Newton's Law of Universal Gravitation

Why does gravity fade with distance in exactly this way — not as $1/r$ , not as $1/r^3$ , but as $1/r^2$ on the nose? Newton's answer was audacious: compare an apple falling in an orchard with the Moon falling around the Earth. The Moon sits about sixty Earth-radii away, and $60^2 = 3600$ — so if gravity dilutes as the square of distance, the Moon should fall toward Earth 3600 times more gently than the apple. Newton worked out the Moon's actual acceleration from its orbit, and the numbers agreed. One law, from orchard to sky. That law needed two more ingredients before it could run the solar system. First, a number: the proportionality constant $G$ , so absurdly small that it took Henry Cavendish a torsion balance, sealed room, and a telescope through the wall to catch gravity acting between lead spheres on a bench — and in measuring it he weighed the Earth. Second, a language: the field, which dissolves the mystery of force reaching across empty space by making gravity local — masses shape a condition in the space around them, and other masses respond to that condition right where they sit. You arrive with Newton's laws and circular motion in hand, which is everything this topic needs. In return you get the master force law of classical mechanics: the equation that runs every orbit in the Kepler chapters ahead, the ratio trick that collapses most gravitation problems to one line, and the field concept that the rest of physics — electromagnetism, relativity, quantum field theory — is written in.

Key takeaways

Every mass attracts every other mass: $F = G m_1 m_2 / r^2$ , directed along the line joining their centres — proportional to each mass, diluted by the square of the centre-to-centre separation.
The exponent 2 is geometry, not decree: influence spreading from a point is shared over spheres of area $4\pi r^2$ , so strength per patch falls as $1/r^2$ .
$G = 6.674 \times 10^{-11}\ \mathrm{N\,m^2/kg^2}$ is universal and tiny — everyday objects barely pull; gravity rules only when one partner has $10^{24}$ kg. Cavendish measured it in 1798 and thereby weighed the Earth: $M_E = gR_E^2/G$ .
Most problems are ratio problems: $F_2/F_1 = (r_1/r_2)^2$ — no calculator, no $G$ .
The field repackages the law locally: a source $M$ fills space with $\vec g = (GM/r^2)$ toward itself; a mass $m$ placed anywhere feels $\vec F = m\vec g$ . Field strength in $\mathrm{N/kg}$ is free-fall acceleration in $\mathrm{m/s^2}$ , and fields from several sources add as vectors.
Key idea: one inverse-square attraction, calibrated by $G$ and spoken in the language of fields, governs apples, moons, planets, and galaxies alike.

The core insight

The gravitational force between two masses is $F = G m_1 m_2/r^2$ along the line joining their centres — proportional to the product of the masses, diluted by the square of the separation because influence from a point spreads over spheres of area $4\pi r^2$ — and it is most powerfully read as a field: each mass fills space with $\vec g = GM/r^2$ toward itself, and every other mass simply responds to the field at its own location.

Where this lives today

Every satellite in the sky is flying on equation (1). GPS constellations, weather satellites, and the interplanetary probes navigated by JPL all have their orbits computed from Newtonian inverse-square gravity, with general relativity added only as a small correction where centimetre precision demands it. Astronomy long ago made peace with $G$ 's stubbornness by sidestepping it: orbits pin down the product $GM$ to ten digits — spacecraft navigation uses $GM$ directly — while $G$ alone remains the least precisely known fundamental constant, still measured Cavendish-style with torsion fibres and, lately, cold-atom interferometers, with the world's best laboratories disagreeing beyond their error bars. Field-mapping, meanwhile, has become climate accounting: the GRACE-FO satellite pair measure their mutual separation to a fraction of a hair's width, and every local surplus of $\vec g$ they fly over — a mountain range, a full aquifer, a melting ice sheet — appears in the data, weighing Greenland month by month. The frontier, remarkably, is testing the exponent itself. Extra-dimension scenarios predict that at tiny separations gravity should steepen away from $1/r^2$ — more directions to dilute into. Torsion-balance experiments descended directly from Cavendish's apparatus (the Eöt-Wash group) confirm pure inverse-square behaviour below a tenth of a millimetre, and lunar laser ranging confirms it at $60\,R_E$ . Three and a half centuries on, the square is still exactly 2 everywhere we can measure.

The physical picture

Point a can of spray paint at a wall from arm's length and it paints a small, dense circle. Step back to double the distance and the same cone of paint covers a circle twice as wide — four times the area — so the coat is four times thinner. Nothing about the paint changed; only geometry did. Gravity plays the same game in all directions at once. Picture a mass $M$ alone in space, its influence radiating outward evenly like light from a bare bulb. At distance $r$ that influence is spread over an imaginary sphere of area $4\pi r^2$ . At $2r$ the sphere is four times bigger, so each patch of it receives a quarter of the strength; at $3r$ , a ninth. The "2" in inverse-square is really the "2" in the area of a sphere — which is really the "3" of three-dimensional space, minus one. And notice what the picture quietly promises: the sphere never stops growing, so gravity never switches off. It only ever thins. The Sun holds planets billions of kilometres out; galaxies hold each other across millions of light-years — all with the same thinning-paint arithmetic.

Figure

Why the square — one pull, spread over a growing sphere

Now the second half of the picture: how does the pull get there? Earth tugs the Moon across 384,000 km of nothing — no rope, no contact. Newton himself called unmediated action at a distance "so great an absurdity" that no competent thinker could accept it, and openly declined to explain the mechanism. Physics' eventual answer is the field: stop saying "Earth pulls the Moon from afar" and say instead that Earth fills the space around itself with a condition — a readiness to push on mass — at every point, near or far, occupied or empty. Draw the map: every point in space carries an arrow aimed at Earth's centre, long nearby, shrinking as $1/r^2$ . The Moon doesn't feel the distant Earth; it feels the arrow right where it is, the way a swimmer feels the local current, not the distant storm that made it.

Figure

The field map — every point in space carries a pull arrow

The zoom panel on the right of the map explains something you have been using all along: the "constant gravity $mg$ , straight down" of every projectile problem is nothing but the local patch of this radial map, viewed so close to the surface that the arrows look parallel and equal. The near-surface article was living inside this one the whole time.

The law, the constant, the field

Take any two point masses $m_1$ and $m_2$ — apple and Earth, Earth and Moon, two lead spheres in a basement — separated by a centre-to-centre distance $r$ . Each attracts the other along the line joining them:

\vec F_{12} = -\,\frac{G\, m_1 m_2}{r^2}\,\hat r_{12} \qquad (1)

where: $\vec F_{12}$ — the force on mass 1 from mass 2; $\hat r_{12}$ — the unit vector from mass 2 toward mass 1, so the minus sign makes the force attractive; $r$ — centre-to-centre separation in m; $G = 6.674 \times 10^{-11}\ \mathrm{N\,m^2/kg^2}$ — the universal gravitational constant. Read the anatomy slowly: the masses sit multiplied together upstairs, so doubling either doubles the force; the distance sits squared downstairs, so doubling it quarters the force; and the law is symmetric in $m_1$ and $m_2$ , so each body pulls the other equally hard — Newton's third law is built in. The apple pulls the Earth with the same 2-newton force the Earth exerts on it; the Earth just responds with $10^{25}$ times less acceleration. When several masses act at once, the net force is the vector sum of the pairwise pulls — gravity superposes, each pair oblivious to the rest. For everyday work the vector dress is overkill, and the magnitude form carries the same physics: $F = Gm_1m_2/r^2$ . Better still, most problems never need $G$ at all. Because the constant and the masses are common to any two configurations of the same pair, dividing kills them:

\frac{F_2}{F_1} = \left(\frac{r_1}{r_2}\right)^2 \qquad (2)

where: the subscripts label two separations of the same two bodies. This is the workhorse. Triple the distance: force drops to $1/9$ . Bring two objects to one-fifth the separation: 25 times stronger. No units, no constants — a ratio squared, done in your head. And what of $G$ itself? It is the exchange rate that converts kilograms-squared-per-metre-squared into newtons, and nature set it spectacularly low: two 1-kg masses 1 m apart attract with $6.674 \times 10^{-11}$ N — the weight of a pollen grain. That is why you do not feel your desk pulling on you, and why gravity nevertheless runs the cosmos: it is the only long-range force that never cancels, so once a partner has $10^{24}$ kg it wins everything. The smallness is also why $G$ resisted measurement for 111 years after the Principia — astronomy is no help, because orbits only ever reveal the product $GM$ , never $G$ alone, and nobody can put a planet on a scale. The constant had to be caught between known masses on a bench, which is exactly what Cavendish's torsion balance did in 1798.

Figure

Cavendish's torsion balance, seen from above — a force microscope for gravity

Derivation AThe Moon test — the check that made it a law

We are verifying the inverse-square exponent the way Newton did: by comparing two falling objects, an apple at the Earth's surface and the Moon in its orbit, sixty Earth-radii away. The strategy is a two-column calculation — column one predicts the Moon's falling acceleration from $g$ diluted by $1/r^2$ ; column two measures it independently, from the Moon's orbital geometry and the calendar, using the centripetal acceleration from the circular-motion article. If the two columns agree, the same law spans orchard and sky. The step to watch is Step 3: nothing in it mentions gravity at all — it is pure kinematics, which is what makes the agreement evidence rather than circularity.

Step 1. Fix the geometry. The Moon's orbital radius is $r = 3.84 \times 10^8\ \mathrm{m}$ , which is $60$ Earth radii ( $R_E = 6.37 \times 10^6\ \mathrm{m}$ ). Why: these distances were known from astronomy — lunar parallax — long before Newton.

Step 2. Predict from the inverse-square hypothesis. If gravity thins as $1/r^2$ , the Moon's falling acceleration should be the apple's, cut down by $60^2$ :

a_{\text{Moon}} = \frac{g}{60^2} = \frac{9.8}{3600} = 2.72 \times 10^{-3}\ \mathrm{m/s^2} \qquad (3)

Why: same law, evaluated at sixty times the distance from Earth's centre.

Step 3. Measure it independently. A body circling at radius $r$ with period $T$ has centripetal acceleration $a = 4\pi^2 r/T^2$ . With $T = 27.3\ \mathrm{days} = 2.36 \times 10^6\ \mathrm{s}$ :

a = \frac{4\pi^2 \,(3.84 \times 10^8)}{(2.36 \times 10^6)^2} = 2.72 \times 10^{-3}\ \mathrm{m/s^2}

Why: the Moon's orbit is a fall that keeps missing — its inward acceleration comes straight from geometry and the calendar, no gravity theory required.

Step 4. Compare. Prediction and measurement agree to three figures: the same gravity that drops apples steers the Moon, diluted by exactly the square of the distance. Why this mattered: before this calculation, "the heavens" and "the Earth" were assumed to run on different rules. One number ended that. *Verified: $9.8/3600 = 2.72 \times 10^{-3}$ ; $4\pi^2 \times 3.84 \times 10^8 = 1.516 \times 10^{10}$ , divided by $(2.36 \times 10^6)^2 = 5.57 \times 10^{12}$ , gives $2.72 \times 10^{-3}\ \mathrm{m/s^2}$ . Matches Halliday–Resnick–Walker Ch. 13 and Kleppner and Kolenkow Ch. 9.*

Go deeper · Derivation B · 1 minWeighing the Earth with G

Derivation BWeighing the Earth with G

The goal is the equation that made Cavendish's bench experiment a planetary measurement: $M_E = gR_E^2/G$ , the mass of the Earth from three measurable numbers. The strategy is to write the weight of an ordinary object two ways — the everyday way, $mg$ , and the universal way, equation (1) with the Earth as one of the masses — and demand they agree, since they describe the same force. Watch the test mass cancel in Step 2: that cancellation is why all objects fall alike, and it is what leaves a formula containing only the planet.

Step 1. Equate the two descriptions of a surface object's weight: $mg = \dfrac{G M_E m}{R_E^2}$ . Why: the centre-to-centre distance for an object on the surface is Earth's radius, and the shell theorem (next article) licenses treating the whole planet as a point at its centre.

Step 2. Cancel $m$ : $g = \dfrac{G M_E}{R_E^2}$ . Why: the object's own mass never mattered — feather and anvil read the same $g$ , the equivalence Galileo saw and this law explains.

Step 3. Solve for the planet:

M_E = \frac{g\,R_E^2}{G} = \frac{9.81 \times (6.371 \times 10^6)^2}{6.674 \times 10^{-11}} = 5.97 \times 10^{24}\ \mathrm{kg} \qquad (4)

Why: $g$ comes from a pendulum, $R_E$ from geodesy (known since Eratosthenes), and $G$ from Cavendish's fibre — the last missing ingredient.

Step 4. Extract the bonus. The mean density is $\rho = M_E/\left(\tfrac43 \pi R_E^3\right) = 5.97 \times 10^{24}/1.08 \times 10^{21} = 5500\ \mathrm{kg/m^3}$ — 5.5 times water. Surface rocks average only about $2700\ \mathrm{kg/m^3}$ , so the deep interior must be far denser. Why this mattered: this was the first physical evidence for Earth's iron core; Cavendish himself reported 5.448 in 1798, within one percent of today's value. *Verified: $(6.371 \times 10^6)^2 = 4.059 \times 10^{13}$ ; $9.81 \times 4.059 \times 10^{13} = 3.98 \times 10^{14}$ ; divided by $6.674 \times 10^{-11}$ gives $5.97 \times 10^{24}\ \mathrm{kg}$ ; sphere volume $1.083 \times 10^{21}\ \mathrm{m^3}$ gives $\rho \approx 5510\ \mathrm{kg/m^3}$ .*

Go deeper · Derivation C · 2 minFrom force to field

Derivation CFrom force to field

We are recasting the force law in the language the rest of physics speaks: the field. The goal is the pair of equations $\vec g = \vec F/m$ (definition) and $g = GM/r^2$ (what a spherical source creates), plus the return trip $\vec F = m\vec g$ . The strategy is one honest division: probe a point in space with a test mass, measure the force, and divide the probe's mass back out — what survives belongs to the place, not the visitor. The step to watch is Step 3, where the test mass vanishes from the formula: that disappearance is the entire conceptual content of the field idea.

Step 1. Define the field operationally. Place a small test mass $m$ at the point of interest, measure the gravitational force $\vec F$ on it, and set

\vec g \equiv \frac{\vec F}{m} \qquad (5)

Why: force depends on what you brought — 20 N on 2 kg, 100 N on 10 kg — but force per kilogram comes out the same. Dividing out the messenger leaves the message.

Step 2. Compute the field of a spherical source. Feed the force law (1) into the definition: a test mass at distance $r$ from mass $M$ feels $F = GMm/r^2$ , so $g = F/m$ gives

g = \frac{GM}{r^2}, \quad \text{directed toward } M \qquad (6)

Why: dividing by the positive scalar $m$ leaves the direction untouched — every arrow on the map aims at the source.

Step 3. Read what survived: $G$ , $M$ , and $r$ — source and location only. A marble and a mountain report the same $\vec g$ at the same point; the field is a fact about space around the source, existing whether or not anything is there to feel it. Why this is not just bookkeeping: locality. Each mass responds to the field where it sits, and the "absurdity" of action at a distance is retired — in Einstein's hands the field itself becomes dynamical, rippling at light speed.

Step 4. Run the definition in reverse and check units. An object of mass $m$ placed where the field is $\vec g$ feels $\vec F = m\vec g$ ; if gravity is the only force, Newton's second law gives $\vec a = \vec F/m = \vec g$ — the free-fall acceleration equals the field vector, and $\mathrm{N/kg} = \mathrm{m/s^2}$ identically. At Earth's surface (6) evaluates to $9.8\ \mathrm{N/kg}$ : the classroom constant, unmasked as a field strength. Why: "9.8 N/kg" and "9.8 m/s²" are one quantity in two outfits — forces on things held still, acceleration of things let go.

Step 5. Add sources as vectors. Several masses each contribute their own $GM_i/r_i^2$ arrow, computed as if the others did not exist, and the total field is the vector sum — which is how one finds the point between Earth and Moon where the two pulls cancel (worked example 2). Why: superposition is inherited directly from the force law's pairwise structure. *Verified against Halliday–Resnick–Walker Ch. 13 and Young and Freedman Ch. 13 — the standard field definition; the near-surface uniform field $g = 9.81\ \mathrm{N/kg}$ recovers the constant used throughout the kinematics chapters.*

Which equation answers which question

you want	use	notes
force between two known masses	$F = Gm_1m_2/r^2$ , eq. (1)	centre to centre; spheres count as points
compare two configurations	$F_2/F_1 = (r_1/r_2)^2$ , eq. (2)	no $G$ , no masses — most exam problems
a planet's mass	$M = gR^2/G$ , eq. (4)	pendulum + geodesy + Cavendish
the field a source makes	$g = GM/r^2$ , eq. (6)	belongs to source and place, not the probe
force on a mass in a known field	$\vec F = m\vec g$	drop anything into the map and multiply
several sources	vector-sum the individual fields	each computed as if alone

Limiting cases

Case	Result	Meaning
$r \to 2r$	$F \to F/4$	doubling distance quarters the pull — the signature move
both masses doubled, $r$ doubled	$F$ unchanged	$2 \times 2$ upstairs cancels $2^2$ downstairs
$r \to \infty$	$F \to 0$ , never exactly	infinite range: gravity thins, never quits
heights $h \ll R_E$	$\vec g$ uniform, $9.8\ \mathrm{N/kg}$ down	the "constant gravity" of projectile problems is the local zoom of the radial map
test mass doubled	$\vec F$ doubles, $\vec g$ unchanged	the field belongs to the source and the place
sphere treated as a point	exact, for uniform spheres	the shell theorem — the next article proves it
$m_1 = m_2 = 1\ \mathrm{kg}$ , $r = 1\ \mathrm{m}$	$F = G$ numerically	$G$ is the pull between unit masses at unit distance

Common trap"Twice as far from the ground means a quarter of the gravity"

The single most common inverse-square error is measuring $r$ from the wrong place. The $r$ in $F = Gm_1m_2/r^2$ runs centre to centre — for anything near Earth it starts $6371\ \mathrm{km}$ down, at the core. Climb a 6 km mountain and $r$ changes by less than $0.1\%$ , so your weight barely budges; a point one Earth-radius above the surface is $2R_E$ from the centre ( $F/4$ ), and "twice as high as that" — $2R_E$ of altitude — is $3R_E$ from the centre, giving $F/9$ , not $F/4$ . The licence to treat a whole planet as a point at its centre is Newton's shell theorem, an exclusive perk of the $1/r^2$ law. A second letter-trap rides along: big $G$ ( $6.674 \times 10^{-11}\ \mathrm{N\,m^2/kg^2}$ , universal, identical in Andromeda) is not little $g$ ( $9.81\ \mathrm{m/s^2}$ , local, different on every floor of every planet); they are linked by $g = GM/R^2$ and interchangeable nowhere.

Worked examples

1 · A student's weight, from scratch (confidence builder)

Earth has mass $M_E = 5.97 \times 10^{24}\ \mathrm{kg}$ and radius $R_E = 6.37 \times 10^6\ \mathrm{m}$ . (a) Find the gravitational force on a 60 kg student standing on the surface. (b) The student rides to an altitude of $2R_E$ . What is the force there?

Step 1. Identify $r$ for part (a): standing on the surface means centre-to-centre distance $r = R_E = 6.37 \times 10^6\ \mathrm{m}$ — not zero, and not the student's height.

Step 2. Apply equation (1). Numerator: $GM_Em = (6.674 \times 10^{-11})(5.97 \times 10^{24})(60) = 2.39 \times 10^{16}$ . Denominator: $r^2 = (6.37 \times 10^6)^2 = 4.06 \times 10^{13}\ \mathrm{m^2}$ .

Step 3. Divide: $F = 2.39 \times 10^{16}/4.06 \times 10^{13} = \boxed{589\ \mathrm{N}}$ . Cross-check: $mg = 60 \times 9.8 = 588\ \mathrm{N}$ ✓ — the familiar weight formula is this law in disguise.

Step 4. Part (b) is a ratio, not a recalculation. Altitude $2R_E$ means $r = 3R_E$ from the centre. By equation (2): $F' = 589 \times (R_E/3R_E)^2 = 589/9 = \boxed{65.4\ \mathrm{N}}$ . The trap dodged here fills exam graveyards: "twice as high" meant three times the $r$ .

2 · The Earth–Moon system, twice (exam level)

(a) Find the gravitational force between Earth ( $M_E = 5.97 \times 10^{24}\ \mathrm{kg}$ ) and the Moon ( $M_M = 7.35 \times 10^{22}\ \mathrm{kg}$ ), centre-to-centre $D = 3.84 \times 10^8\ \mathrm{m}$ apart. (b) On the line joining them, find the point where the total gravitational field vanishes.

Step 1. Force, from equation (1). Numerator: $6.674 \times 5.97 \times 7.35 = 292.9$ with powers $10^{-11+24+22} = 10^{35}$ , so $2.93 \times 10^{37}$ . Denominator: $(3.84 \times 10^8)^2 = 1.47 \times 10^{17}\ \mathrm{m^2}$ .

Step 2. Divide: $F = 2.93 \times 10^{37}/1.47 \times 10^{17} = \boxed{1.99 \times 10^{20}\ \mathrm{N}}$ — two hundred billion billion newtons from a constant with $10^{-11}$ in it: the masses upstairs are simply that enormous. This single force swings the Moon's monthly orbit and drags Earth's oceans into tides.

Step 3. Part (b): between the bodies the two field arrows oppose — Earth's points back toward Earth, the Moon's points ahead. Let the null sit a distance $d$ from Earth's centre, $D - d$ from the Moon's, and set the field strengths (6) equal: $\dfrac{GM_E}{d^2} = \dfrac{GM_M}{(D-d)^2}$ . The $G$ 's cancel immediately.

Step 4. Square-root both sides — cleaner than cross-multiplying: $\dfrac{d}{D-d} = \sqrt{\dfrac{M_E}{M_M}} = \sqrt{81.2} = 9.01$ .

Step 5. Solve the linear equation: $d = 9.01(D-d)$ , so $10.01\,d = 9.01\,D$ , giving $d = 0.900\,D = \boxed{3.46 \times 10^8\ \mathrm{m}}$ — 90% of the way to the Moon. The balance point sits far closer to the lighter body, as it must: the Moon needs a short lever for its $1/r^2$ field to match mighty Earth's.

Step 6. Sanity check by substitution: $g_E = GM_E/d^2 = 3.98 \times 10^{14}/(3.46 \times 10^8)^2 = 3.3 \times 10^{-3}\ \mathrm{N/kg}$ , and $g_M = GM_M/(3.8 \times 10^7)^2 = 4.91 \times 10^{12}/1.47 \times 10^{15} = 3.3 \times 10^{-3}\ \mathrm{N/kg}$ ✓ — equal and opposite. A spacecraft coasting through this point feels Earth's pull hand it over to the Moon's.

Origins · 1 minIsaac Newton and how this was discovered

Isaac Newton

1643–1727 · Woolsthorpe & Cambridge

Newton began the gravity calculation in 1666, when plague closed Cambridge and sent him home to Woolsthorpe — the orchard year. Comparing the fall of an apple with the fall of the Moon, he found the inverse-square dilution "answer pretty nearly," then set the work aside for nearly two decades — partly because he could not yet prove that a whole sphere pulls as if it were a point (the shell theorem, settled in the next article). It took Edmond Halley's visit in 1684, asking what orbit an inverse-square force would produce, to draw the theory out of him: "an ellipse — I have calculated it." The answer became the Principia (1687): one law from apples to planets, Kepler's rules turned into theorems, and the ancient wall between terrestrial and celestial physics demolished. What he refused to supply was the mechanism — "hypotheses non fingo" — leaving the action-at-a-distance puzzle open until Faraday's field idea and, finally, Einstein's dynamical spacetime closed it.

History

1609

— Kepler's Astronomia Nova: planets ride ellipses, sweeping equal areas — precise patterns begging for a cause.

1666

— Plague year at Woolsthorpe: Newton compares the falling apple with the falling Moon and finds a

1/r^2

dilution fits.

1684

— Halley asks Newton what path an inverse-square force gives; the tract De Motu follows, then the Principia (1687) with the Moon test and the shell theorem.

1798

— Cavendish's torsion balance catches gravity between lead spheres, weighs the Earth (density 5.448 times water), and fixes what we now call

G

1846

— Le Verrier predicts Neptune from Uranus's orbital misbehaviour using pure inverse-square arithmetic — a planet discovered at the tip of a pen.

1915

— Einstein's general relativity recasts gravity as the geometry of spacetime; Newton's law survives as its everyday limit.

2015

— LIGO detects gravitational waves: ripples in the gravitational field itself, a century after they were predicted.

Going further

Texts. Halliday, Resnick & Walker, Fundamentals of Physics, Ch. 13 · Young & Freedman, University Physics, Ch. 13 · Kleppner & Kolenkow, An Introduction to Mechanics, Ch. 9 · Feynman, Lectures Vol. I Ch. 7 for the geometric soul of the law, and Vol. II Ch. 1 for what a field is · Cavendish, "Experiments to Determine the Density of the Earth," Phil. Trans. R. Soc. (1798), famously readable.

Provenance. Folded from the v3 pilot trio (inverse-square law, gravitational constant, gravitational field concept); all results are standard textbook physics. A verifier should recompute: the Moon test ( $9.8/3600 = 2.72 \times 10^{-3}\ \mathrm{m/s^2}$ against $4\pi^2 r/T^2$ with $r = 3.84 \times 10^8$ m, $T = 2.36 \times 10^6$ s); Earth's mass $M_E = gR_E^2/G = 5.97 \times 10^{24}$ kg and density $\approx 5500\ \mathrm{kg/m^3}$ ; worked example 1 ( $589$ N, then $65.4$ N at $3R_E$ ); worked example 2 ( $F = 1.99 \times 10^{20}$ N; null point $d/(D-d) = \sqrt{81.2} = 9.01$ , $d = 0.90\,D = 3.46 \times 10^8$ m, both fields $\approx 3.3 \times 10^{-3}$ N/kg there).

Builds on: Gravitational Force Near Earth's Surface · Newton's Second Law · Newton's Third Law · Circular Motion and Centripetal Acceleration. Leads to: Gravitational Potential Energy and Escape Velocity · The Shell Theorem · Central Force Fundamentals · Kepler's Third Law.

Does the apple really pull the Earth upward? Why doesn't the Earth visibly fall toward it?

It really does — equation (1) is symmetric in the two masses, so the forces are an exact third-law pair: the Earth pulls the apple with about 2 N, and the apple pulls the Earth with the same 2 N. What differs is the response. Acceleration is force over mass, and the Earth carries $5.97 \times 10^{24}$ kg, so its acceleration toward the apple is around $3 \times 10^{-25}\ \mathrm{m/s^2}$ — in the time the apple falls a metre, the Earth rises by less than the width of an atomic nucleus. The symmetry is not academic, though: for comparable masses it is everything. The Moon does not orbit a fixed Earth; both orbit their common centre of mass, and detecting exactly this kind of mutual wobble in distant stars is how many exoplanets are found.

If gravity between everyday objects is real, why don't two people standing close together feel it?

Compute it: two 70 kg people half a metre apart attract with roughly $F = 6.674 \times 10^{-11} \times 70 \times 70/0.25 \approx 1.3 \times 10^{-6}\ \mathrm{N}$ — about the weight of a tenth of a milligram. Friction between your shoes and the floor exceeds that by hundreds of millions of times, so the pull is utterly swamped. It is not zero, and this matters: Cavendish's whole achievement was building an instrument — the torsion balance — sensitive enough to see exactly such forces. The wire he hung the rod from could be twisted by a force of $10^{-7}$ N through a visible angle, which is how a pull too feeble to feel became a measured number.

What is the real difference between G and g? They seem to get used interchangeably.

They must never be interchanged. $G = 6.674 \times 10^{-11}\ \mathrm{N\,m^2/kg^2}$ is a universal constant — a fixed property of gravity itself, identical on Earth, on the Moon, in the Andromeda galaxy. Little $g = 9.81\ \mathrm{m/s^2}$ is a local field strength — the value of $GM/R^2$ for one particular planet at one particular radius, and it changes with altitude and with planet ( $1.62$ on the Moon, $24.8$ at Jupiter's cloud tops). The bridge between them is $g = GM/R^2$ : little $g$ is what big $G$ delivers once you specify whose surface you are standing on. A quick self-test: if your formula's answer should be the same in every galaxy, it can only contain $G$ ; if it depends on which planet you are on, $g$ (or $GM/R^2$ ) belongs in it.

How can the field exist at a point where there is nothing to feel it?

This is the conceptual heart of the field idea, and the exam question that probes it asks: what happens to the field at a point if the test mass there is doubled? Answer: nothing — the force doubles ( $F = mg$ ), but force per kilogram is unchanged. The field belongs to the source and the location: Earth's field at the Moon's orbit existed before Apollo, exists between missions, and would exist if the Moon vanished. A test mass only reveals it, the way a flag reveals wind. If "a condition of empty space" sounds like philosophy, note that the field stores energy and carries waves: LIGO detects ripples in the gravitational field arriving from black-hole mergers a billion light-years away — long after the sources stopped existing as separate objects. Something that transports energy across a billion years of vacuum has earned the word "real."

Why can I treat the whole Earth as a point at its centre? Mass near my feet is 6371 km closer than mass on the far side.

You are right to be suspicious — the near rocks pull far harder than the far ones, and it is not obvious the imbalance averages out neatly. Newton was suspicious too; the proof that it does average out delayed his publication for years. The result, the shell theorem, says: a uniform spherical shell attracts an external mass exactly as if all its mass sat at the centre — and a solid sphere is just nested shells. The cancellation is a special perk of the $1/r^2$ exponent (spherical geometry and the force law conspire precisely), so do not assume it for other force laws. The next article does the integral honestly and also answers the stranger companion question: what does a shell do to a mass inside it? (Answer: nothing at all.)

Newton's law says the force acts instantly across space. Is that actually true?

No — and Newton himself was uneasy about the deeper problem of force crossing empty space with no mechanism, calling it an absurdity in a famous letter. In Newtonian mechanics the force is instantaneous by construction, and for solar-system work that approximation is superb. But the field concept contains the seed of the correction: once gravity is a condition of space, that condition can have its own dynamics. Einstein's general relativity (1915) supplies them — changes in the field propagate at the speed of light, as gravitational waves, detected directly by LIGO in 2015. If the Sun vanished this instant, Earth would continue its orbit, and its sunlight, for the same 8.3 minutes before both the light and the field change arrived together. For orbital mechanics at UG level, Newton's instantaneous law remains the working tool; just know it is the low-speed, weak-field limit of a local, light-speed theory.

Is the gravitational field just acceleration with extra words, since g in N/kg equals 9.8 m/s²?

Numerically at a point, yes — the free-fall acceleration equals the field vector, which is why one symbol serves both. Conceptually, no, and the distinction pays off quickly. "Acceleration" describes what a freely falling object does; the field describes what space offers every object, including ones that are not accelerating at all. A book resting on a table sits in a field of $9.8\ \mathrm{N/kg}$ — that field is why the table must push up on it with $mg$ — but its acceleration is zero. The field bookkeeping (sources make fields via $g = GM/r^2$ ; fields make forces via $F = mg$ ) also splits problems cleanly in two: worked example 2 computed the Earth–Moon field map without ever mentioning the spacecraft that would fly through it. That division of labour — source physics separated from probe physics — is the pattern electromagnetism and every later field theory will reuse.

Everything attracts everything — so why doesn't the universe just collapse into one lump?

Sideways motion. Gravity is indeed purely attractive, but an attraction bent perpendicular to a body's velocity does not reel it in — it steers it, and sufficient tangential speed turns falling into orbiting (the Moon has been falling toward Earth for 4.5 billion years and perpetually missing). On the largest scales the same logic applies with expansion in place of orbits: the universe's contents are receding from each other, and gravity decelerates or curves that motion rather than instantly reversing it. Where gravity has won locally — gas clouds collapsing into stars, stars into galaxies — collapse stops when other physics (pressure, rotation) pushes back. Purely attractive does not mean irresistible; it means patient: gravity's infinite range and refusal to cancel let it shape the universe on the timescale of billions of years.

Newton's Law of Universal Gravitation

Where this lives today

The physical picture

The law, the constant, the field

Which equation answers which question

Limiting cases

Worked examples

1 · A student's weight, from scratch (confidence builder)

2 · The Earth–Moon system, twice (exam level)

Going further

Before this

Where this leads