Universal law of gravitation

Drop a pen and it falls. Watch the night sky long enough and the Moon traces the same slow circle around the Earth, month after month, without ever falling into it or drifting away. For most of history these looked like two unrelated facts of nature — one about things near the ground, one about things in the heavens. Newton's great claim, sometime around 1666 and published in the Principia in 1687, was that they are the same fact. The force that pulls the pen down and the force that bends the Moon's path into a circle are one and the same kind of pull, obeying one equation, acting between absolutely any two masses anywhere in the universe. That equation is astonishingly short: the force between two masses grows with the product of the masses and shrinks with the square of the distance between them. Double either mass and the pull doubles. Double the distance and the pull drops to a quarter. This single inverse-square rule, tied together by one fixed constant of nature, $G$ , is powerful enough to predict the orbit of every planet, the fall of every apple, the tides of every ocean, and the trajectory of every satellite humans have ever launched — all from one line of algebra. This article builds that line of algebra carefully: what each symbol means, why the force must be attractive and mutual, how it reduces to the familiar weight $mg$ near Earth's surface, and how Kepler's empirical planetary laws fall out of it almost for free. Two worked examples anchor the numbers — one for two people sitting in a room, one for a person standing on the Earth — so the enormous range of the law, from negligible to essential, becomes concrete rather than abstract.

Key takeaways

One law, every pair of masses. Newton's universal law of gravitation states that any two point masses attract each other with a force $F = \dfrac{Gm_1m_2}{r^2}$ , directed along the line joining them.
$G$ is a fixed constant of nature. $G = 6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ , the same everywhere in the universe — unlike $g$ , which changes with location.
The force is mutual and attractive. By Newton's third law, $m_1$ pulls $m_2$ exactly as hard as $m_2$ pulls $m_1$ , and gravity never pushes.
Spherical bodies act like point masses. For a uniform sphere (like a planet), the whole mass behaves as if concentrated at the centre, so $r$ is measured centre to centre, never surface to surface.
Multiple masses add as vectors. When several bodies act on one object, the net gravitational force is the vector sum of each pairwise pull — the principle of superposition.
It scales from negligible to enormous. Between two people it is about $2\times10^{-7}\ \text{N}$ , immeasurably small; between the Earth and a person it is hundreds of newtons — this is exactly what we call weight.
Key idea: every pair of masses in the universe attracts every other pair with a force F = Gm1m2/r squared, directed along the line joining them, mutual and attractive by Newton's third law, and this one rule explains falling apples, orbiting moons, and planetary motion alike.

The core insight

Every particle of matter attracts every other particle with a force $F = Gm_1m_2/r^2$ along the line joining them — mutual, always attractive, governed by the single universal constant $G$ , and strong enough to matter only when at least one mass is astronomically large.

Where it fits in the syllabus

This is the anchor equation of the CBSE Class 11 Physics chapter on Gravitation, sitting right after Kepler's laws in the prescribed sequence — Kepler described how planets move, and Newton's law explains why. Everything that follows in this chapter (acceleration due to gravity $g = GM/R^2$ , gravitational potential energy, escape speed, orbital velocity) is built directly on this one force law, so the CBSE 2025-26 syllabus expects a student to state it precisely, know $G$ 's value and units cold, and apply it confidently to point masses and spherical bodies. The vector form and the historical Cavendish measurement of $G$ are touched on here only lightly — Cavendish's experiment itself belongs to the next node in this chapter.

The physical picture

Newton's universal law of gravitation says that every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In symbols, for two point masses $m_1$ and $m_2$ separated by distance $r$ :

F = \frac{G\,m_1 m_2}{r^2}

where: $F$ is the magnitude of the mutual attractive force, in newtons; $m_1$ and $m_2$ are the two masses, in kilograms; $r$ is the distance between their centres, in metres; and $G$ is the universal gravitational constant,

G = 6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2.

Read physically, this equation says the pull weakens very fast with distance — twice as far apart means a quarter the force, three times as far means a ninth — while it grows in direct step with how heavy either object is. $G$ itself carries no information about which bodies are involved; it is a fixed conversion factor of nature, the same whether the two masses are protons, planets, or people, which is exactly why the law is called universal. The force is not just a number — it has a direction, and Newton's third law fixes that direction precisely. If $\hat r_{12}$ is the unit vector pointing from mass 2 toward mass 1, the force on mass 1 due to mass 2 is

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

The minus sign is the whole physical content of "attractive": the force on mass 1 points opposite to the direction away from mass 2, i.e. toward mass 2. By Newton's third law, the force on mass 2 due to mass 1 is exactly $\vec F_{21} = -\vec F_{12}$ — equal in size, opposite in direction, both bodies pulled toward each other with identical strength regardless of how unequal their masses are. At Class 11 level, most numerical problems only need the magnitude $F = Gm_1m_2/r^2$ together with the rule "each mass is pulled toward the other," but it is worth remembering that the full vector statement is what actually sits underneath every calculation. Two structural features of this law deserve emphasis before any numbers appear. First, gravity needs no contact and no medium — it is often called an action-at-a-distance force, acting through empty space, unlike friction or a normal push that requires touching surfaces. Second, for extended spherically symmetric bodies such as planets or stars, Newton proved (using integral calculus well beyond this course, so the result is simply quoted here) that the sphere attracts external objects exactly as if its entire mass were concentrated at a single point at its centre. This is the shell theorem result in disguise, and it is the reason a person standing on Earth's surface can be treated as being at distance $r = R_E$ (the Earth's radius) from a point mass $M_E$ sitting at the planet's centre, rather than having to add up the pull of every rock and ocean individually.

Figure

Two circles labelled mass one and mass two connected by a double headed arrow of length r, with equal and opposite force arrows F one two and F two one pointing toward each other along the line joining the centres

When more than two masses act on a single object, the net gravitational force is found by the principle of superposition: calculate each pairwise force separately using $F = Gm_1m_2/r^2$ , then add all of them as vectors — same rule that governs the addition of any other forces in mechanics. There is no special new physics here, only vector addition applied to several inverse-square pulls at once.

Derivation AFrom the inverse-square law to weight and orbital speed

The goal of this derivation is to show that the single universal law $F = Gm_1m_2/r^2$ is not a separate fact from the familiar weight $mg$ that governs everyday falling objects, nor from the circular orbital motion of a satellite — both are the same inverse-square force viewed in two different situations, connected by nothing more than substitution and Newton's second law. This is the calculation that convinces a student the law truly is universal rather than a hypothesis for astronomy alone.

Step 1. Apply the law to an object of mass $m$ resting at the Earth's surface. Treating the Earth (mass $M_E$ , radius $R_E$ ) as a point mass at its centre by the shell-theorem result quoted above, the gravitational force on the object is $F = \dfrac{G M_E m}{R_E^2}$ , since the object sits a distance $r = R_E$ from Earth's centre.

Step 2. Compare this with Newton's second law written for a freely falling object, $F = ma$ , where $a$ is the acceleration produced by that force. Since this acceleration near Earth's surface is what we call $g$ , setting the two expressions for $F$ equal gives $mg = \dfrac{GM_E m}{R_E^2}$ .

Step 3. Cancel the test mass $m$ from both sides — it appears on both sides because both the force and the resulting acceleration depend on $m$ in exactly matching ways, which is itself the deep reason all objects fall at the same rate regardless of mass. What remains is $g = \dfrac{GM_E}{R_E^2}$ , an expression built entirely from properties of the Earth (its mass and radius) and the universal constant $G$ — no trace of the falling object survives. This is the bridge between the universal law and the familiar $g = 9.8\ \text{m/s}^2$ ; a full treatment of how $g$ varies with altitude and depth belongs to a later node in this chapter.

Step 4. Now repeat the same substitution for a satellite of mass $m$ moving in a circular orbit of radius $r$ around a planet of mass $M$ . The gravitational pull provides exactly the centripetal force needed to keep it curving, so $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ , where $v$ is the orbital speed. Cancelling one factor of $m$ and one factor of $r$ from both sides leaves $v^2 = \dfrac{GM}{r}$ , so the orbital speed is $v = \sqrt{GM/r}$ — again built purely from the central mass, the orbit radius, and $G$ , with the satellite's own mass having cancelled out entirely.

Step 5. Read the two results together. Both $g = GM_E/R_E^2$ (a falling object) and $v = \sqrt{GM/r}$ (an orbiting satellite) trace back to one identical starting equation, $F = Gm_1m_2/r^2$ . Nothing separate was assumed for "falling" versus "orbiting" — the only difference between an apple that falls straight down and a Moon that endlessly circles is the object's sideways speed when the pull begins to act, a point Newton illustrated with his famous thought experiment of a cannonball fired ever faster from a mountain top until it never comes down.

Worked examples

Board level - force between two classmates

Find the gravitational force of attraction between two students, one of mass 50 kg and the other of mass 60 kg, sitting 1.0 m apart in a classroom.

Step 1. Identify the known quantities: $m_1 = 50\ \text{kg}$ , $m_2 = 60\ \text{kg}$ , $r = 1.0\ \text{m}$ , and $G = 6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ .

Step 2. Substitute directly into $F = \dfrac{Gm_1m_2}{r^2}$ :

F = \frac{(6.67\times10^{-11})(50)(60)}{(1.0)^2} = \frac{6.67\times10^{-11}\times 3000}{1}

Step 3. Compute the product: $6.67\times10^{-11} \times 3000 = 2.0\times10^{-7}\ \text{N}$ .

Step 4. Interpret the size of the answer. $F \approx 2.0\times10^{-7}\ \text{N}$ , roughly the weight of a single grain of dust. This is exactly why gravity between two everyday-sized objects is never noticed — the constant $G$ is so tiny that only when at least one mass is planet-sized does the force become large enough to feel.

Competitive level - force between Earth and a person, checked against weight

The mass of the Earth is $M_E = 6.0\times10^{24}\ \text{kg}$ and its radius is $R_E = 6.4\times10^6\ \text{m}$ . Find the gravitational force between the Earth and a 70 kg person standing on its surface, and verify this matches the person's weight $mg$ .

Step 1. Since the Earth is spherically symmetric, treat it as a point mass $M_E$ located at its centre, so the person (at the surface) is at distance $r = R_E = 6.4\times10^6\ \text{m}$ from that centre — not the height of the person above the ground, which is negligible next to $R_E$ .

Step 2. Substitute into $F = \dfrac{GM_Em}{R_E^2}$ :

F = \frac{(6.67\times10^{-11})(6.0\times10^{24})(70)}{(6.4\times10^6)^2}

Step 3. Work the numerator in stages: $6.67\times10^{-11} \times 6.0\times10^{24} = 4.0\times10^{14}$ , then $4.0\times10^{14} \times 70 = 2.8\times10^{16}$ .

Step 4. Work the denominator: $(6.4\times10^6)^2 = 4.096\times10^{13}\ \text{m}^2$ .

Step 5. Divide: $F = \dfrac{2.8\times10^{16}}{4.096\times10^{13}} \approx 683.6\ \text{N}$ .

Step 6. Cross-check against the familiar weight formula: $mg = 70 \times 9.8 = 686\ \text{N}$ . The two values agree to within about 0.4 per cent (the small gap comes from rounding $M_E$ and $R_E$ to two significant figures), confirming that a person's weight is precisely the gravitational force the Earth exerts, computed from the exact same universal law used for two classmates in a room. Timing note: in an exam, do not recompute $g$ from $GM_E/R_E^2$ every time weight is needed — that derivation (done once above) is exactly why $g = 9.8\ \text{m/s}^2$ is handed to you as a standard constant. Use $F = GM_Em/R_E^2$ only when the question explicitly gives $M_E$ and $R_E$ and asks you to derive or verify $g$ or weight from first principles; otherwise $mg$ is faster and equally correct.

Common traps

Measuring $r$ from the surface of a spherical body instead of from its centre — for Earth-surface problems this is the difference between $R_E$ and $R_E + h$ , which matters once $h$ is not negligible.
Confusing $G$ (the universal gravitational constant, fixed everywhere) with $g$ (the acceleration due to gravity, which varies with planet, altitude, and depth) — these are different quantities with deliberately similar symbols, and examiners test this confusion directly.
Assuming gravity requires contact or a medium to act through; it is an action-at-a-distance force that works through a vacuum exactly as it works through air.
Forgetting Newton's third law and assuming only the more massive body "does the pulling" — both bodies exert forces of identical magnitude on each other, even though the lighter body accelerates far more visibly.
Adding several gravitational forces as plain numbers instead of vectors when more than two masses are involved — direction cannot be ignored once there are three or more bodies.

Common trapGravity between everyday objects is real but always negligible — never treat it as zero, but never expect to feel it either

Because $G = 6.67\times10^{-11}$ is such an extremely small number, the gravitational force between ordinary human-scale objects (two people, a book and a table, a car and a lamppost) is always many orders of magnitude smaller than other forces around it, such as friction or the normal force. This does not mean the force is zero or that the law fails at small scales — it is simply too small to detect without extraordinarily sensitive apparatus (this is exactly what the Cavendish experiment, met in the next node, was built to do). The mistake to avoid is concluding that gravity "only applies" to planets and stars; it applies to every pair of masses in the universe without exception, and only becomes large enough to notice once at least one of the masses is astronomically large.

You should be able to…

State the universal law of gravitation in words and as the equation $F = Gm_1m_2/r^2$ , with correct SI units for every quantity.
Recall the numerical value and units of $G$ , and explain clearly how $G$ differs from $g$ .
Calculate the gravitational force between two point masses, or between a spherical body and a point mass, given their masses and separation.
Explain why the force must be treated as centre-to-centre for spherical bodies, and why it is both mutual and always attractive.
Combine two or more gravitational forces on a single object using vector superposition.
Connect the universal law to the familiar weight formula $mg$ and to circular orbital speed $v = \sqrt{GM/r}$ by simple substitution.

Ask the tutor: Want to see how this law extends into $g = GM/R^2$ with altitude and depth variation, or work through a three-mass force problem step by step — ask the AI tutor for a guided walkthrough.

Why don't I feel the Earth pulling on the book on my desk with any extra force beyond its weight?

You already do feel it — the book's weight is the Earth's gravitational pull on it, nothing extra needs to happen. What you don't notice is the tiny sideways pull between the book and, say, your pencil case sitting next to it, because that force between two small masses is around $10^{-13}\ \text{N}$ or smaller, utterly swamped by friction and the normal force from the desk. The Earth's pull on the book is large only because the Earth's mass $M_E$ is enormous, not because gravity behaves differently for planets.

Since G is the same everywhere, why is gravity stronger on Jupiter than on Earth?

$G$ never changes — it is a fixed constant of nature. What changes from planet to planet is $g = GM/R^2$ , which depends on that planet's own mass $M$ and radius $R$ . Jupiter has a much larger mass than Earth, and although its radius is also larger (which would reduce $g$ ), the mass effect wins out overall, so $g$ on Jupiter's cloud tops works out larger than Earth's 9.8 m/s². The universal constant $G$ is what lets you calculate $g$ for any planet once you know its mass and radius — it is not itself the thing that differs.

Does the gravitational force ever become exactly zero at very large distances?

Mathematically, $F = Gm_1m_2/r^2$ only ever approaches zero as $r$ grows without bound — it never reaches exactly zero at any finite distance, however large. In practice, once $r$ is large enough the force becomes so small that it is completely swamped by the pull of other, closer masses, so for all practical purposes we treat gravity as negligible beyond a certain range. But strictly, within Newton's theory, every mass in the universe exerts some pull on every other mass, no matter how distant.

How is this law connected to Kepler's laws that I studied just before this topic?

Kepler discovered his three laws purely from careful observation of planetary positions, with no explanation for why planets move that way. Newton's universal law of gravitation is the underlying cause: applying $F = GMm/r^2$ as the centripetal force for a planet in a roughly circular orbit around the Sun leads directly to Kepler's third law, $T^2 \propto r^3$ (the derivation for orbital speed shown above is the first step of that chain). In short, Kepler described the pattern, and Newton's law explains where the pattern comes from.