The Gravitational Field & Variation of g

In 1672 the astronomer Jean Richer sailed from Paris to Cayenne, near the equator, carrying a pendulum clock that kept perfect time at home. In South America it lost two and a half minutes a day. Nothing was wrong with the clock — something was "wrong" with the Earth: gravity at the equator is measurably weaker than in Paris. The classroom constant $9.8\ \mathrm{m/s^2}$ turned out to be not a constant at all but a local reading — and the deviations encode the planet's shape, its spin, and what lies buried beneath your feet. This article builds the two ideas your gravitation chapter runs on. First, the field: the Earth fills the space around it with a condition — $g = GM/r^2$ newtons of pull per kilogram, aimed at the centre — and objects respond to the field where they are. The field is what turns "action at a distance" into local, computable physics, and its unit N/kg is secretly the same as m/s². Second, the variation of g: climb a mountain and $g$ falls off as $1/r^2$ ; descend a mine shaft and it falls off linearly (the shell theorem quietly at work); stand on the spinning equator and rotation skims another third of a percent off your weight. Every board examines this — CBSE derives all three variations, JEE and NEET drill the height-versus-depth comparison, IB and A-Level test the field concept itself. One field formula, three correction stories: this is that article.

Key takeaways

The gravitational field is force per kilogram: $\vec g = \vec F/m$ , in N/kg — identically equal to the free-fall acceleration in m/s². A sphere of mass $M$ makes $g = \dfrac{GM}{r^2}$ toward its centre.
The field belongs to the source and the place, not to the object placed in it: marble and mountain read the same $g$ at the same point.
Surface value: $g_0 = \dfrac{GM_E}{R_E^2} = 9.8\ \mathrm{m/s^2}$ — computable from Cavendish's $G$ , and different on every planet ( $g \propto \rho R$ ).
Altitude: $g(h) = \dfrac{g_0}{(1+h/R)^2} \approx g_0\left(1 - \dfrac{2h}{R}\right)$ for $h \ll R$ — quadratic dilution, factor $2h/R$ .
Depth: $g(d) = g_0\left(1 - \dfrac{d}{R}\right)$ — exactly linear (uniform-Earth model), zero at the centre. Depth $d$ costs half as much as altitude $h = d$ … precisely: equal fractional drops occur at $d = 2h$ .
Rotation: apparent $g$ at latitude $\lambda$ is reduced by $\omega^2 R\cos^2\lambda$ — maximum $0.034\ \mathrm{m/s^2}$ at the equator, zero at the poles.
Key idea: $g$ is a field reading, not a universal constant — it falls as $1/r^2$ above the surface, linearly below it, and is skimmed by the planet's spin.

The core insight

Dividing the gravitational force by the test mass leaves the field $g = GM/r^2$ — a property of source and location measured in N/kg = m/s² — and evaluating how that field changes with altitude ( $-2h/R$ ), depth ( $-d/R$ ) and the planet's rotation ( $-\omega^2R\cos^2\lambda$ ) gives the complete exam story of the variation of g.

Where this lives today

Prospecting companies hunt oil, ore and buried voids with gravimeters sensitive to a few parts per billion of $g$ — a salt dome or a cavity beneath your feet is a dent in the local field, read from the surface without digging. The GRACE-FO satellites run the same idea planet-wide: by tracking micron-level changes in their mutual distance they weigh Greenland's ice loss and India's groundwater draw-down month by month, turning this article's $\Delta g$ into climate accounting. Metrology labs meanwhile need the fine print: the kilogram's modern definition is realized on Kibble balances that must know local $g$ to nine digits — so national labs maintain absolute gravimeters that drop corner-cube mirrors in vacuum and time them with lasers, measuring the variation of g not per planet but per floor of the building. Richer's misbehaving clock has become an instrument industry.

The physical picture

How does the Earth pull the Moon without touching it? Newton computed the pull and called the untouched-touching an absurdity he would not explain. The modern answer splits the transaction in two: the Earth fills the space around it with a field — an arrow at every point, saying "any kilogram placed here gets pulled this hard, this way" — and each object responds to the field at its own location. A local deal replaces spooky reach. The field is defined by an honest division. Bring a small test mass $m$ to the point of interest, measure the gravitational force $\vec F$ on it, and divide the messenger back out:

\vec g = \frac{\vec F}{m} \qquad (1)

where: $\vec g$ — the field, in newtons per kilogram. The force depends on what you brought (20 N on 2 kg, 100 N on 10 kg), but force per kilogram comes out identical — what survives the division belongs to the place, not the visitor. Feed the universal law into the definition and the field of any sphere emerges:

g = \frac{GM}{r^2} \qquad (2)

where: $M$ — the source mass; $r$ — distance from its centre (outside the body). Note what is absent: the test mass. And a lovely unit identity rides along: for an object in free fall, $a = F/m = g$ — the field strength is the free-fall acceleration, so $\mathrm{N/kg} = \mathrm{m/s^2}$ identically. " $9.8$ N/kg" (force per kilogram on things held still) and " $9.8$ m/s²" (acceleration of things let go) are one number in two outfits.

Figure

The field map — every point around the Earth carries a pull-per-kilogram arrow

Evaluate equation (2) at the Earth's surface and the classroom constant assembles itself from Cavendish's ingredients: $g_0 = \dfrac{GM_E}{R_E^2} = \dfrac{6.67\times10^{-11} \times 5.97\times10^{24}}{(6.37\times10^6)^2} = 9.8\ \mathrm{m/s^2}$ . Do the same for any world: $g \propto M/R^2 \propto \rho R$ — Mars, smaller and lighter, posts $3.7$ ; the Moon $1.62$ ; Jupiter $24.8$ . And because the Earth is neither pointlike, nor stationary, nor perfectly rigid history, the surface reading is not one number but a map — which is where the exam's three variation stories begin.

The math: three corrections to one constant

Altitude — the $1/r^2$ story. At height $h$ the centre distance is $R + h$ , so equation (2) gives, exactly,

g(h) = \frac{GM}{(R+h)^2} = \frac{g_0}{\left(1 + \dfrac{h}{R}\right)^2} \qquad (3)

where: $g_0$ — the surface value. For everyday heights ( $h \ll R$ ) the practical form is the linearized one: $g(h) \approx g_0\left(1 - \dfrac{2h}{R}\right)$ — each kilometre of altitude costs about $0.031\%$ of $g$ . The factor 2 is the fingerprint of the square in the inverse square. But the approximation has a licence — $h \ll R$ — and questions about satellites at $h \sim R$ must use the exact form (3); the linearized version at $h = R$ predicts $g < 0$ , which is nonsense advertising its own expiry.

Depth — the shell theorem's story. Descend a mine shaft to depth $d$ and something altitude never did happens: the shell of rock above you stops pulling. The shell theorem (next article proves it; NCERT states it) says a uniform spherical shell exerts zero net gravity on anything inside it — so at radius $r = R - d$ only the inner sphere of radius $r$ still acts. For a uniform-density Earth its mass is $M(r) = M(r/R)^3$ , and

g(d) = \frac{G M (r/R)^3}{r^2} = g_0\,\frac{r}{R} = g_0\left(1 - \frac{d}{R}\right) \qquad (4)

where: $d$ — the depth. Two readings: the fall-off is exactly linear (no approximation — within the uniform model), reaching zero at the centre, where symmetric pulls cancel; and depth is cheaper than altitude — the linear $d/R$ against altitude's $2h/R$ means equal fractional drops occur at $d = 2h$ , a comparison JEE and NEET recycle every second year.

Rotation — the merry-go-round story. A body at rest on the spinning Earth is not in equilibrium: it rides a circle of radius $R\cos\lambda$ (at latitude $\lambda$ ) once per day, which requires a net centripetal force. Part of true gravity is spent providing it; only the remainder presses you into the scale. The apparent field is

g_{\text{app}} = g_0 - \omega^2 R\cos^2\lambda \qquad (5)

where: $\omega = 7.29\times10^{-5}$ rad/s — Earth's spin rate; $\lambda$ — latitude. At the equator ( $\lambda = 0$ ) the correction maxes out at $\omega^2 R = (7.29\times10^{-5})^2 \times 6.37\times10^6 = 0.034\ \mathrm{m/s^2}$ — about $0.35\%$ ; at the poles it vanishes. (The measured pole-equator difference is larger, $\approx 0.052\ \mathrm{m/s^2}$ , because the spinning Earth also bulges: polar dwellers stand closer to the centre. Boards test the rotation term; the bulge is the honest footnote.)

you want	use	licence
$g$ at altitude $h$	$\dfrac{g_0}{(1+h/R)^2}$ ; $\approx g_0(1 - 2h/R)$	approximation only for $h \ll R$
$g$ at depth $d$	$g_0(1 - d/R)$	exact within uniform-Earth model
equal % drop	$d = 2h$	the classic comparison
spin effect at latitude $\lambda$	subtract $\omega^2R\cos^2\lambda$	max $0.034$ at equator, $0$ at poles
another planet	$g = GM/R^2 \propto \rho R$	build from density and size

Derivation Ag at altitude — the exact form and its small-height shortcut

We derive equation (3) and justify the exam-standard approximation $g(h) \approx g_0(1 - 2h/R)$ using algebra alone — no calculus, no binomial theorem taken on faith. The strategy: write the exact ratio $g(h)/g_0$ from the field formula, then show by direct multiplication that $(1 - 2h/R)$ reproduces it up to terms too small to matter when $h \ll R$ . Watch the approximation's expiry date — it is the trap the exact form exists to protect you from.

Step 1. Exact, from $g = GM/r^2$ at $r = R + h$ :

\frac{g(h)}{g_0} = \frac{R^2}{(R+h)^2} = \frac{1}{(1 + x)^2}, \qquad x \equiv \frac{h}{R}

Step 2. The claim: for small $x$ , $\dfrac{1}{(1+x)^2} \approx 1 - 2x$ . Check it by multiplying out:

(1 - 2x)(1 + x)^2 = (1 - 2x)(1 + 2x + x^2) = 1 - 3x^2 - 2x^3

The product misses $1$ only by $3x^2$ (and smaller). For $h = 10$ km, $x = 1.6\times10^{-3}$ : the error term is $\sim 8\times10^{-6}$ — parts per million. The shortcut is honest wherever $x^2$ is negligible.

Step 3. So for $h \ll R$ :

g(h) \approx g_0\left(1 - \frac{2h}{R}\right), \qquad \frac{\Delta g}{g_0} = -\frac{2h}{R}

Why the factor 2: the square in the inverse square. A 1% increase in distance costs 2% in field — the derivative of the exponent, visible without calculus.

Step 4. Expiry audit. At $h = R$ the shortcut predicts $g = -g_0$ — negative gravity, absurd. The exact form gives $g_0/4$ . Rule: heights comparable to $R$ (satellites, "where does g halve?") must use Step 1's exact expression. Verified: Everest, $h = 8.85$ km: $\Delta g/g = 2(8850)/6.37\times10^6 = 2.8\times10^{-3}$ , so $g \approx 9.77$ — matching measured summit values. Exact halving: $(1+h/R)^2 = 2 \Rightarrow h = (\sqrt2 - 1)R = 2640$ km, nowhere near the shortcut's $R/4$ .

Go deeper · Derivation B · 1 ming at depth — why the rock above you stops counting

Derivation Bg at depth — why the rock above you stops counting

We derive the linear law (4) for gravity inside a uniform Earth. The strategy has two moves: the shell theorem removes the outer shell (everything above radius $r$ ), and the $r^3$ scaling of enclosed mass shrinks what remains. Watch the competition that produces the linear answer: the surviving mass falls as $r^3$ while the $1/r^2$ concentration grows — and $r^3/r^2 = r$ wins by exactly one power.

Step 1. Stand at depth $d$ , i.e. radius $r = R - d$ . Split the Earth into (a) the inner sphere of radius $r$ and (b) the shell of thickness $d$ outside you. By the shell theorem, shell (b) contributes zero net field everywhere inside itself — its near side and far side cancel exactly.

Step 2. The inner sphere acts as a point of mass $M(r)$ at the centre. For uniform density:

M(r) = M\,\frac{r^3}{R^3}

Why $r^3$ : mass scales with volume. This is the model's one idealization — the real Earth's dense core makes $g$ actually rise slightly for the first thousand kilometres down; boards examine the uniform model and good papers say so.

Step 3. Field at radius $r$ :

g(r) = \frac{GM(r)}{r^2} = \frac{GM}{R^3}\,r = g_0\,\frac{r}{R}

Step 4. In depth language, $r = R - d$ :

g(d) = g_0\left(1 - \frac{d}{R}\right)

Linear, exact (within the model), and zero at the centre — where you are pulled equally in all directions and net gravity has nothing left to point at. Why depth beats altitude two-to-one: altitude keeps all of $M$ but dilutes it quadratically ( $-2h/R$ ); depth keeps the full $1/r^2$ concentration but discards mass cubically — the net linear loss $-d/R$ is half as steep. Equal drops at $d = 2h$ . *Verified: $d = R/2$ : $g = g_0/2 = 4.9\ \mathrm{m/s^2}$ ✓; centre: $g = 0$ ✓; surface: both formulas meet at $g_0$ , and the inside-linear/outside-inverse-square profile peaks exactly at $r = R$ — the next article's headline graph.*

Go deeper · Derivation C · 1 minRotation — the spin tax on apparent weight

Derivation CRotation — the spin tax on apparent weight

We derive the equatorial reduction $\omega^2 R$ and its latitude generalization, equation (5), by asking what a bathroom scale actually measures on a spinning planet. Strategy: a person at rest on the equator is not force-balanced — they ride a circle of radius $R$ once per sidereal day, so Newton's second law demands a net centripetal force; the scale reading (normal force) is what remains of true gravity after that demand is paid. Watch that this is the lift-in-acceleration logic, worn planet-sized.

Step 1. Equator ledger, taking toward-the-centre as positive. Forces on a person of mass $m$ : true gravity $mg_0$ inward, scale push $N$ outward. The required centripetal acceleration is $\omega^2 R$ :

mg_0 - N = m\,\omega^2 R

Step 2. The scale reading — the apparent weight:

N = m\left(g_0 - \omega^2 R\right) \quad\Longrightarrow\quad g_{\text{app}} = g_0 - \omega^2 R

Step 3. Put in the numbers: $\omega = \dfrac{2\pi}{86164\ \mathrm{s}} = 7.29\times10^{-5}$ rad/s (sidereal day), so

\omega^2 R = (7.29\times10^{-5})^2 \times 6.37\times10^6 = 0.034\ \mathrm{m/s^2}

— about $0.35\%$ of $g_0$ : three grams and a half per kilogram, skimmed by the spin.

Step 4. General latitude $\lambda$ : the daily circle has radius $R\cos\lambda$ , and (resolving the centripetal demand's component along the local vertical — the standard board-level statement) the reduction becomes

g_{\text{app}} = g_0 - \omega^2 R\cos^2\lambda

maximal at the equator, zero at the poles, where the daily circle shrinks to a point. Why a faster Earth would be dramatic: setting $g_{\text{app}} = 0$ at the equator gives $\omega_c = \sqrt{g_0/R} = 1.24\times10^{-3}$ rad/s — a 1.4-hour day. Spin the planet 17 times faster and equatorial objects would float; the condition $\omega_c^2 R = g_0$ is also, not coincidentally, the condition for orbiting at ground level. *Verified: $T_c = 2\pi\sqrt{R/g_0} = 2\pi\sqrt{6.37\times10^6/9.8} = 5{,}060\ \mathrm{s} \approx 84$ min — the famous ground-skimming orbital period ✓. Richer's clock: pendulum period $\propto 1/\sqrt{g}$ ; a $0.35\%$ + bulge deficit slows a seconds pendulum by roughly the two minutes a day he reported from Cayenne ✓.*

Figure

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

Limiting cases

case	result	meaning
$h \ll R$	$\Delta g/g = -2h/R$	the linear shortcut, licenced by $x^2 \approx 0$
$h = R$	$g = g_0/4$ exactly	shortcut illegal here (it predicts $-g_0$ )
$d = R$ (centre)	$g = 0$	symmetric pulls cancel; deepest point of the well, weakest field
$d = 2h$ , both small	equal fractional drops	the height-vs-depth comparison MCQ
poles ( $\lambda = 90^\circ$ )	no rotation effect	daily circle radius zero
$\omega \to \sqrt{g_0/R}$	equatorial weightlessness	the 84-minute critical day
test mass swapped	$\vec F$ changes, $\vec g$ unchanged	the field belongs to source and place

Common trap“Going down a mine, there's more Earth above you pulling up — so g must drop fast”

Two misreadings hide in that sentence, and untangling them is the depth derivation's whole point. First, the rock overhead does not simply "pull up": you are inside a spherical shell of rock, and the shell theorem says its near side (pulling up, close) and far side (pulling down, vast but distant) cancel exactly — the overburden simply switches itself off. Second, what actually reduces $g$ is the shrinking of the sphere still beneath you: mass $\propto r^3$ against concentration $\propto 1/r^2$ nets the gentle linear law $g_0(1 - d/R)$ . Mirror-image trap on the altitude side: the $-2h/R$ shortcut is a small-height licence, not a law — at satellite heights use the exact $1/(1+h/R)^2$ or inherit negative gravity. And the rotation effect reduces apparent weight only: true gravitational pull is untouched; your scale is simply part of a merry-go-round.

Worked examples

1 · Board level — gravity on Mars, from two data (CBSE/IB pattern)

Mars has mass $6.42\times10^{23}\ \mathrm{kg}$ and radius $3.39\times10^6\ \mathrm{m}$ . Find its surface gravitational field strength, and the weight there of an astronaut whose Earth weight is $735\ \mathrm{N}$ . ( $G = 6.67\times10^{-11}$ SI, $g_E = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Field formula at the surface: $g_M = \dfrac{GM}{R^2} = \dfrac{6.67\times10^{-11} \times 6.42\times10^{23}}{(3.39\times10^6)^2}$ .

Step 2. Numerator: $4.28\times10^{13}$ . Denominator: $1.15\times10^{13}\ \mathrm{m^2}$ . So $g_M = \boxed{3.7\ \mathrm{N/kg}}$ — about $0.38\,g_E$ .

Step 3. The astronaut's mass (planet-independent): $m = 735/9.8 = 75$ kg.

Step 4. Mars weight: $W = mg_M = 75 \times 3.7 = \boxed{280\ \mathrm{N}}$ . Mass travelled unchanged; weight is a field reading.

2 · Board level — how much does Everest cost? (ISC/A-Level pattern)

Estimate the fractional decrease in $g$ at the summit of Mount Everest ( $h = 8.85\ \mathrm{km}$ ), and the summit value. ( $R = 6.37\times10^6\ \mathrm{m}$ , $g_0 = 9.80\ \mathrm{m/s^2}$ .)

Step 1. Licence check: $h/R = 1.4\times10^{-3} \ll 1$ — the linear shortcut applies.

Step 2. $\dfrac{\Delta g}{g_0} = \dfrac{2h}{R} = \dfrac{2 \times 8850}{6.37\times10^6} = 2.8\times10^{-3}$ — about $\boxed{0.28\%}$ .

Step 3. $\Delta g = 9.80 \times 2.8\times10^{-3} = 0.027\ \mathrm{m/s^2}$ , so $g_{\text{summit}} \approx \boxed{9.77\ \mathrm{m/s^2}}$ .

Step 4. Read the smallness: a $70$ kg climber is lighter by about $2$ N at the top — the weight of an egg. Altitude within the atmosphere barely dents $g$ ; the $1/r^2$ law needs planetary distances to show its teeth.

3 · Competitive level — height versus depth (JEE-Main/NEET pattern, ~2 minutes)

(a) At what depth is the fractional decrease in $g$ the same as at a height of $40\ \mathrm{km}$ ? (b) At what height does $g$ fall to half its surface value? ( $R = 6400\ \mathrm{km}$ .)

Step 1. (a) Small-fraction regime: height costs $2h/R$ , depth costs $d/R$ . Equate: $d = 2h = \boxed{80\ \mathrm{km}}$ — depth must go twice as far for the same drop.

Step 2. (b) "Half" is not small — the shortcut is banned. Exact: $\dfrac{g_0}{(1+h/R)^2} = \dfrac{g_0}{2}$ .

Step 3. $(1 + h/R)^2 = 2 \Rightarrow 1 + h/R = \sqrt2 \Rightarrow h = (\sqrt2 - 1)R = 0.414 \times 6400 = \boxed{2650\ \mathrm{km}}$ .

Step 4. Audit: the banned shortcut would have said $h = R/4 = 1600$ km — a 40% error, and precisely the distractor option printed under (b). Timing note: part (a) is a 20-second ratio; part (b)'s entire content is recognizing the approximation's expiry. The $d = 2h$ pairing and the $(\sqrt2 - 1)R$ result are both worth carrying into the hall pre-computed.

4 · Competitive level — JEE-Advanced stretch: the day that cancels weight (~3 minutes)

(a) How short would Earth's day have to be for objects at the equator to feel weightless? (b) With the present day, what does a $60\ \mathrm{kg}$ person's scale read at the equator versus at a pole (rotation effect only)? ( $R = 6.37\times10^6\ \mathrm{m}$ , $g_0 = 9.80\ \mathrm{m/s^2}$ .)

Step 1. (a) Weightlessness: all of gravity spent on the circle, $\omega_c^2 R = g_0$ , so $\omega_c = \sqrt{g_0/R} = \sqrt{1.539\times10^{-6}} = 1.24\times10^{-3}$ rad/s.

Step 2. $T_c = \dfrac{2\pi}{\omega_c} = 5060\ \mathrm{s} \approx \boxed{84\ \mathrm{minutes}}$ — spin the planet about 17 times faster and the equator becomes an orbit at ground level.

Step 3. (b) Present spin tax: $\omega^2R = 0.034\ \mathrm{m/s^2}$ . Equator: $N = m(g_0 - \omega^2R) = 60 \times 9.766 = \boxed{586\ \mathrm{N}}$ .

Step 4. Pole: no circle to ride, $N = mg_0 = \boxed{588\ \mathrm{N}}$ . Difference: about $2$ N — real, measurable, and the seed of Richer's lost minutes. (Measured maps show $\sim0.5\%$ total pole-equator difference: rotation plus the equatorial bulge it created.) Timing note: part (a) is the single equation $\omega^2R = g$ — the same condition as a ground-level orbit, and its $84$ -minute answer recurs across this chapter (it is also the tunnel-through-Earth period). Recognize it once, harvest it thrice.

Exam traps

The shortcut's licence. $-2h/R$ requires $h \ll R$ . "Height where g halves/quarters" questions demand the exact $(1+h/R)^{-2}$ ; the linearized answer is the planted distractor (example 3b).

Depth is linear, altitude is quadratic. $d/R$ against $2h/R$ : equal drops at $d = 2h$ . Reversing the factor of two is the most-harvested wrong option in this topic.

Inside is not zero. $g$ vanishes only at the centre. At depth $d < R$ it is $g_0(1 - d/R) > 0$ . ("Field inside a hollow shell is zero" is a different, stronger statement — next article.)

Rotation changes apparent weight, not gravity. True pull $GMm/r^2$ is untouched by spin; the scale reading drops because part of gravity funds the daily circle. At the poles: no effect at all.

Field is not force. $g$ in N/kg is per-kilogram; multiply by the mass to get newtons. Marble and mountain at the same point share the same $g$ , not the same $F$ .

Two lookalike graphs. $g(r)$ rises linearly to a peak at the surface, then falls as $1/r^2$ — maximum at $r = R$ , not at the centre. Sketch it once and the inside/outside MCQs answer themselves.

Which $g$ ? CBSE/ISC 9.8 · JEE/NEET as stated (usually 10) · IB 9.8 · A-Level 9.81. Variation answers are usually ratios, mercifully $g$ -free.

Origins · 1 minMichael Faraday and how this was discovered

Michael Faraday

1791–1867 · London

A bookbinder's apprentice who taught himself science from the volumes he stitched, Michael Faraday rose to the Royal Institution and gave physics its most consequential picture: the field. Working on electricity and magnetism — his laboratory notebooks sketch space filled with "lines of force" — Faraday insisted that the action happens in the space between bodies, not by spooky leaps across it. He lacked the mathematics to formalize it; Maxwell supplied that, and physics never went back. Gravity was re-dressed in Faraday's language soon after: the Earth conditions the space around it with $g = GM/r^2$ , and every object responds locally. When Einstein — who kept Faraday's portrait on his study wall — made the gravitational field itself dynamical in 1915, the bookbinder's picture became the arena of the cosmos.

History

1672

— Jean Richer's pendulum clock loses 2.5 minutes a day in Cayenne: first evidence that g varies over the Earth.

1687

— Newton's Principia explains it: the spinning Earth must bulge, and gravity must weaken with height as 1/r².

1735

— French Academy expeditions to Lapland and Peru measure the bulge, settling the shape of the Earth by geodesy.

1851

— Faraday's "On the physical character of the lines of force" articulates the field as physical reality, not bookkeeping.

1915

— Einstein's general relativity makes the field itself dynamical: g becomes geometry, and Newton its everyday limit.

2002

— the GRACE satellites begin mapping monthly changes in Earth's gravity field — watching groundwater and ice sheets by their pull.

Going further

Texts. NCERT Physics Class 11, Ch. 8 (variation sections) · HC Verma Vol. 1, Ch. 11 · Halliday–Resnick–Walker Ch. 13 · IB/A-Level field-strength chapters for the N/kg framing.

Provenance. All results are standard canon. A verifier should recompute: $g_0 = GM/R^2 = 9.8$ ; the $(1-2x)(1+x)^2 = 1 - 3x^2 - 2x^3$ audit; $g(d) = g_0(1-d/R)$ from shell + $r^3$ ; $\omega^2R = 0.034\ \mathrm{m/s^2}$ ; example 1's $3.7$ N/kg and $280$ N; example 2's $0.28\%$ ; example 3's $80$ km and $2650$ km; example 4's $84$ min and $586$ vs $588$ N.

Prerequisites. Newton's Law of Universal Gravitation · Weight, Normal Force & Apparent Weight. Leads to. The Shell Theorem · Variation-of-g Numericals (archetype) · Satellites: Speed, Period, Energy & Weightlessness.

Is g really different in different cities?

Measurably so. Latitude sets the spin-plus-bulge correction (Singapore ≈ 9.781, London ≈ 9.812, the poles ≈ 9.832 m/s²); altitude subtracts $2h/R$ (Mexico City, at 2.2 km, loses another 0.07%); even local geology shifts the fifth decimal. It's why precision scales are calibrated on site and why "9.8" is an agreed classroom average, not a constant of nature. The constants of nature in the story are $G$ and $M_E$ ; little $g$ is their local quotient.

Down a mine there's rock above me pulling upward. Why doesn't that make g drop faster?

Because you are inside a spherical shell of overburden, and a shell's inside is gravitationally silent: near-side rock pulls up strongly but there's little of it; far-side rock pulls down weakly but there's a hemisphere of it — and the inverse-square law makes the cancellation exact (the next article proves this beautiful bookkeeping). So the overburden neither helps nor hinders; what reduces $g$ is only the shrinking sphere still beneath you. Result: the gentle linear law, not a plunge.

Why is depth "cheaper" than height — the factor of two?

Altitude keeps the whole planet pulling but dilutes it by $1/r^2$ — and near the surface the square costs $2\times$ the fractional distance change. Depth keeps you at close range but removes mass as $r^3$ while the $1/r^2$ focusing partially compensates: $r^3/r^2 = r$ , a single net power. One power against two: depth's $-d/R$ is exactly half as steep as altitude's $-2h/R$ . Hence the exam pairing $d = 2h$ for equal drops.

Does the Earth's spin actually fling things upward at the equator?

Nothing is flung — the effect is bookkeeping, not force. You ride a circle of radius $6370$ km once a day, and circles require centripetal force; gravity pays that $\omega^2R = 0.034\ \mathrm{m/s^2}$ toll before pressing you into the scale, so the scale reads less. Stand at a pole and there's no circle to fund: full reading. The same ledger, run in a lift accelerating downward, is the apparent-weight physics you met in the Newton's-laws chapter — the equator is a very slow lift.

What happens to g at the exact centre of the Earth?

Zero — by symmetry, every direction outward contains equal pulls, and they cancel completely. But be careful with the journey there: $g$ decreases linearly on the way down, not suddenly. And zero field does not mean zero potential — you'd still be at the bottom of Earth's gravity well, needing the full escape budget to leave (the potential article makes this precise). "Weightless at the centre, but profoundly trapped" is the honest summary.

Are N/kg and m/s² really the same thing?

Identically — one quantity, two costumes. Hold an object still and the field manifests as force per kilogram: $9.8$ N on each kilogram, which your arm or a spring balance supplies. Release it and the same field manifests as acceleration: with $F = mg$ the only force, $a = F/m = g$ . Dimensionally, $\mathrm{N/kg} = \mathrm{kg\,m\,s^{-2}/kg} = \mathrm{m/s^2}$ . IB and A-Level mark schemes accept either unit; choose the one matching the question's framing (field ↔ N/kg, kinematics ↔ m/s²).

Why does the uniform-density model get the deep Earth slightly wrong?

Because the real Earth is a layered planet: a dense iron core (≈13,000 kg/m³) under a lighter mantle (≈4,500). Descending through light mantle rock removes little mass while bringing you closer to the massive core — so measured $g$ actually rises a few percent down to the core-mantle boundary before falling to zero at the centre. Exams test the uniform model (linear fall), and stating "for a uniform Earth" in your answer is both honest and mark-safe. The correction is one more example of gravity variation as a probe of what's inside.