Universal law of gravitation

Why does an apple fall to the ground instead of floating away, and why does the Moon stay in orbit instead of drifting off into space? Newton realised both questions share one answer: every object with mass pulls on every other object with mass. This single idea — the Universal Law of Gravitation — ties together falling apples, orbiting moons, ocean tides and the weight you feel every day under one equation. It also explains why every object near Earth's surface, heavy or light, speeds up at exactly the same rate when dropped. This article builds the equation, connects it to your familiar idea of weight and acceleration due to gravity ( $g$ ), and works through the numbers so you can apply it confidently in exams.

The core insight

Every pair of masses attracts with a force directly proportional to the product of the masses and inversely proportional to the square of the distance between their centres, $F = Gm_1m_2/r^2$ , and it is this same force that gives objects near Earth their weight and their constant free-fall acceleration $g$ .

The core idea

The statement. Newton's Universal Law of Gravitation says every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. For two bodies of masses $m_1$ and $m_2$ with centres separated by a distance $r$ , the gravitational force of attraction $F$ is:

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

where: $F$ is the force of attraction (in newtons), $m_1$ and $m_2$ are the two masses (in kg), $r$ is the distance between their centres (in m), and $G$ is the universal gravitational constant, $G = 6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ , the same for every pair of masses anywhere in the universe. You only need to use this value in calculations — you are not expected to derive or measure $G$ experimentally. The inverse-square part of the law means the force falls off sharply as bodies move apart: doubling $r$ cuts the force to a quarter, not a half. This is the same inverse-square pattern you meet with electric and magnetic forces.

Figure

Two spheres of different size pulling on each other with equal and opposite arrows, and a smaller pair of arrows shown when the spheres are moved twice as far apart to show the force weakening with the square of distance

Why it works

Gravity, weight and $g$ . When the two masses in the law are the Earth (mass $M$ , radius $R$ ) and a nearby object (mass $m$ ), the gravitational force between them is exactly the object's weight, and it produces the object's acceleration due to gravity, $g$ :

mg = \frac{GMm}{R^2} \quad \Rightarrow \quad g = \frac{GM}{R^2}

Notice the object's own mass $m$ cancels out — this is why $g$ near Earth's surface does not depend on the mass of the falling object at all; it is treated as a constant, approximately $9.8\ \text{m/s}^2$ .

Free fall is motion under gravity alone, with no other force (such as air resistance) acting. Every object in free fall near Earth accelerates downward at $g$ , whatever its mass — heavier objects are pulled harder, but they also resist acceleration more (greater inertia), and the two effects exactly cancel.

Common syllabus applications

Explains why planets and satellites stay in orbit instead of flying off in a straight line.
Explains ocean tides, caused by the gravitational pull of the Moon and Sun.
Explains why all objects near Earth's surface fall with the same acceleration $g$ , irrespective of their mass.

Derivation AGetting acceleration due to gravity g from Newton's law

The goal is to show, using nothing beyond the Universal Law of Gravitation and Newton's second law, exactly why the acceleration due to gravity $g$ near Earth's surface comes out the same for every object regardless of its mass — a fact that surprises students the first time they meet it, and one that follows directly from the mass of the falling object cancelling out of the equation.

Step 1. Write the gravitational force between the Earth (mass $M$ ) and an object of mass $m$ resting near its surface, treating the distance between centres as the Earth's radius $R$ : $F = \dfrac{GMm}{R^2}$ .

Step 2. This force $F$ is nothing but the object's weight, and by Newton's second law weight equals mass times the acceleration it produces: $F = mg$ , where $g$ is the acceleration due to gravity.

Step 3. Set the two expressions for $F$ equal to each other: $mg = \dfrac{GMm}{R^2}$ .

Step 4. Divide both sides by $m$ . The object's own mass cancels completely, leaving $g = \dfrac{GM}{R^2}$ — an expression that depends only on the Earth's mass $M$ and radius $R$ , not on the mass of the falling object at all. This is exactly why a coin and a hammer, dropped together with no air resistance, hit the ground at the same instant.

Worked example

Two spheres of mass 40 kg and 60 kg have their centres 2 m apart. Find the gravitational force of attraction between them. Take $G = 6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ .

Step 1. Write the formula: $F = \dfrac{Gm_1m_2}{r^2}$ .

Step 2. Substitute the values, $m_1 = 40\ \text{kg}$ , $m_2 = 60\ \text{kg}$ , $r = 2\ \text{m}$ :

F = \frac{6.67\times10^{-11}\times 40 \times 60}{2^2}

Step 3. Compute the numerator and denominator: numerator $= 6.67\times10^{-11}\times 2400 = 1.6008\times10^{-7}$ ; denominator $= 4$ .

Step 4. Divide: $F = \dfrac{1.6008\times10^{-7}}{4} = 4.002\times10^{-8}\ \text{N}$ . The gravitational attraction is about $4\times10^{-8}\ \text{N}$ — vanishingly small, which is why we never feel everyday objects pulling on each other; gravity only becomes noticeable when one mass, like a planet, is astronomically large.

Competitive stretch

A ball is dropped from rest and falls freely for 3 s, with $g = 9.8\ \text{m/s}^2$ . Step 1. Since $u = 0$ , use $v = u + at = 0 + 9.8\times3 = 29.4\ \text{m/s}$ . Step 2. Use $S = ut + \tfrac12at^2 = 0 + \tfrac12\times9.8\times9 = 44.1\ \text{m}$ . The ball hits the ground at 29.4 m/s, having fallen 44.1 m — the same equations of uniformly accelerated motion, applied with $a = g$ , since free fall is just uniform acceleration under gravity.

Common traps

Confusing $G$ (the universal gravitational constant, a fixed number) with $g$ (the acceleration due to gravity, which depends on the mass and radius of the attracting body).
Trying to calculate how $g$ changes with altitude or depth — this is outside the ICSE Class 9 syllabus; treat $g$ as constant near Earth's surface.
Thinking heavier objects fall faster in free fall — without air resistance, all objects near Earth accelerate at the same $g$ , because the extra gravitational pull on a heavier object is exactly cancelled by its extra inertia.
Forgetting that $r$ in $F = Gm_1m_2/r^2$ is the distance between the centres of the two bodies, not between their nearest surfaces.

Ask the tutor: Want to try the gravitation formula with a new pair of masses, or practice combining $g$ with the equations of motion for a different free-fall scenario?

Why don't I feel the gravitational pull of things around me, like a book or another person?

You do feel it, technically, but it is far too small to notice. Gravitational force depends on the product of the two masses, and everyday objects have tiny masses compared to a planet. Even two 40 kg and 60 kg spheres just 2 m apart pull on each other with only about $4\times10^{-8}\ \text{N}$ — millions of times weaker than the friction under your feet. Gravity only becomes a force you can feel when one of the masses is enormous, like the Earth.

Is g the same everywhere on Earth, and can I treat it as one fixed number?

For ICSE Class 9 purposes, yes — treat $g$ as a constant, approximately $9.8\ \text{m/s}^2$ , everywhere near Earth's surface. In reality $g$ varies slightly with altitude and depth because it depends on distance from Earth's centre, $g = GM/R^2$ , but numerical problems on that variation are explicitly outside your syllabus this year, so you only need the constant value.

How is weight related to the gravitational force between me and the Earth?

They are the same thing. Your weight is simply the gravitational force of attraction between your mass $m$ and the Earth's mass $M$ , given by $F = GMm/R^2$ . Because this equals $mg$ , your weight is often written as $mg$ for convenience — but the underlying cause is exactly the Universal Law of Gravitation applied to you and the planet you stand on.