Universal law of gravitation

Every object with mass pulls on every other object with mass — an apple falling to the ground and the Moon staying locked in orbit are the same phenomenon, governed by the same rule. Newton's insight was to unify what looked like two separate worlds: the "earthly" rules that make things fall, and the "heavenly" rules that keep the Moon and planets moving. He showed that a single force law, with a single constant $G$ that is the same everywhere in the universe, explains both. This article gives that law its exact mathematical form, connects it to the everyday acceleration $g\approx9.8\ \text{m/s}^2$ you already use in mechanics, and works through the numbers so the sheer weakness of gravity between ordinary-sized objects becomes concrete.

The core insight

Any two point masses $m_1$ and $m_2$ separated by distance $r$ attract each other along the line joining them with force $F=Gm_1m_2/r^2$ , where $G=6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ is a universal constant, and this same law reduces to the familiar $g=GM/R^2$ for a mass sitting on a planet's surface.

The core idea

Newton proposed that gravity is not a special force reserved for planets and moons — it acts between any two masses, however small, and obeys one precise rule.

F=\dfrac{Gm_1m_2}{r^2}

where: $F$ is the magnitude of the mutual attractive force, $m_1$ and $m_2$ are the two masses, $r$ is the distance between their centres, and $G$ is the universal gravitational constant, $G=6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ . The word "universal" carries two separate meanings worth keeping apart: $G$ itself has the same numerical value everywhere in the universe, and the same inverse-square shape $F\propto1/r^2$ applies whether the two masses are two stones on a table or the Sun and a planet.

The unit and dimensions of $G$ follow directly from rearranging the law: $G=Fr^2/(m_1m_2)$ , so its SI unit is $\text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ . Writing force in base dimensions as $[F]=[MLT^{-2}]$ gives

[G]=\dfrac{[MLT^{-2}][L^2]}{[M^2]}=[M^{-1}L^3T^{-2}]

Newton's third law applies here too. The pull the Earth exerts on you and the pull you exert on the Earth are equal in size and opposite in direction — you simply never notice your own tiny effect on a planet many orders of magnitude more massive than you.

Figure

Two spheres of unequal size pulling on each other with equal and opposite forces along the line joining their centres, labelled with masses m1 and m2 separated by distance r

Why it is called "universal." Before Newton, terrestrial motion (falling apples, thrown stones) and celestial motion (the Moon, the planets) were widely believed to obey completely different rules. Newton's key move was to show that the same force, given by the same $F=Gm_1m_2/r^2$ , pulls an apple down and simultaneously holds the Moon in orbit. That single law unifying the ground and the sky is exactly why it is called universal. The historical measurement of $G$ used a delicate torsion-balance setup (the Cavendish experiment); its procedure is not part of this syllabus — you only need to treat $G$ as a known, measured constant and use it directly.

Why the force is so easy to overlook in daily life. The value of $G$ is extraordinarily small, $6.67\times10^{-11}$ in SI units, so the force between two ordinary masses is tiny unless at least one of $m_1$ or $m_2$ is enormous. This is very different from, say, the electrostatic force between charges, which can be strong even for small everyday objects. Gravity is in fact the weakest of the four fundamental forces, yet it dominates on astronomical scales for a simple reason: mass is always positive and gravity is always attractive, so the pulls of every particle in a large body add up instead of cancelling, the way positive and negative charges cancel in a neutral object. That is why a planet's enormous total mass, built up from countless individually feeble contributions, can hold a whole atmosphere or an orbiting Moon in place.

Derivation AConnecting G to the surface value of g

The goal is to derive the familiar relation $g=GM/R^2$ , linking the universal constant $G$ to the acceleration due to gravity $g\approx9.8\ \text{m/s}^2$ that you already use throughout mechanics, by treating the Earth as a uniform sphere of mass $M$ and radius $R$ and equating the gravitational force on a surface mass with its ordinary weight.

Step 1. Write the gravitational force on a mass $m$ resting at the Earth's surface, treating the Earth as a point mass $M$ located at its centre, a distance $R$ (the Earth's radius) away. By the universal law of gravitation, $F=\dfrac{GMm}{R^2}$ .

Step 2. Write the same force in terms of ordinary mechanics. From everyday kinematics, the weight of a body of mass $m$ is $W=mg$ , where $g$ is the local acceleration due to gravity. Since the gravitational pull is the body's weight, these two expressions describe the same physical force, so $F=W$ .

Step 3. Equate the two expressions and cancel the common factor $m$ . Setting $\dfrac{GMm}{R^2}=mg$ and dividing both sides by $m$ gives $g=\dfrac{GM}{R^2}$ .

Step 4. Read the result. This single line is the bridge between the universal constant $G$ , which never changes anywhere in the universe, and the local acceleration $g$ , which depends on the specific mass and radius of whichever body you are standing on — it is why $g$ differs on the Moon, on Jupiter, and on Earth, even though $G$ is always the same number.

Worked example

Two spheres of mass $50\ \text{kg}$ and $80\ \text{kg}$ have their centres $2.0\ \text{m}$ apart. Find the gravitational force of attraction between them.

Step 1. Identify the quantities and the formula. Here $m_1=50\ \text{kg}$ , $m_2=80\ \text{kg}$ , $r=2.0\ \text{m}$ , and $G=6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ , so $F=\dfrac{Gm_1m_2}{r^2}$ .

Step 2. Substitute and simplify the numerator. $F=\dfrac{(6.67\times10^{-11})(50)(80)}{(2.0)^2}=\dfrac{6.67\times10^{-11}\times4000}{4}=\dfrac{2.668\times10^{-7}}{4}$ .

Step 3. Divide to get the final force. $F=6.67\times10^{-8}\ \text{N}$ . The gravitational attraction between the two spheres is about $6.7\times10^{-8}\ \text{N}$ — roughly the weight of a speck of dust, which is why gravity between two everyday objects is never felt; it only becomes significant when at least one mass is astronomically large.

Competitive stretch

Using $g=9.8\ \text{m/s}^2$ , Earth's radius $R=6.4\times10^6\ \text{m}$ , and $G=6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ , estimate the mass of the Earth.

Step 1. Start from $g=GM/R^2$ and solve for $M$ : $M=\dfrac{gR^2}{G}$ .

Step 2. Compute $R^2=(6.4\times10^6)^2=4.096\times10^{13}\ \text{m}^2$ , then substitute: $M=\dfrac{9.8\times4.096\times10^{13}}{6.67\times10^{-11}}=\dfrac{4.014\times10^{14}}{6.67\times10^{-11}}$ .

Step 3. Divide to get the final mass. $M\approx6.02\times10^{24}\ \text{kg}$ , matching the accepted mass of the Earth and confirming that $g=GM/R^2$ correctly links a surface measurement of $g$ to the planet's bulk properties.

Common traps

Writing $F\propto1/r$ or $F\propto1/r^3$ instead of the correct inverse-square law $F\propto1/r^2$ — always check the power on $r$ .
Confusing $g$ (acceleration due to gravity, depends on location: $9.8\ \text{m/s}^2$ near Earth's surface) with $G$ (universal constant, same value everywhere in the universe).
Forgetting that $r$ in $F=Gm_1m_2/r^2$ is the distance between the centres of the two masses, not the distance between their surfaces.
Assuming gravitational force between two ordinary lab objects is large enough to feel by hand — it is typically only $10^{-8}$ to $10^{-10}\ \text{N}$ , far too small to sense directly, which is why measuring $G$ historically needed a sensitive torsion balance.

Ask the tutor: Want to practice more $F=Gm_1m_2/r^2$ problems with different unit combinations, or see why $r$ is always measured centre-to-centre? Ask the AI tutor about Universal law of gravitation.

If gravity acts between every pair of masses, why do I never feel attracted to a nearby chair or another person?

Because the force is proportional to the product of the two masses and falls off as the square of the distance, and ordinary objects like chairs or people have masses of only tens of kilograms. Plugging typical values into $F=Gm_1m_2/r^2$ gives forces around $10^{-8}$ to $10^{-10}\ \text{N}$ — smaller than the weight of a speck of dust and completely swamped by friction and air resistance. Gravity only becomes noticeable when at least one of the masses is astronomically large, like a planet.

Is g the same thing as G, just written differently?

No, they are related but not the same. $G$ is the universal gravitational constant, $6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ , and it never changes anywhere in the universe. $g$ is the acceleration due to gravity at a particular location, given by $g=GM/R^2$ , and it depends on the mass $M$ and radius $R$ of whichever body you are standing on — that is why $g$ is about $9.8\ \text{m/s}^2$ on Earth but only about a sixth of that on the Moon, even though $G$ is identical in both cases.

Why don't we need to know how the Cavendish experiment works for this syllabus?

The syllabus asks you to state Newton's law, know the unit and dimensions of $G$ , and use $g=GM/R^2$ in problems — not to describe how $G$ was first measured. Historically, $G$ was determined using a torsion balance (the Cavendish experiment), a delicate setup needed precisely because gravitational forces between lab-sized masses are so tiny. You can simply treat $G=6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ as a given, measured constant and plug it directly into calculations.

Universal law of gravitation

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads