ARCHETYPE: Variation-of-g Numericals

In 1672 the French astronomer Jean Richer sailed to Cayenne, near the equator, carrying a pendulum clock that kept perfect time in Paris. In South America it lost about two and a half minutes a day. Nothing was wrong with the clock — something was "wrong" with gravity. At Cayenne, $g$ is smaller than in Paris, for two reasons this archetype will make routine: the equator sits farther from Earth's centre (the planet bulges), and the equator is the fastest-moving part of a spinning planet, so some of gravity's pull is spent keeping you on the merry-go-round rather than pressing you into the floor. The physics behind Richer's clock — how $g$ varies with height, depth and rotation — was built in the field-and-variation-of-g concept module. This archetype is its problem gym. Boards and NEET pose the same handful of numericals every year: percentage change at an altitude, weight at the bottom of a mine, the height where $g$ halves, the spin rate that would make equatorial objects float. Each has a master formula, a licence condition that examiners deliberately test, and a set of planted wrong options. Learn the three knobs — height, depth, spin — and the one discipline that protects your marks: know when the linearized shortcut formulas are legal ( $h \ll R$ , $d \ll R$ ) and when the question secretly demands the exact forms. That single judgement call is where this topic's marks are won and lost.

Key takeaways

Three knobs change the effective $g$ : height $g_h = \dfrac{g_0}{(1 + h/R)^2} \approx g_0\!\left(1 - \dfrac{2h}{R}\right)$ , depth $g_d = g_0\!\left(1 - \dfrac{d}{R}\right)$ (exact for uniform Earth), rotation $g_{\rm app} = g_0 - \omega^2R\cos^2\lambda$ .
The factor-2 asymmetry: a small height $h$ costs $2h/R$ of $g$ ; the same depth costs only $d/R$ — so equal small changes need $d = 2h$ .
The linearized height formula is a licence, valid only for $h \ll R$ ; at satellite heights use the exact square or inherit nonsense.
Rotation edits apparent weight only, maximally at the equator ( $\omega^2R \approx 0.034\ \mathrm{m/s^2}$ ), not at all at the poles.
Anchor results: $g$ halves at $h = 0.414R \approx 2650\ \mathrm{km}$ ; $g \to g/4$ at $h = R$ ; a mine at $d = R/2$ reads $g/2$ .
Key idea: every variation-of-g numerical is one of three master formulas plus a licence check — exact-vs-linearized is the examiner's favourite trap.

The core insight

Variation-of-g numericals reduce to three master results — $g_h = g_0/(1+h/R)^2$ above the surface, $g_d = g_0(1 - d/R)$ below it, and $g_{\rm app} = g_0 - \omega^2R\cos^2\lambda$ on the spinning surface — with the linearized $1 - 2h/R$ shortcut legal only when $h \ll R$ .

Where this lives today

The three knobs are now precision industries. Airborne and marine gravimeters map $\Delta g$ to parts per billion, finding oil domes and ore bodies as local dents in the field; the GRACE-FO satellite pair weighs Greenland's ice loss monthly by feeling the depth-formula's logic from orbit. National metrology labs run the height knob in reverse — an atomic clock or a Kibble balance must know local $g$ so well that labs map its change floor by floor (about 3 parts in $10^7$ per metre of height — the $2h/R$ rule at building scale). And every smartphone accelerometer calibration, every "your weight may vary by location" disclaimer on precision scales, is Richer's Cayenne millimetre industrialized: apparent $g$ really is different in Delhi (28° N) and Singapore (1° N), by almost exactly the $\omega^2R\cos^2\lambda$ this archetype computes.

The setup

One planet, three ways to move the needle on a gravity meter. Go up, and you distance yourself from the whole mass of the Earth — the inverse square dilutes the pull, quickly at first ( $2\%$ per $1\%$ of $R$ climbed, that factor 2 being the fingerprint of the square). Go down, and something cleverer happens: the shell of rock above your head switches itself off (the shell theorem's doing), so only the sphere still beneath you counts, and for a uniform Earth the pull falls linearly to zero at the centre. Stay put and spin: the Earth's rotation converts a slice of true gravity into centripetal supply, so scales — which read the leftover, not the pull — report less, most of all at the fast-moving equator and not at all at the stationary poles.

Figure

The three knobs — height dilutes, depth disables shells, spin taxes the scale

The master formulas, with their licences (the concept module carries the full derivations; here they are working tools):

g_h = \frac{g_0}{\left(1 + \dfrac{h}{R}\right)^2} \;\;\xrightarrow{\;h \ll R\;}\;\; g_0\left(1 - \frac{2h}{R}\right) \qquad (1)

where: $g_0$ — the surface value; $R$ — Earth's radius. The exact form is always legal; the linearized form is a small-height shortcut that goes insane at large $h$ (at $h = R$ it predicts $g < 0$ ). The examiners' line in the sand: percentages and heights up to a few hundred kilometres → shortcut; "where does $g$ halve/quarter?" → exact form, no exceptions.

g_d = g_0\left(1 - \frac{d}{R}\right) \qquad (2)

where: $d$ — depth below the surface. Exact (for a uniform-density Earth) at every depth, straight to zero at the centre — no licence needed. The linearity is the shell theorem's gift: mass below you shrinks as $r^3$ , concentration grows as $1/r^2$ , net one power of $r$ .

g_{\rm app} = g_0 - \omega^2R\cos^2\lambda \qquad (3)

where: $\omega = 7.29\times10^{-5}\ \mathrm{rad/s}$ — Earth's spin; $\lambda$ — latitude. The correction is the centripetal tax: $\omega^2R = 0.034\ \mathrm{m/s^2}$ at the equator (about $0.35\%$ of $g$ ), fading as $\cos^2\lambda$ to zero at the poles. It edits the scale reading, never the true pull.

variant asked	formula + move	trap planted
% change at small $h$	$\Delta g/g_0 = 2h/R$	forgetting the 2
% change at small $d$	$\Delta g/g_0 = d/R$	inserting a spurious 2
same change: $d$ vs $h$	$d = 2h$	offering $d = h$
"where does $g$ halve?"	exact: $(1+h/R)^2 = 2 \Rightarrow h = 0.414R$	shortcut gives $h = R/4$ — wrong
mine/tunnel reading	eq. (2), any depth	using $(1 - 2d/R)$
equator vs pole weigh-in	$\Delta W = m\omega^2R$	adding bulge when not asked
"how fast to float?"	$\omega_c = \sqrt{g_0/R}$ , $T \approx 84$ min	owned by the concept module

Figure

The full g profile — linear ramp inside, inverse-square fall outside, peak at the surface

Derivation AThe d = 2h theorem — and where the factor 2 comes from

We derive the classic comparison result: for small excursions, a depth $d$ and a height $h$ produce the same fractional drop in $g$ precisely when $d = 2h$ . The strategy is to linearize the height formula honestly — exposing the approximation examiners test — and set the two fractional changes equal. Watch the licence: everything here assumes $h, d \ll R$ .

Step 1. Height side. The exact ratio is $g_h/g_0 = (1 + h/R)^{-2}$ . For small $x = h/R$ , the binomial shortcut $(1+x)^{-2} \approx 1 - 2x$ applies (dropping terms of order $x^2$ ):

\frac{\Delta g_h}{g_0} = \frac{2h}{R}

Why the 2: the inverse SQUARE differentiates to a factor 2 — climb 1% of an Earth radius, lose 2% of g. Any inverse-cube law would charge 3. The exponent is visible in the discount rate.

Step 2. Depth side. Equation (2) is already linear, with slope 1:

\frac{\Delta g_d}{g_0} = \frac{d}{R}

Why no 2 here: going down does not distance you from a fixed mass — it removes shells from duty. The two effects (less mass, closer centre) partially offset, leaving the gentle single power.

Step 3. Equate the fractional drops:

\frac{2h}{R} = \frac{d}{R} \quad\Longrightarrow\quad d = 2h

Step 4. State the licence. Step 1's binomial is an $h \ll R$ approximation; push $h$ to a sizeable fraction of $R$ and the exact forms must be compared instead — at large excursions there is no clean $d$ – $h$ correspondence at all (depth runs out at $d = R$ ; height never runs out). Verified: $h = 1\ \mathrm{km}$ : $\Delta g/g_0 = 2/6371 = 0.031\%$ ; $d = 2\ \mathrm{km}$ gives the same $0.031\%$ ✓. Cross-check the licence: at $h = R$ the shortcut claims $-2$ , i.e. $g < 0$ — absurd; the exact form gives $g_0/4$ ✓, exactly the failure mode drilled in worked example 2.

Limiting cases

case	result	meaning
$h \to 0$ or $d \to 0$	$g \to g_0$	all formulas splice at the surface
$h = R$	$g = g_0/4$ (exact)	shortcut illegal here — it predicts $< 0$
$h = 0.414R$	$g = g_0/2$	"where g halves" — exact form territory
$d = R$	$g = 0$	centre of the Earth: all shells surround you
$\lambda = 90°$	no spin correction	poles don't move — scale reads true $g$
$\omega \to \sqrt{g_0/R}$	equatorial $g_{\rm app} \to 0$	the 84-min critical day (concept module's anchor)

Common trap“One formula fits all heights” — the linearized shortcut used outside its licence

The single most-harvested error in this archetype: applying $g_0(1 - 2h/R)$ at satellite altitudes. It is a small- $h$ shortcut, and its failure is not subtle — at $h = R$ it predicts negative gravity, at $h = R/4$ it claims $g/2$ (the true answer to "where does $g$ halve" is $h = 0.414R$ , over 60% higher). The same discipline in reverse: the depth law $g_0(1 - d/R)$ needs no licence — it is exact (uniform Earth) at every depth — and "correcting" it to $1 - 2d/R$ is inventing physics. Rule of thumb before every numerical: percentages at everyday altitudes → shortcut; anything with $h$ comparable to $R$ , or the words "halve/quarter/equal to" → exact square. Choose consciously, and say which you chose.

Worked examples

1 · Board level — up 32 km versus down 32 km (CBSE pattern, 3 marks)

Find the value of $g$ (a) at an altitude of $32\ \mathrm{km}$ , (b) at a depth of $32\ \mathrm{km}$ . Take $g_0 = 9.8\ \mathrm{m/s^2}$ , $R = 6400\ \mathrm{km}$ .

Step 1. Licence check: $h/R = 32/6400 = 0.5\% \ll 1$ — the linear shortcuts are legal for both parts.

Step 2. Altitude, equation (1) linearized: $\Delta g = g_0 \cdot \dfrac{2h}{R} = 9.8 \times 0.01 = 0.098$ , so $g_h = 9.8 - 0.098 = \boxed{9.70\ \mathrm{m/s^2}}$ .

Step 3. Depth, equation (2): $\Delta g = g_0 \cdot \dfrac{d}{R} = 9.8 \times 0.005 = 0.049$ , so $g_d = \boxed{9.75\ \mathrm{m/s^2}}$ .

Step 4. The comparison the examiner wants stated: the same 32 km costs twice as much going up as going down — the $d = 2h$ theorem in the flesh.

2 · Board level — where the shortcut dies (ICSE/NEET pattern)

At what height above the Earth's surface does $g$ fall to (a) $64\%$ of its surface value, (b) half its surface value? ( $R = 6400\ \mathrm{km}$ .)

Step 1. These are large drops — the linear shortcut is out of licence. Use the exact form: $\dfrac{g_h}{g_0} = \dfrac{1}{(1 + h/R)^2}$ .

Step 2. Part (a): $(1 + h/R)^2 = \dfrac{1}{0.64} = 1.5625$ , so $1 + h/R = 1.25$ and $h = 0.25R = \boxed{1600\ \mathrm{km}}$ .

Step 3. Part (b): $(1 + h/R)^2 = 2$ , so $1 + h/R = \sqrt2 = 1.414$ and $h = 0.414R = \boxed{2650\ \mathrm{km}}$ .

Step 4. Audit the trap: the shortcut would have answered (b) with $2h/R = 0.5 \Rightarrow h = 1600\ \mathrm{km}$ — the same number as part (a), and wrong by a thousand kilometres. Options pairing $1600$ and $2650$ are the tell that the examiner is fishing for exactly this.

3 · Competitive level — spin until you're three-quarters as heavy (JEE-Main pattern, ~2.5 minutes)

How long would a day have to be for objects at the equator to weigh three-quarters of their true weight? ( $g_0 = 9.8\ \mathrm{m/s^2}$ , $R = 6.4\times10^6\ \mathrm{m}$ .)

Step 1. Translate the demand into equation (3) at $\lambda = 0$ : $g_{\rm app} = g_0 - \omega^2R = \dfrac34 g_0$ , so $\omega^2R = \dfrac{g_0}{4}$ .

Step 2. Solve for the spin: $\omega = \dfrac12\sqrt{\dfrac{g_0}{R}} = \dfrac12\sqrt{\dfrac{9.8}{6.4\times10^6}} = \dfrac12 \times 1.24\times10^{-3} = 6.2\times10^{-4}\ \mathrm{rad/s}$ .

Step 3. Convert to a day length: $T = \dfrac{2\pi}{\omega} = \dfrac{6.283}{6.2\times10^{-4}} = 1.01\times10^4\ \mathrm{s} \approx \boxed{2.8\ \mathrm{hours}}$ .

Step 4. Sanity anchor: full weightlessness needs $\omega_c = \sqrt{g_0/R}$ , the 84-minute day. Three-quarter weight needs half that spin rate, hence twice that day: $\approx 2.8$ h ✓ — the anchor checks the arithmetic for free. Timing note: recognize the family instantly — "spin changes weight" is always $\omega^2R$ against $g_0$ at the equator. Knowing the 84-minute anchor converts most variants into ratio reasoning: weight fraction $1 - x$ needs $\omega = \sqrt{x}\,\omega_c$ . Budget 90 seconds.

4 · Competitive level — one body, three floors of the planet (NEET/JEE stretch, ~3 minutes)

A body weighs $90\ \mathrm{N}$ on the Earth's surface. Find its weight (a) at a height $h = R/2$ , (b) at a depth $d = R/2$ . (c) State which effect is gentler and why.

Step 1. Part (a): $h = R/2$ is not small — exact form: $\dfrac{g_h}{g_0} = \dfrac{1}{(1 + 0.5)^2} = \dfrac{1}{2.25} = \dfrac49$ . Weight $= 90 \times \dfrac49 = \boxed{40\ \mathrm{N}}$ .

Step 2. Part (b): depth law, exact at any depth: $\dfrac{g_d}{g_0} = 1 - \dfrac12 = \dfrac12$ . Weight $= \boxed{45\ \mathrm{N}}$ .

Step 3. Part (c): halfway down keeps more weight ( $45 > 40$ N): the depth discount runs on one power ( $1 - d/R$ ) while altitude runs on an inverse square. Deep is gentle, high is harsh.

Step 4. Recompute check: $(1.5)^2 = 2.25$ and $90/2.25 = 40$ ✓; the mass never changed — only $g$ did (weight questions are $g$ questions wearing newtons). Timing note: the phrase " $h = R/2$ " is itself the trap flag — anyone auto-writing $1 - 2h/R = 0$ predicts weightlessness at 3200 km, which option (0 N) will be sitting there. Exact form or nothing at large fractions. Total budget: 2 minutes.

Exam traps

The licence trap (the big one). $1 - 2h/R$ at large $h$ : predicts $g = 0$ at $h = R/2$ and $g < 0$ beyond — always wrong, always offered. Exact form whenever $h$ is a stated fraction of $R$ or the drop exceeds ~10%.

A 2 where it doesn't belong. Depth is linear with slope one: $g_0(1 - d/R)$ . Writing $1 - 2d/R$ (symmetry reflex from the height formula) halves the correct depth answers.

$d = 2h$ quoted without its small-print. The theorem is a linearized statement. In exact terms, $g$ at depth $R/2$ ( $= 0.5g_0$ ) does not equal $g$ at height $R/4$ ( $= 0.64g_0$ ) — the correspondence is strictly a small-excursion result.

Rotation edits the scale, not the planet. $g_{\rm app} = g_0 - \omega^2R\cos^2\lambda$ changes weigh-ins and pendulum clocks; the true field $GM/r^2$ is untouched. Assertion–reason items probe exactly this wording.

$\cos^2$ , not $\cos$ . The centripetal radius at latitude $\lambda$ is $R\cos\lambda$ and only the component of that centripetal demand along the local vertical counts — two cosines, hence $\cos^2\lambda$ . NEET options with $\cos\lambda$ appear regularly.

Pole–equator conflation. Two separate reasons the equator reads lighter: the spin tax ( $0.034\ \mathrm{m/s^2}$ , this article) and the equatorial bulge (extra $\sim 0.018$ , geometry — the equator is farther from the centre). Quote only what the question asks; "uniform spherical Earth" means spin only.

Which $R$ , which $g$ . Boards mix $R = 6400$ and $6371\ \mathrm{km}$ , $g = 9.8$ and $10$ . Percentage answers barely care; absolute answers (metres of height for a given drop) shift visibly. Use the paper's data, and don't round $\sqrt2 = 1.414$ to 1.4 if options sit close.

Origins · 1 minJean Richer and how this was discovered

Jean Richer

1630–1696 · Paris & Cayenne

Jean Richer was a working astronomer of the young French Academy of Sciences, sent to Cayenne (5° N) in 1672 to observe Mars and help fix the scale of the solar system. He took along a pendulum clock regulated in Paris — and made the trip's accidental great discovery: at the equator the clock lost about two and a half minutes a day, and a pendulum beating true seconds had to be shortened by more than a millimetre. Gravity itself was weaker there. Newton seized on Richer's numbers in the Principia as field evidence that the Earth spins and bulges: rotation taxes the equatorial scale reading, and the bulge lifts the equator farther from the centre — the two effects this archetype's equation (3) and height formula quantify. Richer's reward was modest (Cassini, his superior, reportedly resented the sensational result), but his millimetre made him the first human to measure the variation of g — with a clock, two continents, and no theory at all.

History

1672

— Richer's Cayenne expedition: the Paris-regulated pendulum clock runs slow at the equator — g varies.

1687

— Newton's Principia explains Richer's clock by rotation plus the equatorial bulge.

1735

— Bouguer and La Condamine haul pendulums up the Andes, measuring g's decrease with altitude and the pull of mountains.

1798

— Cavendish weighs the Earth, turning every measured g into a statement about what lies beneath.

1930

s — gravimeter prospecting begins: ore bodies and salt domes found as local dips and bumps in g.

2002

— the GRACE twin satellites start mapping month-by-month changes in Earth's gravity — watching ice sheets and groundwater by their pull.

Going further

Texts. NCERT Physics Class 11, Ch. 8 (Gravitation) · HC Verma, Concepts of Physics Vol. 1, Ch. 11 · the concept module "The Gravitational Field & Variation of g" for all three derivations in full.

Provenance. All results are standard canon; drill values recomputed here. A verifier should recompute: example 1's $9.70/9.75\ \mathrm{m/s^2}$ pair; example 2's $1600$ vs $2650\ \mathrm{km}$ (from $1/0.64 = 1.5625$ and $\sqrt2 = 1.414$ ); example 3's $\omega = 6.2\times10^{-4}$ rad/s and $T \approx 2.8$ h against the 84-min anchor; example 4's $40$ N vs $45$ N; the equatorial tax $\omega^2R = (7.29\times10^{-5})^2 \times 6.37\times10^6 = 0.034\ \mathrm{m/s^2}$ .

Prerequisites. The Gravitational Field & Variation of g · The Shell Theorem. Leads to. The Tunnel Through the Earth (archetype) · Satellites: Speed, Period, Energy & Weightlessness.

Why does going up reduce g twice as fast as going down?

Because the two moves do physically different things. Going up keeps all of Earth's mass acting on you and stretches the distance — an inverse-square penalty, whose small-change rate is $2h/R$ . Going down doesn't distance you from anything meaningfully; instead the shell theorem retires the rock above your head from duty, and the shrinking active mass ( $\propto r^3$ ) beats the shrinking distance ( $1/r^2$ concentration) by exactly one net power — slope $d/R$ . Same planet, two different mechanisms, factor 2 between them.

Is the depth formula really exact when the height formula needs an approximation?

Within the uniform-density model, yes — $g_d = g_0(1 - d/R)$ holds at every depth, straight to zero at the centre; no small- $d$ licence is involved. The honesty note: the real Earth is not uniform (the core is far denser), and measured $g$ actually rises slightly for the first couple of thousand kilometres down. Exams that say "assume uniform density" — which is all of them at this level — want the linear law. The height formula's exact version $(1+h/R)^{-2}$ is also exact; it's only its convenient linearization that carries a licence.

Does the rotation effect mean gravity is actually weaker at the equator?

The pull is the same (for a spherical Earth); the reading is smaller. Your scale measures the normal force — what's left of gravity after $\omega^2R\cos^2\lambda$ of it has been diverted to keep you circling with the planet. That's why the honest name is apparent weight. Two things do genuinely reduce the pull at the real equator, though: the bulge (you stand ~21 km farther from the centre) — together with spin giving the full ~0.5% pole-to-equator difference geographers quote. Exam questions on a "uniform spherical rotating Earth" want the spin term only.

Why cos²λ and not cosλ in the rotation formula?

Two geometric discounts multiply. First, at latitude $\lambda$ you circle a smaller ring of radius $R\cos\lambda$ , so the centripetal demand is $\omega^2R\cos\lambda$ . Second, that demand points at the ring's centre (into the Earth's axis), not straight down — only its component along the local vertical, another factor $\cos\lambda$ , reduces the scale reading. The perpendicular remainder slightly deflects a plumb line instead. One $\cos\lambda$ for the smaller circle, one for the tilt: $\cos^2\lambda$ .

The "84-minute day" keeps appearing — here, in orbits, in the tunnel problem. What is it really?

It is the period $2\pi\sqrt{R/g_0}$ , and it marks the same physical boundary each time: the spin rate at which required centripetal acceleration at the Earth's surface equals available gravity. Spin the planet that fast and equatorial scales read zero (this article) — which is identical to saying equatorial objects are in orbit at ground level (satellites article), and the same $\omega^2 = g_0/R$ reappears as the oscillation rate of a body dropped through the planet (the tunnel archetype next door). One number, three costumes; recognize it and several exam parts collapse into "that's the 84-minute condition."

Do these formulas work on other planets?

The structures transfer perfectly; only the numbers change. Any (uniform, spherical) planet has $g_h = g_s/(1+h/R_p)^2$ , $g_d = g_s(1 - d/R_p)$ and a spin tax $\omega_p^2R_p\cos^2\lambda$ — with its own surface gravity $g_s = GM_p/R_p^2$ . Exams exploit this for transfer questions: "on a planet with twice Earth's radius and the same density, at what height does g halve?" (answer: $0.414 \times 2R = 0.83R_E$ -worth of kilometres — the fraction of the planet's radius is universal). Learn the fractions, not the kilometres.