Acceleration due to gravity at Earth's surface

Drop a coin and a car key together from the same height, in the same room, at the same instant. Air resistance aside, they hit the floor together — not because they weigh the same, but because gravity does not care what they weigh at all. Every object near the Earth's surface, heavy or light, dense or fluffy, accelerates downward at the same rate, roughly $9.8\ \text{m/s}^2$ . That single fact — universal, mass-independent free fall — is one of the oldest results in physics, reportedly tested by Galileo from the Leaning Tower of Pisa centuries before Newton explained why it must be true, and dramatically confirmed on live television when Apollo 15 astronaut David Scott dropped a feather and a hammer together on the airless Moon and watched them land at the same instant. This article builds that "why" from the ground up, using nothing more than Newton's law of gravitation and Newton's second law, already familiar from earlier in the Gravitation chapter. We will derive the compact formula $g = GM_E/R_E^2$ , see exactly which quantity cancels and why, plug in the real numbers for Earth to recover the everyday value $g \approx 9.8\ \text{m/s}^2$ , and then reuse the same one-line formula to size up gravity on the Moon and to time a falling stone — carrying forward, and extending, the Pass-1 worked numbers. Along the way we will separate two ideas that get blurred in casual speech: $G$ , the universal gravitational constant, which never changes anywhere in the universe, and $g$ , the local acceleration due to gravity, which depends on exactly which planet — or which patch of that planet — you happen to be standing on.

Key takeaways

Newton's law of gravitation gives the force on a mass $m$ at Earth's surface as $F = GM_Em/R_E^2$ ; Newton's second law says that same force equals $mg$ .
Equating the two and cancelling $m$ gives $g = GM_E/R_E^2$ — the acceleration due to gravity depends only on the planet's mass and radius, never on the falling object's own mass.
Because $m$ cancels, all objects in free fall (no air resistance) share exactly the same acceleration $g$ — the reason a feather and a hammer land together in vacuum.
Plugging in $G = 6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ , $M_E \approx 6.0\times10^{24}\ \text{kg}$ and $R_E \approx 6.4\times10^6\ \text{m}$ gives the familiar surface value $g \approx 9.8\ \text{m/s}^2$ .
$g$ is not a universal constant like $G$ — it is planet-specific (the Moon's $g$ is about one-sixth of Earth's) and even varies slightly across Earth's own surface with latitude, because Earth bulges at the equator and spins.
The cancellation of mass reflects a deep coincidence: "gravitational mass" (how strongly a body is pulled) and "inertial mass" (how strongly a body resists acceleration) are experimentally equal, though the full significance of that equality belongs to a later, more advanced treatment of gravity.
Key idea: setting the gravitational force $GM_Em/R_E^2$ equal to $ma=mg$ and cancelling the falling object's mass $m$ gives $g = GM_E/R_E^2$ , a formula fixed only by the planet, which is why every object at a given location falls with the identical acceleration.

The core insight

Equating Newton's law of gravitation with Newton's second law for a body at a planet's surface cancels the falling object's mass and leaves $g = GM/R^2$ , so acceleration due to gravity depends only on the planet, never on what is falling.

Where it fits in the syllabus

This topic sits directly inside the CBSE Class 11 Physics chapter Gravitation, immediately after the universal law of gravitation and the determination of $G$ , and it is fully within the prescribed 2025-26 board scope: $g = GM/R^2$ is one of the named formulas in the syllabus. It is the bridge between the abstract inverse-square force law you have just met and the everyday $g = 9.8\ \text{m/s}^2$ used throughout kinematics and mechanics since Class 9. Mastering this derivation also sets up the very next topic in the chapter — how $g$ changes with altitude and depth — so getting the "why $m$ cancels" reasoning solid here pays off immediately afterward.

The physical picture

Consider a small object of mass $m$ sitting at (or falling near) the Earth's surface. Treat the Earth as a uniform sphere of mass $M_E$ and radius $R_E$ , with the object essentially at distance $R_E$ from Earth's centre. Newton's universal law of gravitation gives the attractive force between the two:

F = \frac{GM_Em}{R_E^2}

where: $G$ is the universal gravitational constant, $M_E$ is Earth's mass, $m$ is the mass of the falling object, and $R_E$ is Earth's radius (measured from the centre, not from the object to the surface). Physically, this is just the same inverse-square attraction that holds between any two masses anywhere in the universe — Earth and the object are no different from any other pair of masses in Newton's law; the only special thing here is that one of the masses is enormous and close. This gravitational force is the only force acting on a freely falling object (ignoring air resistance), so by Newton's second law it must produce an acceleration $g$ in that object:

F = mg

where: $g$ is the acceleration due to gravity — exactly the quantity we want to find — and $m$ here is the same mass $m$ as before, now playing the role of "resistance to acceleration" rather than "amount of gravitational pull." That double appearance of $m$ is the crux of the whole topic and is unpacked below.

Figure

Two identical balance-scale style diagrams comparing gravitational force and Newtons second law, showing the Earth pulling a small mass with force GMm over R squared on one side, and the same mass accelerating at g under that force on the other side, with mass m appearing in both expressions and cancelling to leave g equals GM over R squared

The derivation

Derivation AWhy acceleration due to gravity does not depend on the falling object's mass

The goal here is to combine the two force expressions above — one from the universal law of gravitation, one from Newton's second law — into a single formula for $g$ , and to see precisely why the falling object's own mass drops out of the answer entirely, leaving an acceleration that depends only on the planet doing the pulling.

Step 1. Write the gravitational force on the object using Newton's universal law of gravitation. For a mass $m$ at the Earth's surface, distance $R_E$ from Earth's centre, the force is $F = \dfrac{GM_Em}{R_E^2}$ . This is the pull, and only the pull — it says nothing yet about how the object responds to that pull.

Step 2. Write the same force using Newton's second law. Because gravity is the only force acting on a freely falling body, it must equal mass times acceleration: $F = ma$ . Since the acceleration produced by gravity near Earth's surface is by definition called $g$ , this reads $F = mg$ .

Step 3. Equate the two expressions for $F$ , since both describe the identical force on the identical object at the identical instant: $\dfrac{GM_Em}{R_E^2} = mg$ .

Step 4. Cancel the mass $m$ from both sides — it appears once as "the property gravity acts on" (left side) and once as "the property that resists acceleration" (right side), and these two roles happen to be numerically equal for any object, so the $m$ 's cancel exactly, regardless of what $m$ actually is. This leaves $g = \dfrac{GM_E}{R_E^2}$ , a quantity that involves only the planet's mass $M_E$ , the planet's radius $R_E$ , and the universal constant $G$ — never the falling object at all. That is the entire reason a feather and a hammer, or a coin and a car key, share the same free-fall acceleration. Verified: with $G=6.67\times10^{-11}$ , $M_E=6.0\times10^{24}\ \text{kg}$ , $R_E=6.4\times10^6\ \text{m}$ , $g = GM_E/R_E^2 = 9.77\ \text{m/s}^2 \approx 9.8\ \text{m/s}^2$ , matching the accepted surface value.

This cancellation is worth pausing on, because it is not a mathematical accident so much as a deep physical coincidence. Mass shows up in physics in two conceptually different jobs. In Newton's law of gravitation, mass measures how strongly a body participates in gravitational attraction — call this gravitational mass. In Newton's second law, mass measures how strongly a body resists being accelerated — call this inertial mass. Every experiment ever performed finds these two masses to be equal to extraordinary precision, and it is exactly this equality that lets the $m$ in $F=GMm/R^2$ cancel against the $m$ in $F=ma$ . If gravitational mass and inertial mass were even slightly different for different materials, a lead ball and a wooden ball would fall at measurably different rates. They do not — and this equivalence later became one of the founding ideas of Einstein's general theory of relativity, though building that connection is well beyond the Class 11 syllabus and is mentioned here only as a signpost for later. With the formula $g = GM_E/R_E^2$ in hand, substituting Earth's accepted values — $G = 6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ , $M_E \approx 6.0\times10^{24}\ \text{kg}$ , and $R_E \approx 6.4\times10^6\ \text{m}$ — gives the number every student already knows from experience:

g \approx 9.8\ \text{m/s}^2

It is worth being honest that this is a slightly idealised number. Earth is not a perfect uniform sphere: it bulges very slightly at the equator and is flattened at the poles (an effect of its own rotation), and its density is not perfectly uniform inside. Because $g$ depends on $R_E$ through an inverse square, and the equatorial radius is very slightly larger than the polar radius, $g$ is a touch smaller at the equator ( $\approx 9.78\ \text{m/s}^2$ ) than at the poles ( $\approx 9.83\ \text{m/s}^2$ ). Earth's rotation also subtracts a small extra centrifugal effect at the equator that further lowers the equator's effective $g$ . For essentially every Class 11 problem, however, $g = 9.8\ \text{m/s}^2$ is the value to use unless a question states otherwise — the latitude variation is a qualitative fact to know, not a number to compute here.

Worked examples

Board level - acceleration due to gravity on the Moon

Estimate the acceleration due to gravity on the surface of the Moon, given its mass $M_m = 7.4\times10^{22}\ \text{kg}$ and radius $R_m = 1.74\times10^6\ \text{m}$ . Take $G = 6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ .

Step 1. Write down the same formula derived above, but for the Moon instead of Earth — the derivation did not use any property special to Earth, so it applies to any planet or moon: $g_m = \dfrac{GM_m}{R_m^2}$ .

Step 2. Compute the numerator: $G M_m = (6.67\times10^{-11})(7.4\times10^{22}) = 4.936\times10^{12}$ .

Step 3. Compute the denominator: $R_m^2 = (1.74\times10^6)^2 = 3.028\times10^{12}$ .

Step 4. Divide: $g_m = \dfrac{4.936\times10^{12}}{3.028\times10^{12}} \approx 1.63\ \text{m/s}^2$ . Answer: $g_m \approx 1.63\ \text{m/s}^2$ , about one-sixth of Earth's $9.8\ \text{m/s}^2$ — consistent with the famously large, slow-motion leaps of astronauts on the Moon, and with why Commander Scott's feather and hammer fell together but noticeably more gently than they would have on Earth.

Competitive level - free fall timing and the mass-independence check

A stone is dropped from rest from the top of a tower $78.4\ \text{m}$ tall on Earth, where $g = 9.8\ \text{m/s}^2$ . Find (a) the time it takes to reach the ground, (b) its speed just before impact, and (c) briefly justify why the answers would be identical for a stone twice as heavy.

Step 1. Use $h = \tfrac12 g t^2$ for a body starting from rest, and solve for $t^2$ : $78.4 = \tfrac12(9.8)t^2 \implies t^2 = \dfrac{78.4\times2}{9.8} = \dfrac{156.8}{9.8} = 16$ .

Step 2. Take the square root: $t = 4\ \text{s}$ .

Step 3. Find the impact speed using $v = gt$ (starting from rest): $v = 9.8\times4 = 39.2\ \text{m/s}$ .

Step 4. Justify mass-independence: nowhere in Steps 1–3 did the stone's mass $m$ appear — $h=\tfrac12gt^2$ and $v=gt$ are both built entirely from $g$ , which the earlier derivation showed depends only on Earth's $M_E$ and $R_E$ . A stone of twice the mass would experience twice the gravitational force, but Newton's second law says it would also need twice the force to produce the same acceleration, so the two effects cancel exactly and $t=4\ \text{s}$ , $v=39.2\ \text{m/s}$ regardless. Answer: $t = 4\ \text{s}$ , $v = 39.2\ \text{m/s}$ , and both values are unchanged for any mass. Timing note: this pattern (drop height given, find time and speed, then justify mass-independence) is a common short-answer combination; the arithmetic itself should take under a minute once $h=\tfrac12gt^2$ is set up — the trap is spending time "computing" a mass-dependence check instead of simply pointing to the cancellation already proved.

Common traps

Students frequently lose marks in this exact topic for a small, repeatable set of reasons. First, believing — often from casual intuition about heavy versus light objects — that heavier bodies fall faster; in the absence of air resistance $g$ is completely independent of the falling object's mass, precisely because $m$ cancels in the derivation, and any answer suggesting otherwise (for identical drop conditions) should be treated as a red flag. Second, confusing $g$ with $G$ : $G$ is a genuine universal constant, the same number everywhere in the universe, while $g$ is not a constant at all in that sense — it is a derived quantity, $GM/R^2$ , that changes from planet to planet (and even slightly across one planet's own surface). Writing " $g$ is a universal constant" on an exam is a guaranteed mark loss. Third, using the wrong radius: $R_E$ in $g=GM_E/R_E^2$ must be measured from the Earth's centre, not from the object down to the ground — a subtlety that matters once altitude or depth enters later problems, but is easy to get wrong even here if a student subtracts instead of using the full radius. Fourth, attributing a feather's slow fall in ordinary air to "weaker gravity" rather than to air resistance — gravity itself pulls the feather at the same $g$ as everything else; it is only the drag force from air that visibly slows a feather in atmosphere, which is exactly why the Apollo 15 feather-and-hammer demonstration needed the airless Moon to make its point cleanly.

Common trapg is not a universal constant — do not confuse it with G

The single biggest misconception in this topic is treating $g \approx 9.8\ \text{m/s}^2$ as if it were as fundamental and unchanging as $G = 6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ . It is not. $G$ is the same number for any two masses anywhere in the cosmos — a genuine constant of nature. $g$ , by contrast, is computed from $G$ , a specific planet's mass, and that planet's radius: $g = GM/R^2$ . Change the planet (Earth to Moon, a factor of about six drop) or even change your latitude on the same planet (equator to pole, a much smaller shift), and $g$ changes while $G$ never does. On an exam, always be ready to state that $g$ is location-dependent and $G$ is not — this distinction is tested directly and often.

You should be able to…

Derive $g = GM/R^2$ starting from Newton's universal law of gravitation and Newton's second law, showing clearly where and why mass cancels.
Explain, in words, why $g$ is independent of the mass of the object experiencing it, and connect that cancellation to the equivalence of gravitational and inertial mass.
Calculate $g$ for any planet or moon given its mass and radius, and compare it meaningfully to Earth's $9.8\ \text{m/s}^2$ .
State clearly the distinction between $G$ (a universal constant) and $g$ (a planet- and location-dependent quantity computed from $G$ , $M$ , and $R$ ).
Use $g$ correctly inside the standard free-fall kinematics equations ( $v=u+gt$ , $h=ut+\tfrac12gt^2$ ) for board-level and competitive numericals.
State, qualitatively, why $g$ varies slightly with latitude on Earth, without needing to compute the variation.

Ask the tutor: Curious how this same $g = GM/R^2$ formula changes once you go up a mountain or down a mine shaft — ask the AI tutor to walk through the variation of g with altitude and depth, the very next topic in this chapter.

If g cancels out the falling object's mass, why do heavier things feel like they hit the ground harder?

They do hit harder, but that is about force and energy, not acceleration. A heavier stone still falls at the same $9.8\ \text{m/s}^2$ as a lighter one and reaches the ground at the same speed from the same height, but the force needed to stop it on impact is bigger (from $F=ma$ with a larger $m$ ), and it delivers more kinetic energy ( $\tfrac12mv^2$ ) into whatever it lands on. So "falls at the same rate" and "hits with more impact" are both true at once — they are answers to different questions.

Why does the Moon have a smaller g than Earth if the Moon's density is actually a bit less than Earth's, not six times less?

Because $g = GM/R^2$ depends on both mass and radius, not density alone. The Moon has roughly $1/81$ of Earth's mass but only about $1/4$ of Earth's radius, so its $R^2$ shrinks much more than its $M$ does. Working through $g_m = GM_m/R_m^2$ with the real numbers gives about $1.63\ \text{m/s}^2$ , roughly one-sixth of Earth's $9.8\ \text{m/s}^2$ — the small radius partly compensates for the small mass, but not enough to bring $g$ back up to Earth's value.

Is g really exactly 9.8 everywhere on Earth, or should I worry about which value to use in a problem?

No, it is not exactly the same everywhere — it is about $9.78\ \text{m/s}^2$ at the equator and about $9.83\ \text{m/s}^2$ at the poles, mainly because Earth bulges slightly at the equator (making $R_E$ a little larger there) and because Earth's rotation subtracts a small effective amount from the pull felt at the equator. For essentially every Class 11 numerical, though, use whatever value the question states, and if none is given, use the standard $g = 9.8\ \text{m/s}^2$ — do not try to correct for latitude unless a problem explicitly asks you to.

Does this formula work for any planet, or only for Earth?

It works for any spherical planet or moon — nothing in the derivation used a property special to Earth. The formula $g = GM/R^2$ needs only that planet's own mass $M$ and radius $R$ in place of $M_E$ and $R_E$ . That is exactly what the Moon worked example above does: swap in the Moon's mass and radius, and the same two-line derivation gives the Moon's surface gravity instead of Earth's.