Gravitational constant

Drop a pencil and it falls. Watch the Moon and it does not fall into us, yet something is unmistakably pulling it our way, bending its path into a circle instead of letting it fly off in a straight line. Newton's genius was to say these are the same pull — one law, $F = Gm_1m_2/r^2$ , governs the pencil, the Moon, and every pair of masses in the universe. But a law with a blank in it is not yet a number you can use. That blank is $G$ , the gravitational constant, and until someone measured it, nobody could actually calculate how hard two masses pull on each other, nor how much the Earth itself weighs. $G$ turns out to be almost embarrassingly small: $6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ . Two 1 kg masses held 1 metre apart tug on each other with a force of about $6.67\times10^{-11}$ newton — roughly the weight of a single bacterium. That is why you never feel the gravitational pull of your chair, your laptop, or the person sitting next to you: gravity is by a huge margin the weakest of nature's four fundamental forces, and it takes planet-sized masses before its effect becomes obvious. Yet this feeble constant, once pinned down, unlocked the mass of the Earth, then the Sun, then every other planet and star whose orbit we can watch. The man who caught this elusive number was Henry Cavendish in 1798, using nothing but two small lead spheres, two large lead spheres, and a fibre thin enough to twist under a force smaller than a housefly's footstep. This article follows that story: what $G$ actually is, why it is so punishingly hard to measure precisely even today, how its dimensional formula is built from first principles, and how one small experimental number let scientists finally weigh the world.

Key takeaways

$G$ is the universal constant of proportionality in Newton's law $F = Gm_1m_2/r^2$ , with accepted value $G = 6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ .
$G$ never changes — same value for any two masses, anywhere in the universe, regardless of what lies between them. This is unlike $g$ , the acceleration due to gravity, which depends on the mass and radius of whichever body you are standing on.
Its dimensional formula is $[\text{M}^{-1}\text{L}^3\text{T}^{-2}]$ , derivable equally from $F = Gm_1m_2/r^2$ or from $g = GM/R^2$ — both routes must agree, and checking that agreement is a classic exam question.
Henry Cavendish measured $G$ in 1798 with a torsion balance: two small lead spheres on a suspended rod, attracted by two large fixed lead spheres, twisting a fibre by a tiny, measurable angle.
Once $G$ was known, the Earth could finally be weighed. Combining the measured $G$ with the already-known surface value $g = 9.8\ \text{m/s}^2$ and Earth's radius $R_E$ in $g = GM_E/R_E^2$ gives $M_E \approx 6.0\times10^{24}\ \text{kg}$ — the calculation Cavendish's contemporaries called "weighing the world."
$G$ is known far less precisely than most fundamental constants — only to about 1 part in $10^4$ — because gravity is so weak that stray forces (air currents, nearby vibrations, electrostatic effects) easily swamp the tiny signal in any laboratory measurement.
Every other formula in this chapter quietly leans on $G$ — orbital velocity, escape speed, gravitational potential energy, and Kepler's third law all carry it, so a firm grip on $G$ underlies the whole topic.
Key idea: $G$ is the fixed, universal, but stubbornly small proportionality constant in Newton's law of gravitation, first measured by Cavendish's torsion balance, and it is precisely this number that converts an abstract force law into a tool for weighing planets and stars.

The core insight

The gravitational constant $G = 6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ is a fixed, universal number — the same for any two masses anywhere — that fixes the strength of gravity in $F = Gm_1m_2/r^2$ , and once Cavendish measured it experimentally in 1798, it became possible for the first time to calculate the mass of the Earth from $g = GM_E/R_E^2$ .

Where it fits in the syllabus

This topic sits inside the CBSE Class 11 Physics chapter on Gravitation, immediately after the universal law of gravitation is introduced, and it is squarely within the prescribed 2025-26 board scope. You are expected to know the numerical value, SI unit, and dimensional formula of $G$ ; to explain clearly why $G$ is a universal constant while $g$ is not; and to describe, at a conceptual level, how Cavendish's torsion-balance experiment measured $G$ and thereby allowed the mass of the Earth to be calculated for the first time. The syllabus does not expect you to derive the torsion-balance formula itself, model the fibre's restoring torque mathematically, or discuss modern high-precision determinations of $G$ — those sit beyond the Class 11 ceiling and are not board-examinable.

The physical picture

Newton's law of gravitation states that any two point masses attract each other with a force directed along the line joining them, proportional to the product of the masses and inversely proportional to the square of the distance between them:

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 gravitational constant that makes the equation dimensionally and numerically correct. Read physically, $G$ is simply the force, in newtons, between two 1 kg masses placed exactly 1 metre apart — and that force is a minuscule $6.67\times10^{-11}\ \text{N}$ . Every other pair of masses at every other distance just scales up or down from that one anchor number. The accepted value of $G$ , quoted to the precision the CBSE syllabus expects, is:

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

where: the unit $\text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ falls straight out of rearranging $F = Gm_1m_2/r^2$ for $G$ — force times distance-squared divided by mass-squared — and it is a common exam slip to misremember this as merely $\text{N/kg}^2$ , dropping the area factor entirely.

Figure

Two spheres of unequal size separated by a distance r, connected by a double headed arrow labeled force F equals G m one m two over r squared, with a small inset panel showing two one kilogram spheres one metre apart producing a force of about 6.67 times ten to the minus 11 newtons

The same physical constant $G$ reappears, unchanged, in every other formula this chapter builds: acceleration due to gravity $g = GM/R^2$ , gravitational potential energy $U = -GMm/r$ , gravitational potential $V = -GM/r$ , orbital velocity $v_o = \sqrt{GM/r}$ , and escape speed $v_e = \sqrt{2GM/R}$ . None of those relations can be turned into an actual number without first knowing $G$ — which is exactly why Cavendish's 1798 measurement was treated as such a landmark result, not a footnote.

Derivation AFinding the dimensional formula of G and confirming it two independent ways

The goal here is to extract the dimensional formula of $G$ purely from the structure of Newton's law of gravitation, and then to confirm that the very same formula falls out of the completely different-looking relation $g = GM/R^2$ — because both equations describe the same physical constant, they had better agree, and checking that agreement is exactly the kind of consistency argument examiners like to test. No new physics is introduced; this is pure algebra applied to dimensions.

Step 1. Start from $F = Gm_1m_2/r^2$ and isolate $G$ by simple rearrangement: $G = \dfrac{Fr^2}{m_1 m_2}$ . This says $G$ has whatever dimensions are needed to convert a force and a squared distance into something divided by mass-squared.

Step 2. Substitute the base dimensions of each quantity. Force has dimensions $[\text{MLT}^{-2}]$ (mass times acceleration), distance squared has dimensions $[\text{L}^2]$ , and each mass has dimension $[\text{M}]$ , so $m_1 m_2$ contributes $[\text{M}^2]$ . Putting these into the rearranged formula:

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

Step 3. Now repeat the exercise starting from a completely different defining relation, $g = GM/R^2$ , which links $G$ to the acceleration due to gravity rather than to force directly. Rearranging for $G$ gives $G = gR^2/M$ . Here $g$ carries the dimensions of acceleration, $[\text{LT}^{-2}]$ , $R^2$ carries $[\text{L}^2]$ , and $M$ carries $[\text{M}]$ , so:

[G] = \frac{[\text{LT}^{-2}][\text{L}^2]}{[\text{M}]} = [\text{M}^{-1}\text{L}^3\text{T}^{-2}]

Step 4. Compare the two results. Both routes — one starting from the force between two arbitrary masses, the other starting from the acceleration a large mass produces at its surface — land on the identical dimensional formula $[\text{M}^{-1}\text{L}^3\text{T}^{-2}]$ . This is not a coincidence; it is a required consistency check, because $g = GM/R^2$ is itself just $F = GMm/R^2$ divided through by the test mass $m$ , so the two equations were never really independent. Reporting mismatched dimensions from the two routes is a sure sign of an algebra slip, not a sign that $G$ has two different natures.

Worked examples

Board level - dimensional-formula consistency check

State the SI unit of $G$ and verify, by dimensional analysis, that the value of $G$ obtained from $F = Gm_1m_2/r^2$ is dimensionally consistent with the value obtained from $g = GM/R^2$ .

Step 1. Write the SI unit of $G$ directly from its defining equation: rearranging $F = Gm_1m_2/r^2$ gives $G = Fr^2/(m_1m_2)$ , so the unit is newton times metre-squared divided by kilogram-squared, written $\text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ .

Step 2. Find the dimensional formula from the force law. Substituting $[F] = [\text{MLT}^{-2}]$ , $[r^2] = [\text{L}^2]$ , and $[m_1m_2] = [\text{M}^2]$ gives $[G] = [\text{M}^{-1}\text{L}^3\text{T}^{-2}]$ .

Step 3. Find the dimensional formula from $g = GM/R^2$ , so $G = gR^2/M$ . Substituting $[g] = [\text{LT}^{-2}]$ , $[R^2] = [\text{L}^2]$ , and $[M] = [\text{M}]$ gives $[G] = [\text{M}^{-1}\text{L}^3\text{T}^{-2}]$ again.

Step 4. Compare and conclude. Both routes give exactly $[\text{M}^{-1}\text{L}^3\text{T}^{-2}]$ , confirming that $G$ is dimensionally consistent regardless of which defining relation is used. The dimensional formula of $G$ is $\mathbf{[M^{-1}L^3T^{-2}]}$ , and its SI unit is $\mathbf{\text{N}\,\cdot\,\text{m}^2/\text{kg}^2}$ .

Competitive level - weighing the Earth, then finding its average density

Using $G = 6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ , $g = 9.8\ \text{m/s}^2$ at Earth's surface, and $R_E = 6.4\times10^6\ \text{m}$ : (a) estimate the mass of the Earth, and (b) use that mass to estimate Earth's average density, treating it as a uniform sphere of volume $\tfrac{4}{3}\pi R_E^3$ .

Step 1. Rearrange $g = GM_E/R_E^2$ for the mass: $M_E = gR_E^2/G$ . First compute $R_E^2 = (6.4\times10^6)^2 = 4.096\times10^{13}\ \text{m}^2$ .

Step 2. Substitute: $M_E = \dfrac{9.8 \times 4.096\times10^{13}}{6.67\times10^{-11}}$ . The numerator is $9.8 \times 4.096\times10^{13} = 4.014\times10^{14}$ , so $M_E = \dfrac{4.014\times10^{14}}{6.67\times10^{-11}} \approx 6.02\times10^{24}\ \text{kg}$ — matching the accepted mass of the Earth, $\approx 6.0\times10^{24}\ \text{kg}$ , closely.

Step 3. Now find Earth's volume as a uniform sphere: $V = \tfrac{4}{3}\pi R_E^3$ . With $R_E^3 = (6.4\times10^6)^3 \approx 2.621\times10^{20}\ \text{m}^3$ , the volume is $V \approx \tfrac{4}{3}\times 3.1416 \times 2.621\times10^{20} \approx 1.098\times10^{21}\ \text{m}^3$ .

Step 4. Divide mass by volume to get average density: $\rho = M_E/V \approx \dfrac{6.02\times10^{24}}{1.098\times10^{21}} \approx 5480\ \text{kg/m}^3$ , or about $5.5\ \text{g/cm}^3$ . The Earth's mass is $\mathbf{\approx 6.02\times10^{24}\ \text{kg}}$ and its average density is $\mathbf{\approx 5.5\ \text{g/cm}^3}$ — roughly five times the density of ordinary surface rock (about $2.7\ \text{g/cm}^3$ ), which is itself the classic clue that Earth's deep interior must be far denser than its crust, made of iron and nickel rather than silicate rock. Timing note: in a competitive exam this is a two-mark arithmetic chain once the two formulas are recalled — budget about 90 seconds for part (a) and another 60 seconds for part (b); the density step is just one more division once $M_E$ is already found, so do not re-derive anything from scratch.

Common traps

Writing the unit of $G$ as $\text{N/kg}^2$ instead of the correct $\text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ — the missing area factor is a very common one-mark loss.
Confusing $G$ with $g$ on sight because the symbols and even the sound of the words are so close — remember $G$ is universal and constant everywhere, while $g$ depends on the mass and radius of whichever planet or altitude you are on.
Believing Newton himself measured $G$ — he proposed the law of gravitation and showed how $G$ must enter it, but the actual numerical value was only pinned down by Cavendish more than a century later, in 1798.
Assuming $G$ changes if the two masses are separated by water, glass, or any other medium instead of vacuum — gravity, unlike electric or magnetic forces, is completely unaffected by whatever lies between the masses.
Carrying too many significant figures in a final numerical answer involving $G$ — since $G$ itself is known only to about four significant figures, answers should be rounded sensibly, not reported to six or seven digits of false precision.

Common trapDo not confuse G, the universal gravitational constant, with g, the acceleration due to gravity

$G$ is a single fixed number, $6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ , that never changes no matter where in the universe you measure it. $g$ is completely different in kind: it is an acceleration, it depends on the mass and radius of whichever body you happen to be standing on, and it even varies with altitude and depth on the very same planet. The relation between them is $g = GM/R^2$ — $g$ is what you get once you plug a specific planet's mass and radius into the universal $G$ . Every year, students lose marks writing " $G = 9.8$ " or reasoning as though $G$ itself is different on the Moon; it is $g$ that is different on the Moon, precisely because $G$ stays the same.

You should be able to…

State the numerical value, SI unit, and dimensional formula of the gravitational constant $G$ .
Derive the dimensional formula of $G$ from either $F = Gm_1m_2/r^2$ or $g = GM/R^2$ , and show the two derivations agree.
Explain, in your own words, why $G$ is called universal, and clearly distinguish it from $g$ .
Describe, in outline, how Cavendish's torsion-balance experiment measured $G$ and why the result mattered.
Use $G$ , together with a measured $g$ and $R$ , to estimate the mass of a planet, and go one step further to estimate its average density.
Explain, qualitatively, why gravity between everyday objects is never felt even though the gravitational force law applies to them too.

Ask the tutor: Want a step-by-step walkthrough of exactly how Cavendish's torsion balance turned a tiny fibre-twist into a number, or a few more practice problems using G to weigh other planets?

If G is so important, why is its value only known to four significant figures when other constants like the speed of light are known to many more digits?

Because gravity is by far the weakest of the four fundamental forces, and $G$ itself is a very small number. Any laboratory attempt to measure $G$ has to detect an extremely feeble attractive force between two modest masses, and that tiny signal is easily swamped by stray effects — air currents, tiny vibrations from traffic or footsteps outside the lab, even faint electrostatic charges on the equipment. Constants like the speed of light $c$ can be measured with extraordinary precision because light and electromagnetic effects are comparatively easy to isolate and measure sharply. $G$ has resisted that kind of precision for over two centuries, and even modern torsion-balance-style experiments only manage about 1 part in $10^4$ , which is why textbooks still round it to $6.67\times10^{-11}$ .

Is G exactly the same on the Moon, inside a star, or anywhere else in the universe, or does it get weaker or stronger somewhere?

As far as every experiment and observation to date can tell, yes — $G$ is exactly the same everywhere. That is precisely what makes it "universal": the same $6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ that Cavendish measured in his London laboratory also governs the pull between the Earth and Moon, between binary stars in a distant galaxy, and between two protons momentarily near each other. What changes from place to place is $g$ , the acceleration due to gravity, because $g = GM/R^2$ depends on the mass $M$ and radius $R$ of whichever specific body you are near — the Moon has a smaller mass and radius than Earth, so its $g$ is about one-sixth of Earth's, even though the underlying $G$ is identical in both places.

How exactly did knowing G let scientists calculate the mass of the Earth — surely you cannot just put the Earth on a scale?

Exactly right, you cannot weigh a planet on a scale, but you do not need to. Long before Cavendish, scientists already knew $g$ , the acceleration due to gravity at Earth's surface, very precisely from simple pendulum and falling-body experiments, and they knew Earth's radius $R_E$ reasonably well from geometric surveys. The one missing piece in $g = GM_E/R_E^2$ was $G$ itself — everything else in that equation was already measured. The moment Cavendish supplied a laboratory value for $G$ , the equation could be rearranged to $M_E = gR_E^2/G$ and solved directly, giving $M_E \approx 6\times10^{24}\ \text{kg}$ . This is why Cavendish's result was celebrated as "weighing the world" even though his actual apparatus only ever touched two small lead spheres.

Does the same trick work for finding the mass of the Sun or other planets, using the same G?

Yes, and this is really the payoff of $G$ being universal. For the Sun, instead of using $g = GM/R^2$ at a surface, scientists use the orbital motion of a planet: combining Newton's law of gravitation with the circular-orbit condition gives Kepler's third law in the form $T^2 = \dfrac{4\pi^2}{GM_\odot}r^3$ , where $T$ is a planet's orbital period and $r$ its orbital radius. Since $T$ and $r$ are both directly observable by telescope, and $G$ is already known from Cavendish's experiment, this equation can be solved for $M_\odot$ , the Sun's mass. The exact same logic, applied to any moon orbiting any planet, gives that planet's mass too — which is how astronomers have weighed Jupiter, Saturn, and even planets around other stars, without a scale in sight.