Free fall

Drop a coin and a feather in ordinary air, and the coin wins the race to the ground every time — but seal them in a tube with the air pumped out, and they land in the exact same instant. That single demonstration overturns the everyday intuition that heavier objects fall faster, and it is the doorway into one of Galileo's most important discoveries: freed from air resistance, gravity accelerates every object by exactly the same amount, no matter how light or heavy, wide or narrow it is. This article defines free fall precisely, shows algebraically why an object's mass drops out of its own falling acceleration, and separates that clean physics from the everyday clutter of air resistance that makes feathers flutter and stones plummet. The equations of motion that describe how far and how fast an object falls are built on this foundation and covered separately.

The core insight

Free fall is motion under gravity alone, and because both the gravitational force on an object and its resistance to acceleration scale with the same mass, that mass cancels out — so every object in free fall speeds up at the identical rate, $g \approx 9.8\ \text{m/s}^2$ .

The core idea

Free fall is the motion of an object falling under the influence of the Earth's gravitational pull alone, with no other force — no air resistance, no friction, no push or throw — acting on it.

Where the acceleration comes from. The Earth attracts any object of mass $m$ near its surface with the gravitational force

F = \dfrac{GMm}{R^2},

where: $G$ is the universal gravitational constant, $M$ is the Earth's mass, $R$ is the Earth's radius (treated as constant for objects near the surface), and $m$ is the mass of the falling object. By Newton's second law, this force produces an acceleration $a = F/m$ .

Why the mass cancels. Substituting $F$ into $a = F/m$ gives $a = GM/R^2$ — an expression with no $m$ left in it at all. The falling object's own mass has vanished from its acceleration. This happens because $F$ is directly proportional to $m$ (heavier objects are pulled harder) while the inertia that resists acceleration is also proportional to $m$ (heavier objects are harder to speed up) — the two effects are equal in size and cancel exactly when you divide one by the other.

The everyday exception is air, not gravity. A stone beats a feather to the ground in ordinary air only because air resistance pushes back much harder, relative to its weight, on the light, wide feather than on the compact stone — gravity is not treating them differently. Strip the air away, as in a vacuum tube, and only gravity remains: both fall together, landing at the same instant.

The constant has a name. This shared, mass-independent acceleration is called the acceleration due to gravity, written $g$ , with $g \approx 9.8\ \text{m/s}^2$ near the Earth's surface — its full derivation and use in the equations of motion belongs to the next article. During free fall from rest, velocity grows uniformly because $g$ stays constant, directed straight down toward the Earth's centre.

Figure

Two side by side chambers, a stone and a feather dropped together landing apart when air resistance acts and landing together when air is removed in a vacuum

Derivation AWhy mass cancels out of the acceleration of a freely falling object

The goal is to start from Newton's law of gravitation and Newton's second law of motion, both already familiar, and show algebraically — with nothing beyond substitution and cancellation — that the acceleration of a freely falling object cannot depend on that object's own mass, which is the precise reason a heavy stone and a light coin accelerate identically once air resistance is removed.

Step 1. Write the gravitational force on a falling object of mass $m$ near the Earth's surface: $F = \dfrac{GMm}{R^2}$ , where $G$ , $M$ and $R$ are all constants (the gravitational constant, the Earth's mass, and the Earth's radius) that do not depend on which object is falling.

Step 2. Write Newton's second law for that same object: $F = ma$ , so the acceleration produced is $a = F/m$ .

Step 3. Substitute the force from Step 1 into the expression for $a$ : $a = \dfrac{1}{m}\left(\dfrac{GMm}{R^2}\right) = \dfrac{GM}{R^2}$ . The $m$ in the numerator (from the gravitational force, which grows with the object's mass) exactly cancels the $m$ in the denominator (from the inertia the object offers against being accelerated, which also grows with its mass).

Step 4. Read off the result. $a = GM/R^2$ contains only Earth's properties, $G$ , $M$ and $R$ — nothing about the falling object survives. So every object, regardless of its own mass, accelerates at this same rate in free fall; this constant value is what is called $g \approx 9.8\ \text{m/s}^2$ .

Worked example

A 100 g coin and a 500 g stone are both released from rest at the same height, with air resistance removed. Using $G = 6.67\times10^{-11}\ \text{N m}^2/\text{kg}^2$ , Earth's mass $M = 5.97\times10^{24}\ \text{kg}$ and Earth's radius $R = 6.37\times10^{6}\ \text{m}$ , verify numerically that both reach the same acceleration, and state what that means for when they land.

Step 1. Find the acceleration formula that applies to both, from the derivation above: $a = GM/R^2$ . Notice it does not contain $m$ , so the same calculation applies to the coin and the stone without any change.

Step 2. Compute the numerator: $GM = (6.67\times10^{-11})(5.97\times10^{24}) = 3.98\times10^{14}\ \text{m}^3/\text{s}^2$ .

Step 3. Compute the denominator and divide: $R^2 = (6.37\times10^{6})^2 = 4.06\times10^{13}\ \text{m}^2$ , so $a = \dfrac{3.98\times10^{14}}{4.06\times10^{13}} = 9.8\ \text{m/s}^2$ .

Step 4. Check both masses give the identical number. For the 100 g coin, $F = GMm/R^2 = (9.8)(0.100) = 0.98\ \text{N}$ , and $a = F/m = 0.98/0.100 = 9.8\ \text{m/s}^2$ . For the 500 g stone, $F = (9.8)(0.500) = 4.9\ \text{N}$ , and $a = F/m = 4.9/0.500 = 9.8\ \text{m/s}^2$ . Both accelerate at exactly $9.8\ \text{m/s}^2$ , so starting from rest at the same height, they reach the ground at the same instant — neither one "wins."

Common traps

Believing heavier objects fall faster because gravity pulls harder on them. Gravity does pull harder on a heavier object (larger $F$ ), but that object also has proportionally more inertia resisting the pull, and the two effects cancel exactly, leaving the same acceleration for every mass.
Thinking "free fall" describes any object falling through air. Strictly, free fall means falling under gravity alone; a feather or parachute drifting through air where resistance is significant is not undergoing true free fall.
Assuming a vacuum has "no gravity." A vacuum only has no air, hence no air resistance — gravity still acts fully inside it, which is exactly why a vacuum is where free fall can be observed cleanly.
Mixing up mass and force when explaining the cancellation. It is the acceleration that is independent of mass, not the force: the gravitational force on a heavier object genuinely is larger, but its inertia is larger by the same factor.

Ask the tutor: Want to see the classic feather-and-hammer Moon experiment explained, or move on to how this constant acceleration $g$ is used with the equations of motion? Ask the AI tutor for more on free fall.

If mass cancels out, why does a heavy object feel like it would "obviously" fall faster?

That intuition comes from everyday experience with air, not with pure gravity. In real air, a heavy compact object (like a stone) is barely slowed by air resistance, while a light, spread-out object (like paper or a feather) is slowed a lot, so we associate "heavier" with "faster." But the cancellation in $a = GM/R^2$ is about gravity alone — remove the air, as in a vacuum, and the "obviously faster" heavy object no longer has any advantage at all.

Does crumpling a sheet of paper actually change the physics, or is it just a classroom trick?

It genuinely changes the physics, but only by changing air resistance, not gravity. A flat sheet has a large surface area pushing against air as it falls, so air resistance nearly balances gravity and it flutters down slowly. Crumpling the same sheet into a ball keeps its mass and its gravitational acceleration unchanged, but shrinks its surface area sharply, so air resistance drops and it falls noticeably faster — closer to how a stone of similar mass would fall.

Is free fall the same thing as weightlessness, like astronauts experience?

They are closely related but not identical in wording. An object in free fall — falling under gravity with nothing else acting on it — feels weightless because there is no supporting surface pushing back on it; a skydiver before the parachute opens, or an astronaut orbiting Earth, is technically in continuous free fall. The astronaut in orbit is still falling under Earth's gravity, but moving sideways fast enough that the curve of that fall matches the curve of the Earth, so they fall around the planet rather than into it.

Free fall

The core idea

Worked example

Common traps

Before this

Where this leads