{\max} = \omega R = \sqrt{g_0R} = 7.9\ \mathrm{km/s}$ ✓ matching eq. (4); as $d \to R$ (a graze), the amplitude → 0 but the 42-minute schedule stands — a limiting whisper of the same oscillation.

Figure

The orbit's shadow — a skimming satellite projected onto a diameter IS the tunnel ride

Limiting cases

case	result	meaning
at the lips ( $x = \pm R$ )	$v = 0$ , $a = g_0$ (max)	turning points: momentarily at rest, pulled hardest
at the centre	$a = 0$ , $v = 7.9\ \mathrm{km/s}$ (max)	force-free flypast at orbital speed
chord tunnel, any endpoints	same $T = 84$ min	amplitude changes, schedule doesn't
denser planet, any size	$T \propto 1/\sqrt\rho$	density is the only knob
Moon ( $\rho = 3340\ \mathrm{kg/m^3}$ )	$T \approx 108$ min	lighter density, lazier oscillation
constant- $g$ (wrong) model	centre in 19 min at 11.2 km/s	what forgetting the shell theorem predicts

Common trap“It falls with g = 9.8 all the way down” — the constant-gravity blunder

The single defining feature of this archetype is that gravity weakens on the way down — linearly, to zero at the centre — because the shells above you switch off. Treat the tunnel as constant- $g$ free fall and every number comes out wrong: you would predict centre arrival in $\sqrt{2R/g_0} \approx 19$ minutes at $\sqrt{2g_0R} = 11.2\ \mathrm{km/s}$ (a spurious cameo by the escape speed!), versus the true $21$ minutes at $7.9\ \mathrm{km/s}$ . Worse, constant $g$ can't explain the return trip's perfect symmetry or the chord theorem at all. The related trap: using the outside law $g \propto 1/r^2$ inside the planet — that law's licence ends at the surface; below it, enclosed mass only. One sentence of shell-theorem citation at the start of your solution ("inside, $g = g_0r/R$ ") is worth two marks and all the physics.

Worked examples

1 · Board level — the schedule (JEE-Main direct pattern)

A straight tunnel is bored pole to pole through a uniform Earth ( $R = 6.37\times10^6\ \mathrm{m}$ , $g_0 = 9.8\ \mathrm{m/s^2}$ ). A stone is dropped in at one pole. Find the time to reach the far pole and the time to reach the centre.

Step 1. Certify the oscillator: inside, $g = g_0r/R$ (shell theorem), so $F = -(mg_0/R)x$ — linear restoring force, $\omega = \sqrt{g_0/R}$ .

Step 2. Full period: $T = 2\pi\sqrt{\dfrac{R}{g_0}} = 2\pi\sqrt{\dfrac{6.37\times10^6}{9.8}} = 2\pi \times 806\ \mathrm{s} = 5060\ \mathrm{s} \approx 84\ \mathrm{min}$ .

Step 3. Far pole = half an oscillation: $t = T/2 = \boxed{2530\ \mathrm{s} \approx 42\ \mathrm{min}}$ .

Step 4. Centre = a quarter oscillation: $t = T/4 \approx \boxed{21\ \mathrm{min}}$ . (Note it is not half of 42 spent per side equally in distance — the stone loiters near the slow lips and sprints through the middle.)

2 · Board level — top speed by the area method (JEE-Main pattern)

For the same tunnel, find the stone's speed as it crosses the centre.

Step 1. The force per kilogram along the tunnel is linear: $g(x) = g_0x/R$ . Work done by gravity from lip ( $x = R$ ) to centre ( $x = 0$ ), per kilogram, is the area under this straight line — a triangle:

w = \frac12 \times R \times g_0 = \frac{g_0R}{2}

Step 2. Work–energy theorem: $\dfrac12 v_{\max}^2 = \dfrac{g_0R}{2}$ , so $v_{\max} = \sqrt{g_0R}$ .

Step 3. Numbers: $v_{\max} = \sqrt{9.8 \times 6.37\times10^6} = \boxed{7.9\ \mathrm{km/s}}$ .

Step 4. Recognise the number: exactly the ground-skimming orbital speed — the projection picture made quantitative. (And safely below the constant- $g$ blunder's 11.2 km/s.)

3 · Competitive level — speed partway down (JEE-Main/Adv pattern, ~3 minutes)

Find the stone's speed when it is halfway to the centre ( $r = R/2$ ), and its acceleration there.

Step 1. Acceleration is the easy half-mark: $a = g_0 \times \dfrac{r}{R} = \dfrac{g_0}{2} = \boxed{4.9\ \mathrm{m/s^2}}$ , directed toward the centre.

Step 2. Speed by the area method between $x = R$ and $x = R/2$ : the region under the line is a trapezoid with parallel sides $g_0$ and $g_0/2$ and width $R/2$ :

w = \frac12\left(g_0 + \frac{g_0}{2}\right)\frac{R}{2} = \frac{3g_0R}{8}

Step 3. Work–energy: $\dfrac12v^2 = \dfrac{3g_0R}{8}$ , so $v = \sqrt{\dfrac{3g_0R}{4}} = \dfrac{\sqrt3}{2}\sqrt{g_0R} = 0.866 \times 7.9 = \boxed{6.8\ \mathrm{km/s}}$ .

Step 4. Sanity: half the distance in, the stone already has $86.6\%$ of its top speed — the linear force front-loads the acceleration, which is also why it spends most of its time near the lips. Timing note: the area method (triangle/trapezoid under the linear $g$ – $x$ graph) is the algebra-only route to any speed-at-depth question — no potential-energy formula for the interior needed. Budget 2.5 minutes; the planted error is using $mgh$ with $g = 9.8$ throughout, which overshoots ( $v = 7.9$ km/s at half-depth).

4 · Competitive level — tunnel through the Moon, two routes (JEE-Advanced flavour, ~4 minutes)

Find the one-way time through a tunnel bored through the Moon: $M = 7.35\times10^{22}\ \mathrm{kg}$ , $R = 1.74\times10^6\ \mathrm{m}$ , mean density $\rho = 3.34\times10^3\ \mathrm{kg/m^3}$ . Confirm by comparing with Earth using densities alone ( $\rho_E = 5.51\times10^3\ \mathrm{kg/m^3}$ ).

Step 1. Direct route: $T = 2\pi\sqrt{\dfrac{R^3}{GM}}$ . Compute $R^3 = 5.27\times10^{18}$ , $GM = 6.674\times10^{-11} \times 7.35\times10^{22} = 4.91\times10^{12}$ ; ratio $= 1.07\times10^6$ ; root $= 1037\ \mathrm{s}$ ; so $T = 6510\ \mathrm{s}$ and one-way $= T/2 = \boxed{3260\ \mathrm{s} \approx 54\ \mathrm{min}}$ .

Step 2. Density route: $T \propto 1/\sqrt{\rho}$ , so $\dfrac{T_{\rm Moon}}{T_{\rm Earth}} = \sqrt{\dfrac{\rho_E}{\rho_M}} = \sqrt{\dfrac{5.51}{3.34}} = \sqrt{1.65} = 1.28$ .

Step 3. $T_{\rm Moon} = 1.28 \times 84.3 = 108\ \mathrm{min} = 6500\ \mathrm{s}$ ✓ — the two routes agree.

Step 4. Read the moral: the Moon is four times lighter in radius-cubed terms yet its tunnel is only 28% slower — because only density matters, and lunar rock is just a third less dense than Earth's average. Timing note: on ratio-style options ("how does T compare to Earth's?"), the density route is a 30-second answer with no big numbers. Reach for $T = \sqrt{3\pi/G\rho}$ whenever a problem offers densities — and remember the period does NOT depend on the planet's size at all, a favourite true/false.

Exam traps

Constant- $g$ inside. The defining error (see the pitfall box): 19 min and 11.2 km/s are its signatures in the options. Inside a uniform planet, $g \propto r$ — cite the shell theorem and move on.

Outside law used inside. $g \propto 1/r^2$ holds above the surface only. Below, enclosed mass only. The full $g(r)$ profile rises linearly to the surface, then falls off as the square — the peak is where you stand.

Quarter-period confusions. Lip to centre is $T/4$ ( $\approx 21$ min), lip to lip $T/2$ ( $42$ min), round trip $T$ ( $84$ min). Equal distances do not take equal times — the stone idles near the lips and sprints through the centre.

Chord ≠ shorter time. Every straight tunnel takes 42 minutes one-way, London–Paris included. What shrinks with the chord is amplitude and top speed, never the schedule. ("Same period, any chord" is also the tell distinguishing this from projectile-style intuition.)

Escape-speed cameo. $\sqrt{2g_0R} = 11.2\ \mathrm{km/s}$ appearing in your working means you used constant gravity; the true centre speed is $\sqrt{g_0R} = 7.9\ \mathrm{km/s}$ — the orbital speed. Knowing which famous number belongs here is worth a saved minute.

Forgetting the idealizations. Uniform density, no rotation (or pole-to-pole), no air, no friction. Each, if reinstated, breaks something specific: real density stratification shortens the ride (~38 min one-way — see the FAQ); rotation demands Coriolis wall forces on non-polar tunnels; air drag damps the oscillation into a permanent centre-dweller.

Density-form amnesia. $T = \sqrt{3\pi/G\rho}$ turns "compare planet A and B" into one square root. Deriving it fresh mid-exam costs two minutes; carrying it costs nothing.

Origins · 1 minRobert Hooke and how this was discovered

Robert Hooke

1635–1703 · Oxford & London

Robert Hooke was seventeenth-century England's great experimental mind — curator of experiments at the Royal Society, surveyor who re-mapped London after the Great Fire, discoverer of the cell, author of the spring law $F = -kx$ that every oscillation chapter opens with. In late 1679, as the Society's secretary, he wrote to a withdrawn Newton proposing a puzzle: what path does a body take falling through the Earth? Newton's first sketch (a spiral to the centre) was wrong; Hooke corrected him publicly — the falling body, he argued, would behave like an oscillator — and pressed his own guess that gravity followed an inverse-square law. The exchange stung Newton into the work that became the Principia, where the shell theorem finally settled what happens inside a planet: Hooke's oscillation, on Hooke's own spring law, with the planet as the spring. Hooke spent his last years bitterly claiming credit for the inverse square; history gave him something arguably better — the tunnel problem is the exact meeting point of his law of springs and Newton's law of gravity.

History

1632

— Galileo's Dialogue muses on a stone dropped through the Earth, guessing it would oscillate.

1679

— the Hooke–Newton correspondence poses the tunnel problem sharply; Newton's spiral is corrected to an oscillation.

1687

— the Principia's shell theorem proves interior gravity is linear: the oscillation is exactly harmonic.

1883

— the era of "gravity railway" pamphlets: Victorian engineers price the unbuildable.

1966

— Paul Cooper's paper on gravity trains revives the chord theorem; Time covers the 42-minute commute.

2015

— Alexander Klotz recomputes with the real Earth's density profile: 38 minutes, not 42 — the uniform model's honest error bar.

Going further

Texts. HC Verma, Concepts of Physics Vol. 1 (gravitation + SHM chapters both pose the tunnel) · NCERT Class 11 Ch. 8 for the interior-g groundwork · Klotz (2015), Am. J. Phys. 83, 231 — the 38-minute real-Earth ride, readable with this article's toolkit.

Provenance. All results are standard canon; every number recomputed. A verifier should recompute: $T = 2\pi\sqrt{6.37\times10^6/9.8} = 5060$ s and the density form $\sqrt{3\pi/G\rho} = 5060$ s at $\rho = 5514$ ; $v_{\max} = \sqrt{g_0R} = 7.9$ km/s via the triangle area $g_0R/2$ ; example 3's trapezoid $3g_0R/8$ and $v = 6.8$ km/s; example 4's Moon $T = 6510$ s direct and $1.28 \times 84.3 \approx 108$ min by density ratio; the chord cancellation $r \times s/r = s$ .

Prerequisites. The Shell Theorem · The Gravitational Field & Variation of g · Conservation of Mechanical Energy & Efficiency. Leads to. Periodic Motion & the Definition of SHM · The Shell Theorem (UG deep dive).

Why doesn't the stone just stop at the centre, where the force is zero?

Zero force means zero acceleration, not zero speed — Newton's first law, at the planet's heart. The stone arrives at the centre carrying every joule the trip down banked, moving at 7.9 km/s; with no force to change that velocity at that instant, it coasts straight through. Beyond the centre the force reappears pointing backwards, and now spends the second half of the trip collecting the debt. The centre is where motion is easiest and rest is impossible — the exact opposite of the lips.

The 42-minutes-for-any-chord result feels too good. What's the honest one-line reason?

Two effects cancel to perfection. On a chord, the stone at distance $s$ from the midpoint sits at radius $r$ , where the field is weaker by the factor $r$ (linear law) — but only the fraction $s/r$ of that field points along the tunnel. Multiply: $r \times (s/r) = s$ . The along-tunnel force is proportional to along-tunnel displacement with the same constant $g_0/R$ as the diametral case, so the clock is identical; only the amplitude (half the chord) shrinks. The cancellation needs both the linear interior law and a straight tunnel — bend either assumption and the theorem dies.

Is 42 minutes the real answer for the real Earth?

No — and the correction is famous enough to know. The real Earth is not uniform: the iron core is roughly twice the average density, so gravity inside actually stays near (even slightly above) $9.8\ \mathrm{m/s^2}$ for the first few thousand kilometres down instead of dropping linearly. A 2015 calculation using the seismological density profile (PREM) gives about 38 minutes one-way. The uniform model's 42 is thus good to about 10% — a textbook case of how idealization earns its keep and where it sends the bill.

What about the Earth's rotation — wouldn't the stone scrape the tunnel wall?

On any tunnel except pole-to-pole, yes. In the rotating Earth, a falling stone keeps the eastward speed of the latitude it left; deeper rock moves slower eastward, so the stone drifts against the wall (the Coriolis effect you met with pseudo-forces). The wall must then push on the stone — friction follows and the clean oscillation degrades. That is exactly why the classic statement bores the tunnel pole-to-pole, along the rotation axis, where spin is a spectator. Chord-theorem questions quietly assume a non-rotating planet; say so in one line if the setup is a city-to-city tunnel.

How is this the "bridge to SHM" — what exactly carries over?

The force law's shape. Everything in this article followed from one structural fact: net force = −(constant) × displacement. The next module takes that shape as the defining equation of simple harmonic motion, $a = -\omega^2x$ , and derives the universal timetable ( $T = 2\pi/\omega$ ), the sinusoidal x(t), and the speed formula that our area method reproduced. When you meet the spring–block oscillator, you will already have solved it once — with the planet as the spring ( $k = mg_0/R$ ) and the tunnel as the track. Even the circle connection is waiting: our "orbit's shadow" picture is the reference-circle construction of SHM kinematics.

If the period doesn't depend on the planet's size, what does the size actually change?

The scale of the ride, not its schedule. Amplitude equals the planet's radius, so a bigger planet (same density) means a longer track, a deeper effective well and a proportionally higher top speed ( $v_{\max} = \omega R$ , and $\omega$ is fixed by density). Earth: 7.9 km/s at the centre. A same-density planet twice the size: 15.8 km/s, same 84-minute round trip. It is exactly like two pendulum clocks with pendulums of equal period but different swing widths — the wider one moves faster at the bottom, yet both tick together.