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The Cavendish experiment makes the attraction predicted by Newton's law of universal gravitation measurable between laboratory masses. Two small test masses hang at the ends of a horizontal rod in a Cavendish torsion balance; nearby larger source masses exert a tiny torque. A thin suspension fiber supplies an opposing torque as it twists. Measuring its twist and calibrating its torsion constant lets an experimenter infer the gravitational force and, in a modern analysis, the gravitational constant. ^[ref]

The beautiful feature is that the two horizontal forces can cancel as vectors while their torques add. The instrument converts weak attraction into a rotation, with sensitivity set by the fiber's torsion constant. Earth's much larger downward gravitational weight is supported by the suspension rather than supplying the intended torque about its vertical axis. A laboratory measurement can therefore probe the same Newtonian gravitational field that governs the two-body problem.

Henry Cavendish's original paper was Experiments to determine the Density of the Earth, published in Philosophical Transactions, volume 88 (1798), pp. 469–526, and read on 21 June 1798. On p. 469 he credits John Michell with the apparatus idea and explains that Michell died before using it; the instrument reached Cavendish through Francis John Hyde Wollaston. Cavendish's stated aim was Earth's mean mass density. Calling the experiment a determination of the gravitational constant is a modern interpretation, not Cavendish's own formulation of his result. ^[ref]

The connection between Earth's mean mass density and the gravitational constant follows in a spherical, nonrotating model of Earth. With Earth mass $M_E$, radius $R_E$, and surface gravitational acceleration $g$, the spherical shell theorem and Newton's law of universal gravitation giveThus a local comparison of gravitational attractions, together with knowledge of Earth's size and surface gravitational acceleration, reaches a property of the entire planet. These are modern equations explaining the connection, not equations quoted from Cavendish's paper.

For a worked example, use an explicitly simplified two-near-pair model. Each test point mass is $m$, each source point mass is $M$, and each lever arm is $L$, so the rod's full length is $2L$. Neglect the rod's mass. At zero twist put the test masses at $(-L,0)$ and $(L,0)$, and the sources at $(-L,-r)$ and $(L,r)$. Here $r$ is the near-pair center separation. Ignore each source's attraction on the test mass at the opposite end. Evaluate the two near-pair forces at zero twist and treat their torque as constant for a small twist, requiring $L|\theta |\ll r$. This is a teaching model, not a precision model of Cavendish's actual apparatus.

By Newton's law of universal gravitation, each retained force has magnitude $F=GmM/r^2$. The left force points down and the right force points up. Using torque about the suspension axis, both contributions are counterclockwise when viewed from above:The net retained horizontal force is zero, yet the net torque is positive. Moving the sources to $(-L,r)$ and $(L,-r)$ reverses both forces and hence the torque. In a real apparatus, the opposite source also attracts each test mass; that correction is explicitly included in the university derivation. ^[ref]

For illustrative parameters, take $m=0.010\mathrm{kg}$, $M=1.0\mathrm{kg}$, $r=0.050\mathrm{m}$, $L=0.10\mathrm{m}$, and torsion constant $\kappa =1.0\times 10^{-8}\mathrm{N}m/rad$. Use $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, the central value in NIST's 2022 CODATA table; that table gives standard uncertainty $0.00015\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$. This is an input for the example, not a measurement made with these assumed parameters. ^[ref]This sub-nanonewton force produces a measurable twist in the Cavendish torsion balance. The displayed digits make the toy calculation reproducible; they do not describe experimental precision.

A uniform Newtonian gravitational field inside one spherical void can be cancelled at a point by the repulsive field representing another removed sphere. The latter field has a Jacobian matrix proportional to $I-3\hat{r}{\hat{r}}^T$: its axial eigenvalue is negative and both transverse eigenvalues positive. The equilibrium is restoring along the axis but unstable transversely. A zero force is thus not enough for stable rest.

A circular orbit maintains constant distance from its centre of attraction. Its velocity changes direction even when its speed stays constant. The inward centripetal acceleration has magnitude $v^2/r$, so the radial force must satisfy $F_r=-m{v}^2/r$. For a Newtonian gravitational field with standard gravitational parameter $\mu$, this gives $v=\sqrt{\mu /r}$.

For a fixed spherical gravitating source, place a test particle at $(a,0)$ with initial velocity $(0,\sqrt{\mu /a})$, where $a$ is outside the source. Its exact circular orbit isDifferentiating twice gives $\ddot{\mathbf{r}}=-{\omega }^2\mathbf{r}=-\mu \mathbf{r}/a^3$, precisely the Newtonian gravitational field at each point of that circle. There is no outward physical force to cancel gravity in this inertial frame. Gravity supplies the inward centripetal acceleration that continually changes the velocity's direction.

Over a short interval, the displacement from the straight tangent path begins inward with magnitude $\frac{1}{2}(\mu /a^2)t^2$. The curvature of the orbit matters at the same order: the particle moves sideways while continually falling away from its instantaneous tangent. The full circular solution remains at radius $a$ rather than striking the source.

Its speed is $\sqrt{\mu /a}$, its oscillation period is $2\pi \sqrt{a^3/\mu }$, and its specific orbital energy is $-\mu /(2a)$. Removing gravity while keeping the initial velocity would give the straight path $(a,a\omega t)$, rather than a stationary particle. This contrast explains why orbital weightlessness is compatible with a substantial gravitational force: a freely coasting spacecraft and its contents fall together. ^[ref]

Suppose the shell $2^{-n}<r<2^{1-n}$ has mass density $2^{n-1}{\rho }_0$. Its mass is $(14\pi {\rho }_0/3)4^{-n}$, by integrating $4\pi r^2\rho dr$. Summing the geometric series gives total mass $14\pi {\rho }_0/9$. At $r=2^{-N}$ the enclosed shells have indices $n\ge N+1$, so the enclosed mass is the total mass times $r^2$. The magnitude of the Newtonian gravitational field at every such boundary is therefore the same, $14\pi G{\rho }_0/9$. This does not assert that the field is constant throughout each shell.

Stipulate a fixed source point mass $M=10^{12}\mathrm{kg}$, two equal test particles of mass $m=1\mathrm{kg}$, and $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$. This is a chosen model, not a measurement of a particular object. Place the test particles on the outward radial axis at $R-L/2$ and $R+L/2$, with midpoint distance $R=100\mathrm{m}$ and separation $L=2\mathrm{m}$. Neglect their mutual attraction, other sources and all supporting forces.

Each signed radial gravitational acceleration is inward:Their separation has outward relative accelerationThe exact value is $2.6702540\times 10^{-4}\mathrm{m}{\mathrm{s}}^{-2}$. The midpoint tidal tensor predicts $2\mu L/R^3=2.6697200\times 10^{-4}\mathrm{m}{\mathrm{s}}^{-2}$, an underestimate of $0.019999\%$ relative to the exact value. For these equal $1\mathrm{kg}$ masses, the difference of external forces has the same numerical value in the SI unit of force, the newton. It is not an additional mutual force between the test particles.

The exact formula requires $L<2R$ and describes the instantaneous relative acceleration. Because the two sample positions are symmetric about the midpoint, the Taylor expansion error relative to the linear result starts at order $(L/R{)}^2$; a one-sided comparison with a reference point generally has relative error of order $L/R$. If either test particle enters an extended source, use that source's interior Newtonian gravitational field instead of extrapolating the singular point mass model.

Free fall is motion under gravity alone: there is no supporting normal force, engine thrust or appreciable drag force. In the Newtonian mechanics model, a test particle of mass $m$ obeys $m\ddot{\mathbf{r}}=m\mathbf{g}(\mathbf{r})$, hence $\ddot{\mathbf{r}}=\mathbf{g}(\mathbf{r})$. The mass cancels. Bodies released at the same position with the same velocity follow the same path in this model, even if their gravitational weights differ. ^[ref]

A dropped stone is one example; a coasting satellite is another. Free fall does not require motion straight down or decreasing distance to the attracting body. An ideal circular orbit has constant radius while its velocity continually turns under centripetal acceleration. The circular orbit as continuous free fall connects the familiar drop with orbital motion.

Near a surface, treating the gravitational acceleration as constant gives a useful first prediction. Over larger heights, the inverse-square law changes the acceleration during the fall. Radial free fall from rest uses conservation of mechanical energy to account for that change exactly, within the specified Newtonian gravitational field. The contrast matters: the simplest formula is an approximation to a physical model, rather than a law with the same numerical $g$ everywhere.

History also makes the idealization tangible. Galileo's Two New Sciences (1638), Third Day, Theorem II, derives distance proportional to the square of time for uniformly accelerated motion from rest; its first corollary gives successive distances in the ratios $1,3,5,7,\dots$ over equal time intervals. This is a kinematic statement, predating the gravitational inverse-square law. ^[ref]

On Apollo 15's final lunar surface excursion, David Scott released a hammer and a feather together. The near-vacuum removed the large difference in drag force familiar on Earth. NASA's report describes their arrival together within the accuracy of the release; this demonstration illustrates the free fall prediction without establishing infinite experimental precision. ^[ref]

Galactic disk flaring is an outward increase of vertical thickness. A constant vertical velocity dispersion in an approximately thin-sheet stellar Newtonian gravitational field gives $h\propto 1/{\Sigma }_{\ast }$. An exponential galactic disk then predicts $h\propto e^{R/R_d}$ for a gas tracer. Inside a thick stellar layer the potential is instead approximately harmonic, and the tracer width is ${\sigma }_z/{\nu }_z$ with vertical epicyclic frequency ${\nu }_z$.

The differential Newtonian gravitational field of a galaxy across a smaller bound system. Its tidal tensor can slowly change the angular momentum of distant comets, changing their periapses and their access to planetary encounters.

The outward flux of the Newtonian gravitational field through a closed surface is $-4\pi G$ times the enclosed mass. A sphere around a point mass establishes the coefficient; superposition extends the integral law to general mass distributions. The minus sign expresses attractive gravity. A symmetric Gaussian surface can turn the flux relation into a local field calculation, but the integral law alone does not imply that the field is uniform on an arbitrary surface.

Gravitational acceleration is the acceleration of a test particle in free fall in a Newtonian gravitational field. It is a vector pointing along that field; $g$ commonly denotes its positive magnitude. Near Earth's surface, the magnitude is approximately constant over shallow vertical scales. For a fixed spherical source it instead varies as $g(r)=\mu /r^2$, where $\mu$ is the standard gravitational parameter.

This is distinct from gravitational weight, the force $m\mathbf{g}$, and from apparent weight, the support force a scale measures. An orbiting body can have substantial gravitational acceleration while its apparent weight is zero.

The gravitational constant $G$ sets the strength of Newton's law of universal gravitation. Two point masses $M$ and $m$ a center-to-center distance $r$ apart attract with force magnitudeFor fixed masses and separation, doubling $G$ would double the force. For bodies with spherical symmetry, the same exterior formula follows from the spherical shell theorem; for arbitrary extended bodies, add the attractions of their mass density elements instead. The gravitational constant connects a laboratory attraction to the Newtonian gravitational field controlling a Kepler orbit.

The gravitational constant $G$ and local gravitational acceleration $g$ are different quantities. Applying Newton's second law to a test particle gives $g=F/m=GM/r^2$ for one spherical source. The source's mass and the observer's position change $g$; $G$ remains the same parameter of the law. A local falling-body experiment therefore measures a field, not $G$ alone. On rotating Earth, an instrument fixed to the surface also responds to centrifugal acceleration, while surrounding mass density and altitude change the gravitational contribution.

Dimensional analysis explains the units. Denote the base physical dimensions of length, mass and time by $\mathsf{L},\mathsf{M},\mathsf{T}$. Since force has dimension $\mathsf{ML}{\mathsf{T}}^{-2}$,By contrast, $[g]=\mathsf{L}{\mathsf{T}}^{-2}$ and the standard gravitational parameter $\mu =GM$ has dimension ${\mathsf{L}}^3{\mathsf{T}}^{-2}$. A small numerical value of a dimensional constant depends on the units chosen: it is the resulting force for specified physical masses and separation that conveys the laboratory scale.

NIST's current complete listing givesfor the 2022 CODATA adjustment, the latest adjustment listed when checked on 14 October 2026. Parentheses indicate a standard uncertainty $u(G)=0.00015\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, approximately $22$ parts per million relative to $G$. The adjustment year is 2022; NIST's May 2024 data update is a different date.^{[ref] [ref]}

For an original scale calculation, stipulate two $1\mathrm{kg}$ point masses separated by $1\mathrm{m}$ and insert this recommended $G$: $F=6.67430\times 10^{-11}\mathrm{N}$. The corresponding acceleration of either mass is $6.67430\times 10^{-11}\mathrm{m}{\mathrm{s}}^{-2}$ at that instant. Reducing the separation to $0.10\mathrm{m}$ multiplies both by $100$. These are ideal model predictions, not observations; their tiny scale explains why stray forces matter.

Being a constant of a physical law does not make the numerical value exact. The SI fixes the numerical values of seven defining constants, including the speed of light and Planck constant; $G$ is determined experimentally. Exact unit definitions still require physical realizations with measurement uncertainty: calibrating a length or a mass does not become error-free.^[ref]

A laboratory determination of the gravitational constant must connect a sensor's reading to a gravitational signal from independently characterized masses. For an ideal two-point mass force measurement, invert the model:These are signed small changes obtained by taking the logarithmic derivative, not an instruction to add standard uncertainties with signs. For independent small input uncertainties, linear propagation of uncertainty instead givesCorrelated inputs require their covariances too. In particular, common geometric calibration errors must not be treated as independent simply because several distances were recorded.

For a concrete model check, hold the actual force and masses fixed but overestimate a true $r=0.100\mathrm{m}$ by $1\mu \mathrm{m}$. The exact inferred ratio isThe resulting bias is $20.0001$ parts per million, already comparable with the relative standard uncertainty of the recommended gravitational constant. A sensor can resolve very small changes while its inferred answer remains limited by source geometry. The example concerns one idealized separation; it is not an uncertainty estimate for any actual experiment.

Making the source mass larger or bringing it closer strengthens the signal, but also makes detailed mass density and geometry important. Nearby apparatus supplies its own Newtonian gravitational field; electromagnetic shielding does not remove that attraction. Sensor offsets, changing temperature, vibration and source motion can imitate or alter a signal. The Cavendish experiment illustrates calibration of a torsion constant; a different sensor requires its own calibrated response and a full source model.

Repeating readings can reduce independent measurement noise, but a fixed scale error survives averaging. A useful experiment deliberately changes the source configuration, checks whether the signal follows the predicted change, and characterizes what moves with the source. A systematic error that changes with that configuration also survives background subtraction. Comparing methods with different sensors and different dominant errors is therefore valuable.

Passive gravitational mass is the coefficient in $\mathbf{F}=m_{\mathrm{g}}\mathbf{g}$ for an imposed Newtonian gravitational field. Active gravitational mass instead measures a body's contribution to the source of that field in Newton's law of universal gravitation. Universality of the ratio of passive gravitational mass to inertial mass is the Newtonian content of the weak equivalence principle. In general relativity, the source is the full stress-energy tensor, rather than a single scalar gravitational mass.

The Newtonian gravitational field is the gravitational force per unit mass of a test particle. For a point mass $M$ at the origin,The test particle's mass cancels: different small masses have the same gravitational acceleration at the same position. In SI, the Newtonian gravitational field has units $\mathrm{N}\mathrm{k}{\mathrm{g}}^{-1}=\mathrm{m}{\mathrm{s}}^{-2}$. The same exterior expression holds for a spherically symmetric source, by the spherical shell theorem.

Choose a reference radius $r_0>0$, set $g_0=GM/r_0^2$, and use nondimensionalization with $z=r/r_0$. Then $g/g_0=z^{-2}$: at $z=1,2,4$, the magnitudes are $1,1/4,1/16$. This makes the spatial variation visible without specifying a planet. ^[ref]

The Newtonian gravitational potential $\Phi$ is potential energy per unit mass of a test particle. Its gradient determines the Newtonian gravitational field:In SI, $U$ has units $\mathrm{J}$, while $\Phi$ has units $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}={\mathrm{m}}^2{\mathrm{s}}^{-2}$; its gradient has units of acceleration, not force. Adding a constant to $\Phi$ leaves the Newtonian gravitational field unchanged. For an isolated point mass, the convenient reference is $\Phi (\infty )=0$. ^[ref]

Increasing the radius from $r_0$ to $r_1>r_0$ raises the Newtonian gravitational potential energy:An external agent supplies this positive work if the initial and final kinetic energies agree; gravity does work $-\Delta U$. For $r_1=2r_0$, the required work is $GMm/(2r_0)$. ^[ref]

Write $h=r^2\dot{\theta }$ for the conserved specific angular momentum. Zero specific orbital energy in the inverse-square Newtonian gravitational field givesDefine $r_0=h^2/(2GM)$, the periapsis radius for a nonradial parabolic trajectory. ThenOn the outgoing branch, integrate from the periapsis time $t_0$ and put $s=\sqrt{r-r_0}$:The incoming branch has the opposite sign. Squaring gives the radius-time relation for a parabolic Kepler orbit, a version of the Barker equation:Here $r_0$ and $t_0$ are fixed by angular momentum and the time origin at periapsis. The relation applies on both branches, with $r\ge r_0$; it is not necessary to solve for the polar angle.

For approximately circular motion in a spherically dominated Newtonian gravitational field, $v_c^2(r)=GM(<r)/r$. If most gravitating mass followed the concentrated luminous galaxies, then beyond their main light distribution the enclosed mass would nearly saturate, giving a declining Keplerian galaxy rotation curve, $v_c\propto r^{-1/2}$.

Instead, extended galaxy rotation curves commonly remain roughly flat. The spherical mass profile for a flat rotation curve then requiresThe inferred mass continues growing where the luminous contribution is small, so the mass-to-light ratio increases outward. Under the assumed gravitational dynamics this is evidence for an extended dark matter halo. Measurements of neutral-gas motion can probe radii beyond the bright stellar disk. A realistic disk contribution must be computed using its flattened geometry; the spherical formula describes a halo-dominated region and does not turn every disk rotation curve into a spherical mass profile.

For positive masses and distinct positions, Newtonian gravitational field givesInsert ${\mathbf{x}}_i=a(t){\mathbf{a}}_i$ with $a>0$. The separation numerator contributes a factor $a$ and the denominator $a^3$, soHence the time-dependent equations are satisfied wheneverThe fixed-shape condition is a Newtonian central configuration, and the resulting motion is a homothetic gravitational motion. It applies away from the collision scale $a=0$; a negative scale would require absolute values in the force scaling rather than the same positive-scale formula.

Multiplying the scalar equation by $\dot{a}$ givesThus its first integral is ${\dot{a}}^2/2-\Lambda /a=k/2$ for some real constant $k$.

In the pair potentialdifferentiating a pair with respect to ${\mathbf{a}}_i$ produces $G{m}_i{m}_j({\mathbf{a}}_i-{\mathbf{a}}_j)/|{\mathbf{a}}_i-{\mathbf{a}}_j{|}^3$. Every pair incident on $i$ contributes exactly once, so ${\nabla }_{i}W={\mathbf{G}}_i$. Moreover $W(\lambda {\mathbf{a}}_1,\dots ,\lambda {\mathbf{a}}_n)={\lambda }^{-1}W$ for $\lambda >0$. Differentiating with respect to $\lambda$ at one proves the relevant Euler homogeneous function theorem identity directly:Dot each fixed-shape equation with ${\mathbf{a}}_i$ and sum to obtainFor a nontrivial collision-free configuration with at least two positive masses, $W<0$ and $I>0$, and therefore $\Lambda >0$. Positivity requires an interacting pair; an isolated one-particle system is the degenerate zero-force exception rather than a positive-$\Lambda$ gravitational configuration. Pairwise cancellation also gives ${\sum }_i{\mathbf{G}}_i=0$, so the nondegenerate configuration has ${\sum }_i{m}_i{\mathbf{a}}_i=0$: its scaling origin is the centre of mass.

Finally, the system's kinetic energy is $I{\dot{a}}^2/2$ and its potential energy is $W/a$. ConsequentlyThis establishes the energy relation for every fixed configuration satisfying the algebraic condition, without assuming any particular particle arrangement.

In an axisymmetric cylindrical coordinate system, the steady Collisionless Boltzmann equation isLet $\nu =\int f{d}^{3}v$ and $\overline{A}={\nu }^{-1}\int Af{d}^{3}v$. Multiply by $v_R$ and integrate over velocities. The spatial terms give ${\partial }_R(\nu \overline{v_R^2})+{\partial }_z(\nu \overline{v_R{v}_z})$. Integration by parts in $v_R$ supplies $\nu {\Phi }_R-\nu \overline{v_{\varphi }^2}/R$, and the $v_{\varphi }$ derivative supplies $\nu \overline{v_R^2}/R$. The $v_z$ force term vanishes because $v_R$ does not depend on $v_z$. This assumes sufficient decay of the galactic distribution function to discard all velocity boundary terms. The radial Jeans equation is consequentlyReflection symmetry about $z=0$ makes $\nu$ even and $\overline{v_R{v}_z}$ odd. At the midplane the latter vanishes, and ${\nu }^{-1}{\partial }_z(\nu \overline{v_R{v}_z})={\partial }_z\overline{v_R{v}_z}$. Multiplying by $R/\nu$ provesAll quadratic velocities here are population moments, not individual stellar velocities. In particular, symmetry does not make the midplane tilt contribution to asymmetric drift disappear: an odd mixed moment can have a nonzero derivative at zero.

For application to the Milky Way, assume negligible mean radial and vertical flows, and define the circular speed by $v_c^2=R{\Phi }_R(R,0)$. Write $\overline{v_R^2}={\sigma }_R^2$ and $\overline{v_{\varphi }^2}={\overline{v}}_{\varphi }^2+{\sigma }_{\varphi }^2$. Rearranging gives the exact stress-support relationThus the stellar asymmetric drift $v_a=v_c-{\overline{v}}_{\varphi }$ is the right-hand side divided by $v_c+{\overline{v}}_{\varphi }$. For $v_a\ll v_c$, this denominator is approximately $2v_c$. If $\nu \propto e^{-R/h_{\nu }}$ and ${\sigma }_R^2\propto e^{-R/h_{\sigma }}$ locally, the bracket becomes $R/h_{\nu }+R/h_{\sigma }-1+{\sigma }_{\varphi }^2/{\sigma }_R^2$, making the density and dispersion-gradient contributions explicit.

A radially declining random-motion stress supplies some of the force needed to support a galactic disk, so its stars need less ordered azimuthal motion than a population on circular orbits. Hotter, often older, stellar populations therefore generally rotate more slowly than cold populations in the same gravitational potential; this is the observed stellar asymmetric drift. It is a stress-support effect, not a frictional slowing of each orbit. For a velocity ellipsoid aligned approximately with spherical coordinates, $\overline{v_R{v}_z}\simeq ({\sigma }_R^2-{\sigma }_z^2)z/R$, and the midplane tilt term contributes $-({\sigma }_R^2-{\sigma }_z^2)$ rather than zero.

Local measurements of mean rotation and all components of the velocity dispersion, combined with tracer-density and dispersion gradients, can therefore constrain $v_c$ and the radial Newtonian gravitational field. A hotter sample's mean rotation must not be identified directly with the circular speed. The tracer number density need not be the gravitating mass density. Nonaxisymmetric streaming, a nonstationary disk, poorly measured radial gradients, or a neglected tilt term can bias this inference; these are assumptions of the derivation rather than extra free corrections to the circular force.