Newton's Law of Universal Gravitation

In 1666, with Cambridge closed by plague, a twenty-three-year-old sitting in a Lincolnshire orchard asked the most productive question in the history of physics: does the force that drops an apple reach all the way to the Moon? Everyone before him had kept two rulebooks — one for earthly falling, one for heavenly circling. Newton's suspicion was scandalous: one rule, one force, stretching from the orchard grass to the Moon and beyond, weakening with the square of the distance. Twenty years later he could prove it. The Moon, he showed, is perpetually falling toward the Earth — by about 1.4 millimetres every second — and that falling rate is exactly the apple's $9.8\ \mathrm{m/s^2}$ diluted by the square of sixty, the Moon's distance in Earth-radii. The same law then delivered the planets to Kepler's ellipses, the tides to the Moon's pull, and — once Cavendish caught the constant $G$ on a basement torsion balance in 1798 — the mass of the Earth itself to four honest digits. For your examinations, this law is the gateway to an entire chapter: fields, orbits, escape, satellites. The good news is its exam craft is compact: one formula, one ratio trick that answers most questions without touching $G$ , one rule about measuring distances centre-to-centre, and a superposition principle for multiple masses. This article builds all four, with the two greatest calculations in the law's history as your worked derivations.

Key takeaways

Every pair of point masses attracts: $F = \dfrac{G\,m_1 m_2}{r^2}$ , along the line joining them, with $G = 6.67 \times 10^{-11}\ \mathrm{N\,m^2/kg^2}$ .
$r$ is centre-to-centre: a person "on the ground" is $6370$ km from Earth's $r$ -origin. Uniform spheres act as points at their centres (the shell theorem's gift).
The exam workhorse is the ratio: $\dfrac{F_2}{F_1} = \left(\dfrac{r_1}{r_2}\right)^2$ — double the distance, quarter the force; no $G$ , no masses needed.
$G$ is universal and tiny (two 1-kg balls 1 m apart: $6.67 \times 10^{-11}$ N); little $g = GM/R^2 = 9.8\ \mathrm{m/s^2}$ is local. They are different letters and different ideas.
The law is symmetric: the apple pulls the Earth exactly as hard as the Earth pulls the apple — Newton's third law is built in.
Multiple masses: net force = vector sum of pairwise pulls (superposition), each pair computed as if alone.
Key idea: one inverse-square attraction between every pair of masses runs the apple, the Moon, the tides and the galaxies — and $mg$ is just its surface abbreviation.

The core insight

Any two point masses attract each other along the line joining their centres with force $F = Gm_1m_2/r^2$ — doubling either mass doubles the force, doubling the centre-to-centre distance quarters it — and the net gravity from several masses is the vector sum of the pairwise pulls.

Where this lives today

The purest modern descendants of Cavendish's bench are flying two satellites called GRACE-FO, chasing each other around the Earth 220 km apart and measuring their separation to a fraction of a hair's width. Every mountain range, aquifer and melting ice sheet below tugs the leading satellite first, opening and closing the gap by microns — and from those microns, mission scientists rebuild monthly maps of Earth's gravity field, watching groundwater vanish under drought-stricken regions and Greenland shed its hundreds of gigatonnes a year. The same $Gm_1m_2/r^2$ bookkeeping navigates every interplanetary probe (JWST's orbit, Mars landings, the gravitational slingshots that flung the Voyagers outward), sets the masses of exoplanets from the wobble of their stars, and — via the product $GM$ — weighs black holes at galactic centres. Not bad for a formula whose constant was caught between two lead balls in a basement.

The physical picture

Why the square? Not decoration — geometry. Picture a mass broadcasting its influence equally in all directions, like a bulb radiating light. At distance $r$ that influence is spread over a sphere of area $4\pi r^2$ . Double the distance and the same total influence must cover four times the area; each square metre — and each kilogram sitting on it — receives a quarter of the pull. The exponent 2 in the law is the 2 in "area goes as radius squared." Space itself dilutes gravity.

Figure

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

Two more readings before the formula. First, gravity is mutual: the law contains $m_1 m_2$ symmetrically, so the apple hauls on the Earth with the same newtons the Earth exerts on the apple. Nothing paradoxical follows — the Earth, $10^{25}$ times more massive, responds with $10^{25}$ times less acceleration. Second, gravity is universal and unscreenable: every pair of masses in the universe participates, and no material blocks it. You are right now being tugged by your desk, this page, and the Andromeda galaxy. You fail to notice only because the constant of proportionality is spectacularly small — and that smallness, with its consequences, is half the story of this chapter.

The law, the constant, the ratio trick

Two point masses $m_1$ and $m_2$ , centres a distance $r$ apart, attract each other along the joining line:

F = \frac{G\,m_1 m_2}{r^2} \qquad (1)

where: $G = 6.67 \times 10^{-11}\ \mathrm{N\,m^2/kg^2}$ — the universal gravitational constant; $r$ — the centre-to-centre separation. Read the anatomy: masses multiplied upstairs (double either, double the force), distance squared downstairs (double it, quarter the force). For real bodies the law strictly applies to points — but a uniform sphere gravitates exactly as if all its mass sat at its centre (the shell theorem, next article), which is why planets and billiard balls alike enter with their centre positions. Feel the size of $G$ with the definitional case: two 1-kg masses 1 m apart pull with $6.67 \times 10^{-11}$ N — roughly the weight of a bacterium colony. Two students, 60 and 70 kg, seated a metre apart: $F = 6.67\times10^{-11} \times 4200 \approx 2.8 \times 10^{-7}$ N — a thousand times less than the weight of a hair. Gravity between everyday objects is real, measurable (Cavendish measured it), and utterly negligible. It runs the cosmos anyway, because it has infinite range, never cancels (mass has no negative sign), and the masses upstairs can be $10^{24}$ kg. Most exam questions never need $G$ at all. For the same pair of masses at two separations, divide the law by itself — $G$ and the masses vanish:

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

where: subscripts label the two configurations. Triple the distance: $1/9$ the force. Halve it: 4 times. This ratio, plus the centre-to-centre discipline, dispatches the majority of board and NEET gravitation MCQs in under thirty seconds each. When several masses act on one, gravity superposes: compute each pairwise pull by equation (1) as if the others did not exist, then add them as vectors:

\vec F_{\text{net}} = \vec F_1 + \vec F_2 + \cdots \qquad (3)

where: each $\vec F_i$ points from the test body toward source $i$ . The vector character matters: two equal masses on either side of you pull equally and cancel; along a line between unequal masses there is a null point closer to the lighter one — a favourite two-mass exam setup.

you want	use	notes
force between two masses	eq. (1)	centre-to-centre $r$ ; spheres count as points
compare two separations	ratio (2)	no $G$ , no masses — the 30-second weapon
several sources	vector sum (3)	each pair as if alone
surface gravity of a planet	$g = \dfrac{GM}{R^2}$	equate $mg$ with eq. (1) — next article's headline
a planet's mass	$M = \dfrac{gR^2}{G}$	Cavendish's bench → the Earth on a scale

Derivation AThe Moon test — the calculation that unified the heavens and the Earth

We verify the inverse-square exponent the way Newton did: compare two falling objects — an apple at the Earth's surface, and the Moon, sixty Earth-radii away. The strategy is a two-column audit. Column one predicts the Moon's falling acceleration by diluting $g$ with $1/60^2$ ; column two measures that acceleration independently from the Moon's orbital geometry and the calendar, with no gravity theory anywhere in it. If the columns match, one law spans orchard and sky. Watch Step 3: it is pure circular-motion kinematics — that independence is what makes the agreement evidence rather than circular reasoning.

Step 1. The geometry, known since antiquity from lunar parallax: the Moon's orbital radius is $r = 3.84 \times 10^8$ m $= 60\,R_E$ , with $R_E = 6.37 \times 10^6$ m.

Step 2. The prediction. If gravity thins as $1/r^2$ , then at $60$ Earth-radii the falling acceleration should be

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

Step 3. The independent measurement. A body circling at radius $r$ with period $T$ has centripetal acceleration $a = \dfrac{4\pi^2 r}{T^2}$ — geometry and a clock, nothing else. With $T = 27.3$ days $= 2.36 \times 10^6$ 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 doesn't crash despite falling: it falls about 1.4 mm every second — and in that second it also moves a kilometre sideways, so the ground curves away beneath it by the same 1.4 mm. An orbit is a fall that keeps missing.

Step 4. Compare: prediction and measurement agree to three figures. The apple's gravity, diluted by exactly the square of the distance, steers the Moon. Before this number, terrestrial and celestial physics were separate subjects; after it, they were one. Verified: $9.8/3600 = 2.72\times10^{-3}$ ; $4\pi^2 \times 3.84\times10^8 = 1.52\times10^{10}$ ; $(2.36\times10^6)^2 = 5.57\times10^{12}$ ; quotient $2.72\times10^{-3}\ \mathrm{m/s^2}$ ✓. Treatment as in HRW Ch. 13; NCERT Ch. 8 presents the same test.

Go deeper · Derivation B · 1 minWeighing the Earth — what Cavendish's fibre really bought

Derivation BWeighing the Earth — what Cavendish's fibre really bought

We derive $M_E = gR_E^2/G$ : the mass of the Earth from three benchtop-and-map numbers. The strategy is to describe the weight of an ordinary object twice — the everyday way ( $mg$ ) and the universal way (equation 1, with the Earth as partner) — and demand the two descriptions agree, since they are the same force. Watch the object's own mass cancel in Step 2: that cancellation is why all objects fall alike, and it is what leaves behind a formula containing only the planet.

Step 1. One force, two languages. For a mass $m$ on the surface (centre-to-centre distance $= R_E$ ):

mg = \frac{G M_E\, m}{R_E^2}

Step 2. Cancel $m$ :

g = \frac{G M_E}{R_E^2}

Why this line matters beyond the algebra: the falling acceleration is independent of what falls — feather and hammer agree — because the falling body's mass appears on both sides. Galileo's observation becomes Newton's theorem.

Step 3. Solve for the planet:

M_E = \frac{g R_E^2}{G} = \frac{9.8 \times (6.37 \times 10^6)^2}{6.67 \times 10^{-11}} = 5.97 \times 10^{24}\ \mathrm{kg}

Why $G$ was the bottleneck: $g$ comes from any pendulum and $R_E$ from geodesy (known since Eratosthenes), but orbits only ever reveal the product $GM$ — never $G$ alone. Someone had to measure the raw pull between known masses on a bench. Cavendish's 1798 torsion balance — two lead spheres tugging two small balls through a twist of a fibre — did exactly that, which is why his paper could honestly be titled an experiment "to determine the density of the Earth."

Step 4. The bonus that reshaped geology: mean density $\rho = M_E / \left(\tfrac43\pi R_E^3\right) \approx 5500\ \mathrm{kg/m^3}$ — but surface rocks average only $\sim 2700$ . The deep Earth must be far denser than anything in the crust: the first physical evidence for an iron core. Verified: $(6.37\times10^6)^2 = 4.06\times10^{13}$ ; $\times\,9.8 = 3.98\times10^{14}$ ; $\div\,6.67\times10^{-11} = 5.97\times10^{24}$ kg ✓; volume $1.08\times10^{21}\ \mathrm{m^3}$ gives $\rho = 5.5\times10^3\ \mathrm{kg/m^3}$ ✓. Cavendish's own 1798 value, 5.448 relative to water, sits within 1% of the modern figure.

Figure

Cavendish's torsion balance — a force microscope catches G on a bench

Limiting cases

case	result	meaning
$r \to 2r$	$F \to F/4$	the signature move — memorize the ratio reflex
both masses doubled, $r$ doubled	$F$ unchanged	$2\times2$ upstairs cancels $2^2$ downstairs
$r \to \infty$	$F \to 0$ but never exactly	infinite range: gravity thins, never quits
heights $h \ll R_E$	$F \approx mg$ , uniform	projectile physics is the local zoom of this law
$m_1 = m_2 = 1$ kg, $r = 1$ m	$F = G$ numerically	$G$ is the unit-mass, unit-distance pull
equal masses either side of a body	net force zero	superposition is a vector sum — direction counts
sphere treated as a point	exact for uniform spheres	the shell theorem's licence (next article)

Common trap“Twice as high means a quarter of the gravity”

The most reliable mark-loser in this chapter is measuring $r$ from the wrong place: the $r$ in equation (1) runs centre to centre, so it starts $6370$ km beneath your feet. Climbing a 6 km mountain changes $r$ by under $0.1\%$ — your weight barely notices. An astronaut at altitude $R_E$ is at $r = 2R_E$ (force $/4$ ); at altitude $2R_E$ she is at $r = 3R_E$ (force $/9$ , not $/4$ ). Every board recycles this trap annually. Its sibling rides along: big $G$ is not little $g$ . $G = 6.67\times10^{-11}\ \mathrm{N\,m^2/kg^2}$ is universal — identical on the Moon, in Andromeda, forever. $g = 9.8\ \mathrm{m/s^2}$ is local shorthand for $GM/R^2$ , different on every planet and every altitude. Swapping the letters in a formula is an automatic zero.

Worked examples

1 · Board level — a weight, from the universal law (CBSE/ISC pattern)

Earth has mass $6.0 \times 10^{24}\ \mathrm{kg}$ and radius $6.4 \times 10^6\ \mathrm{m}$ . (a) Compute the gravitational force on a $60\ \mathrm{kg}$ student standing on the surface. (b) Find the force when the student is at an altitude of $2R_E$ . ( $G = 6.67 \times 10^{-11}$ SI.)

Step 1. Identify $r$ : standing on the surface means $r = R_E = 6.4 \times 10^6$ m — centre-to-centre.

Step 2. Numerator of (1): $GMm = 6.67\times10^{-11} \times 6.0\times10^{24} \times 60 = 2.40 \times 10^{16}$ . Denominator: $r^2 = 4.10 \times 10^{13}\ \mathrm{m^2}$ .

Step 3. $F = \dfrac{2.40\times10^{16}}{4.10\times10^{13}} = \boxed{586\ \mathrm{N}}$ . Cross-check: $mg = 60 \times 9.8 = 588$ N ✓ — the everyday weight formula is this law wearing its surface disguise.

Step 4. (b) is a ratio, not a recomputation. Altitude $2R_E$ ⇒ $r = 3R_E$ . By (2): $F' = 586 \times \left(\tfrac{1}{3}\right)^2 = \boxed{65\ \mathrm{N}}$ . Dodged: the "twice as high = one quarter" reflex.

2 · Board level — the Earth–Moon bond (ISC/IB pattern)

Find the gravitational force between the Earth ( $5.97 \times 10^{24}\ \mathrm{kg}$ ) and the Moon ( $7.35 \times 10^{22}\ \mathrm{kg}$ ), centres $3.84 \times 10^8\ \mathrm{m}$ apart, and compute the Moon's resulting acceleration.

Step 1. Numerator: $6.67\times10^{-11} \times 5.97\times10^{24} \times 7.35\times10^{22} = 2.93 \times 10^{37}$ .

Step 2. Denominator: $(3.84\times10^8)^2 = 1.47\times10^{17}\ \mathrm{m^2}$ .

Step 3. $F = \boxed{1.99 \times 10^{20}\ \mathrm{N}}$ — from a constant with $10^{-11}$ in it: the masses upstairs are simply that enormous. This one force runs the monthly orbit and drags the ocean tides.

Step 4. The Moon's acceleration: $a = F/M_{\text{Moon}} = 1.99\times10^{20}/7.35\times10^{22} = 2.7\times10^{-3}\ \mathrm{m/s^2}$ — precisely the Moon-test number, closing the loop.

3 · Competitive level — superposition on a line (JEE-Main/NEET pattern, ~2.5 minutes)

Point masses of $4.0\ \mathrm{kg}$ and $1.0\ \mathrm{kg}$ are fixed $0.90\ \mathrm{m}$ apart. Where, on the line joining them, should a third mass be placed so it feels no net gravitational force?

Step 1. The null must lie between the masses (outside, both pulls point the same way) and closer to the lighter one (its weaker field needs a shorter lever). Let it sit $x$ from the 4 kg mass.

Step 2. Equal magnitudes: $\dfrac{G(4)m}{x^2} = \dfrac{G(1)m}{(0.9 - x)^2}$ — $G$ and the third mass cancel instantly.

Step 3. Square-root both sides (cleaner than cross-multiplying): $\dfrac{x}{0.9 - x} = \sqrt{4} = 2$ .

Step 4. $x = 2(0.9 - x) \Rightarrow 3x = 1.8 \Rightarrow x = \boxed{0.60\ \mathrm{m}}$ from the 4 kg mass ( $0.30$ m from the 1 kg). Sanity: nearer the lighter body ✓, and the position is independent of the third mass ✓. Timing note: square-rooting the balance equation converts every null-point problem into one linear equation — 2.5 minutes total. Cross-multiplying into a quadratic doubles the time and invites sign slips; the distance ratio is always $\sqrt{m_1/m_2}$ .

4 · Competitive level — JEE-Advanced stretch: the ratio gauntlet (~3 minutes)

A planet has twice the Earth's density but half its radius. (a) Find its surface gravity in terms of $g_E$ . (b) A body weighs $600\ \mathrm{N}$ on Earth's surface; find its weight at one Earth-radius altitude above THIS planet's surface.

Step 1. Build surface gravity from density: $g = \dfrac{GM}{R^2} = \dfrac{G \cdot \rho\,\tfrac43\pi R^3}{R^2} = \tfrac43\pi G\rho R$ — surface gravity is proportional to $\rho R$ .

Step 2. (a) With $\rho' = 2\rho_E$ , $R' = R_E/2$ : $g' = g_E \times 2 \times \tfrac12 = \boxed{g_E}$ — same surface gravity, different planet.

Step 3. (b) The body's mass: $600/g_E = 61.2$ kg (at $g_E = 9.8$ ). At altitude $R_E$ above the planet of radius $R_E/2$ , the centre distance is $r = R_E/2 + R_E = \tfrac32 R_E = 3R'$ .

Step 4. Ratio from the planet's own surface: $W = 600 \times \left(\dfrac{R'}{3R'}\right)^2 = \dfrac{600}{9} = \boxed{67\ \mathrm{N}}$ . Timing note: the two moves are $g \propto \rho R$ (build) and the $(r_1/r_2)^2$ ratio (compare) — neither needs $G$ 's value. Three minutes; the planted error is using altitude over the EARTH's radius instead of the planet's own.

Exam traps

Centre-to-centre, always. Surface separation, altitude, "height above the ground" — all must be converted to centre distances before equation (1) or ratio (2) touches them. Two spheres of radii $R_1$ , $R_2$ "touching": $r = R_1 + R_2$ .

Big $G$ versus little $g$ . $G$ : universal constant, $6.67\times10^{-11}$ , set by nature. $g$ : local field, $GM/R^2$ , different per planet and altitude. Formulas mixing them ( $F = gm_1m_2/r^2$ ) score zero on sight.

The ratio reflex — used and audited. Most numerical MCQs are equation (2) in costume. But audit which distance doubled: altitude doubling is not r doubling (the classic $F/9$ vs $F/4$ ambush).

Superposition is a vector sum. Adding magnitudes when two sources pull from different directions is the standard multi-mass error. Draw the two arrows first; then add components (or exploit symmetry).

" $F$ is huge, so $G$ can't be small" confusions. Force between planets is $10^{20}$ N with $G = 10^{-11}$ — because $10^{24}$ -kg masses sit upstairs. Between people, the same law gives $10^{-7}$ N. Let the masses, not intuition, set the scale.

Third-law symmetry. "The Earth pulls the apple harder than the apple pulls the Earth" is false by the law's own structure — the forces are equal; the accelerations differ by the mass ratio. Boards love this as a one-mark conceptual.

Which $g$ ? CBSE/ISC: $9.8\ \mathrm{m/s^2}$ ; JEE/NEET: as stated (usually 10); IB: 9.8; A-Level: 9.81. And $G$ is always supplied — never memorize it wrong.

Origins · 1 minIsaac Newton and how this was discovered

Isaac Newton

1643–1727 · Woolsthorpe & Cambridge

The plague years of 1665–66 sent Cambridge's students home, and in a Lincolnshire farmhouse garden Isaac Newton, twenty-three, laid the foundations of calculus, optics and gravitation more or less simultaneously. The apple story is real — he told it himself in old age — but the apple was a question, not an answer: if gravity reaches the treetop, why not the Moon? The hard part took twenty years: proving that a sphere gravitates as a point (he needed to invent the mathematics first), extracting the inverse square from Kepler's laws, and running the Moon test. The Principia of 1687, prodded into existence by Halley, derived the tides, the flattening of the Earth, and the comets' paths from one formula. Newton never learned what $G$ was — that took Cavendish, 111 years later — and never claimed to know what gravity is: "I frame no hypotheses." He computed what it does, and that was enough to run celestial mechanics for two centuries.

History

1609

— Kepler publishes his first two laws: planets ride ellipses, sweeping equal areas — patterns in search of a cause.

1666

— the plague-year insight at Woolsthorpe: the apple's gravity, diluted by distance squared, might be the Moon's leash.

1687

— the Principia: one inverse-square law derives Kepler's laws, the tides and the shape of the Earth.

1798

— Cavendish's torsion balance measures the pull between lead spheres, pinning

G

and weighing the Earth at 5.45 times water's density.

1846

— Le Verrier predicts Neptune from Uranus's orbital misbehaviour — the law discovers a planet sight unseen.

1915

— Einstein's general relativity explains why the law holds (and where it bends: Mercury's orbit, light, black holes); Newton survives as its everyday limit.

Going further

Texts. NCERT Physics Class 11, Ch. 8 (Gravitation) · HC Verma Vol. 1, Ch. 11 · Halliday–Resnick–Walker Ch. 13 · Feynman Lectures Vol. I, Ch. 7 for the story told unforgettably.

Provenance. All results are standard gravitation canon. A verifier should recompute: the Moon test ( $9.8/3600 = 2.72\times10^{-3} = 4\pi^2 r/T^2$ ); $M_E = gR^2/G = 5.97\times10^{24}$ kg and $\rho \approx 5500\ \mathrm{kg/m^3}$ ; example 1's $586$ N and $65$ N; example 2's $1.99\times10^{20}$ N and $2.7\times10^{-3}\ \mathrm{m/s^2}$ ; example 3's $0.60$ m via the $\sqrt{4}=2$ ratio; example 4's $g' = g_E$ and $67$ N.

Prerequisites. Newton's Second Law: F = ma · Weight, Normal Force & Apparent Weight · Circular Motion: Centripetal & Tangential Acceleration. Leads to. The Gravitational Field & Variation of g · The Shell Theorem · Gravitational Potential Energy & Potential.

If everything attracts everything, why don't I feel objects around me pulling?

You are being pulled — by your desk, your phone, the person beside you — but run the numbers: two 60-kg people a metre apart attract with about $2 \times 10^{-7}$ N, a hundred-thousandth of the weight of a paperclip. $G$ 's smallness makes everyday gravity utterly negligible next to friction, air currents, even sound pressure. Gravity only becomes the boss when one partner is planet-sized — and then, because it has infinite range and never cancels, it becomes the only boss. Cavendish needed a torsion fibre of heroic delicacy precisely because the desk-scale force is that faint.

Why is the exponent exactly 2 and not 2.01?

Geometrically, because influence spreading through three-dimensional space dilutes over spherical areas $\propto r^2$ — an exponent other than 2 would mean space itself leaks or concentrates gravity. Experimentally, the exponent is tested ferociously: planetary orbits close on themselves only for $1/r^2$ (any other power makes ellipses precess — and Mercury's tiny leftover precession was explained by general relativity, not by amending the exponent), and laboratory tests confirm the law down to fractions of a millimetre. For your exams: it is exactly 2, and the shell theorem you'll meet next survives only because of it.

The Moon is falling toward the Earth — why doesn't it get any closer?

Because it also moves sideways, fast. Each second the Moon drops about 1.4 mm toward the Earth relative to straight-line motion — and each second its kilometre of sideways travel carries it around the curve, so the Earth's surface "falls away" by the same amount. The fall and the miss repeat forever: that is an orbit. Newton's own thought experiment — cannonballs fired ever faster from a mountaintop until one circles the Earth — remains the best mental picture ever drawn for it.

Why do a feather and a hammer fall at the same rate if gravity depends on mass?

Gravity pulls the hammer harder — but the hammer also resists acceleration proportionally harder. The force is $GMm/r^2$ (proportional to $m$ ); the response is $a = F/m$ , and the $m$ cancels: $a = GM/r^2$ , identical for feather, hammer and astronaut. The Apollo 15 hammer-and-feather drop on the airless Moon showed it live. This "gravitational mass = inertial mass" coincidence runs deeper than Newton could say — it became the founding principle of general relativity.

Is there really zero gravity on the International Space Station?

No — at 400 km altitude, $r$ has grown only from $6370$ to $6770$ km, so gravity is still about $89\%$ of its surface strength. Astronauts float not because gravity quit but because station and astronaut are falling together — a permanent shared free fall that happens to keep missing the Earth (see the previous FAQ). "Weightlessness" is the absence of the support forces that usually oppose gravity, not the absence of gravity. The satellites article makes this quantitative.

How can the Earth pull the Moon without touching it?

Newton was bothered too — he called action at a distance an "absurdity" and computed with it anyway. The modern resolution is the field: the Earth fills the space around it with a gravitational field $g = GM/r^2$ , and the Moon responds to the field where the Moon is — a local transaction, no spooky reach. The next article builds this idea properly. Deeper still, Einstein recast the field as curved spacetime, and changes in it travel at light speed — confirmed when LIGO heard spacetime ripple in 2015.

Do I ever need to plug in the actual value of G in an exam?

Rarely — and that is a strategy, not trivia. Most questions compare two configurations, and the ratio (2) kills $G$ and the masses before arithmetic starts. Questions that do need $G$ (computing an absolute force, weighing a planet) always supply its value in the stem. What you must carry in memory is not $6.67\times10^{-11}$ but the structure: masses multiply, distance squares, centre-to-centre, vectors add. Spend your memorization budget on the ratio reflex — it pays every single paper.