Newton's law of universal gravitation ($\mathbf{F}=-GMm\mathbf{r}/|\mathbf{r}{|}^3$)

Newton's law of universal gravitation gives the attractive force between two point masses. With source mass $M$ at the origin and relative position $\mathbf{r}$ of mass $m$, write $r=|\mathbf{r}|>0$:Here $G$ is the gravitational constant. The minus sign points toward $M$; the magnitude follows an inverse-square law, so doubling the separation quarters the force. The opposite force on $M$ satisfies Newton's third law. ^[ref]

Both masses accelerate. Applying Newton's second law to each in the isolated two-body problem givesThe relative standard gravitational parameter includes both masses; replacing it by $GM$ requires the test particle approximation $m\ll M$.

Historically, Newton's Book III, Proposition VII argues that gravitational attraction applies to all bodies and scales with their masses; its second corollary states the inverse-square law for individual particles. ^[ref]

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]

Take downward displacement $s$ as positive, assume a constant gravitational acceleration $g>0$, and release a point mass from rest at $s=0$. Integrating $\ddot{s}=g$ twice givesFor a downward distance $h>0$, the arrival time and speed areFor the model values $h=20\mathrm{m}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, these are approximately $2.0193\mathrm{s}$ and $19.8091\mathrm{m}{\mathrm{s}}^{-1}$. Doubling the height multiplies both quantities by $\sqrt{2}$; it does not double them. The same speed follows from the work-energy theorem, $mgh=m{v}^2/2$.

For intervals of length $\Delta t$, the successive downward displacements areThus equal spacing in time produces increasingly separated positions. This is the odd-number pattern in Galileo's result. It assumes constant acceleration and release from rest; a nonzero initial velocity adds a linear term to $s(t)$.

The approximation can be estimated for a spherical source of radius $R$ and standard gravitational parameter $\mu$. Writing $g_0=\mu /R^2$, the gravitational magnitude at height $h$ isFor $h/R\ll 1$, a Taylor expansion gives $g(R+h)/g_0=1-2h/R+O((h/R{)}^2)$. Small height relative to the source's radius makes nearly constant gravity plausible; negligible drag force remains a separate requirement.

Let a test particle start at rest at radius $r_0$ outside a fixed spherically symmetric source. The spherical shell theorem gives the exterior Newtonian gravitational potential $\Phi (r)=-\mu /r$, where $\mu =GM$ is the source's standard gravitational parameter. Zero initial transverse velocity makes the motion radial. Conservation of mechanical energy per unit mass givesThe minus sign selects inward motion; the positive square root is its speed. For a fall from $r_0=2R$ to the surface $r=R$, that speed is $\sqrt{\mu /R}$. In the limit of release from rest infinitely far away, the familiar free-fall speed is $\sqrt{2\mu /r}$, equal in magnitude to the local escape velocity. A finite starting radius gives a smaller speed.

The arrival time follows from the same first integral. Set $r=r_0{\cos }^2\eta$. ThenFor arrival at $R>0$ this is finite, and the field formula is used only outside the source. For $r_0=2R$, $\eta =\pi /4$ givesKeeping the smaller initial gravitational magnitude $g(r_0)=\mu /(4R^2)$ constant would predict $\sqrt{8}\sqrt{R^3/\mu }$, about $10.0\%$ longer. The actual acceleration increases during the fall.

The curves compare a shallow fall, $h/R=0.02$, with a fall from $r_0=2R$, so $h/R=1$. Each comparison holds the release height and initial gravitational magnitude fixed. The constant-gravity model is close for the shallow fall and increasingly inaccurate for the larger one. This is also a practical use of nondimensionalization: the normalized trajectories depend on $h/R$, while $\sqrt{R^3/\mu }$ supplies the dimensional time scale.

In an Earth-fixed rotating reference frame, the horizontal displacement of a freely falling body receives contributions from Coriolis acceleration and centrifugal acceleration. A short-fall approximation substitutes the unperturbed downward velocity into the Coriolis term and the Earth's surface position into the centrifugal term. If gravitational acceleration excludes centrifugal acceleration, both contributions must be included explicitly.

A body released from rest relative to Earth at the equator has unperturbed downward velocity $v_U=-gt$. The Coriolis acceleration in local Earth coordinates gives eastward acceleration $a_E=-2{\Omega }_E{v}_U=2{\Omega }_{E}gt$. Integration gives $x_E={\Omega }_{E}g{t}^3/3$ and impact time $t_f=\sqrt{2h/g}$, proving the title formula. This is first order in Earth's rotation, neglecting air resistance and the height dependence of the effective local gravitational acceleration. For a short fall at latitude $\phi$, its leading eastward term is multiplied by $\cos \phi$.

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]

Spread mass $M$ uniformly along $-a<z<a$ on an axis, with linear mass density $\lambda =M/(2a)$. In its perpendicular bisector plane, axial field components cancel, and the radial component isThis is an exact line mass model and a thin-rod approximation for a physical cylinder. A centred circular orbit in this symmetry plane has $v^2=GM/\sqrt{R^2+a^2}$. Thus $v\to \sqrt{GM/a}=\sqrt{2G\lambda }$ for $R/a\to 0$, and $v\sim \sqrt{GM/R}$ for $R/a\to \infty$. For a finite-width rod, the first limit is useful only while its transverse size remains much smaller than $R$.

A circular infinite cylinder of radius $b$ and uniform mass density $\rho$ has linear mass density $\lambda =\pi b^2\rho$. A coaxial Gaussian surface of length $L$ has inward radial gravitational field and no end-cap flux. Gauss's law for gravity gives $2\pi rL{g}_r=-4\pi G\lambda L$ outside, hence the displayed result. Inside, enclosed linear mass density is $\pi r^2\rho$, so $g_r=-2\pi G\rho r$. An exterior circular orbit has speed $\sqrt{2G\lambda }$ independent of radius. The exterior formula presupposes circular axial symmetry; area alone would not specify the near field of a noncircular rod.

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.

Remove a spherical void centered at $c$ from a uniform sphere centered at the origin. At points inside the void, superposition combines the sphere's interior field $-4\pi G\rho x/3$ with the removed sphere's field $+4\pi G\rho (x-c)/3$. Their sum is the displayed constant Newtonian gravitational field. A second removed mass outside this void adds a nonuniform contribution.

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.

For a Newtonian gravitational potential obeying the Poisson equation for Newtonian gravity, define $T_{ij}=[{\partial }_i\Phi {\partial }_j\Phi -{\delta }_{ij}|\nabla \Phi {|}^2/2]/(4\pi G)$. Commutation of mixed partial derivatives gives ${\partial }_j{T}_{ij}=\rho {\partial }_i\Phi$. Thus the gravitational force density can be put in conservative form as $-{\partial }_j{T}_{ij}$. The sign is chosen so that $T_{ij}$ is added to the momentum flux; reversing the stress convention reverses the divergence sign.

In cylindrical coordinates, the Newtonian gravitational stress tensor gives radial angular-momentum flux $r{T}_{r\varphi }=r({\partial }_r\Phi )(r^{-1}{\partial }_{\varphi }\Phi )/(4\pi G)$. Nonaxisymmetric structure can correlate the field components and transport angular momentum. The isotropic negative-pressure part has zero off-diagonal component, so it gives no direct radial transport. This separates a shear flux from a contribution to normal force balance.

Self-gravity is the gravitational acceleration produced by a system's own mass density. It is obtained by solving the Poisson equation for Newtonian gravity for that density, rather than treating the Newtonian gravitational potential as an externally fixed field. It competes with pressure and rotational support in gravitational instability of an astrophysical disk.

The product $\mu =GM$ of the gravitational constant and a central mass sets the strength of the exterior Newtonian gravitational field. Its dimensions are length cubed divided by time squared. In the exact relative two-body problem, the parameter is $G(M_1+M_2)$; using $GM$ for one source assumes the test particle approximation.

The combination is useful because orbital and falling-body observations constrain $\mu$ directly. A gravitational circular orbit of radius $a$ has oscillation period $T=2\pi \sqrt{a^3/\mu }$, so measuring $a$ and $T$ gives $\mu =4{\pi }^2{a}^3/T^2$. That measurement alone does not separately determine $G$ and $M$. Similarly, radial free fall from rest has a natural time scale $\sqrt{r_0^3/\mu }$.

Other interactions require a different model. Inverse-square radiation pressure can reduce the effective attraction for a dust Kepler orbit; calling that effective coefficient a gravitational parameter must not hide the added force.

The acceleration tidal tensor is $T_{jk}={\partial }_k{g}_j=-{\partial }_j{\partial }_k\Phi$ for gravitational acceleration $\mathbf{g}=-\nabla \Phi$. It describes differential acceleration $\Delta \mathbf{g}\simeq T\Delta \mathbf{r}$. Some authors instead call the positive Hessian matrix ${\partial }_j{\partial }_k\Phi$ the tidal tensor, reversing this sign convention. For a point mass and the positive potential convention $\mathbf{g}=\nabla \psi$, the tensor is $Gm(3\mathbf{R}{\mathbf{R}}^{\mathsf{T}}/R^5-1/R^3)$.

The tidal tensor is the Jacobian matrix of the Newtonian gravitational field, rather than the gravitational acceleration itself. Its entries have units ${\mathrm{s}}^{-2}$: multiplying by a displacement produces an acceleration. For a smooth Newtonian gravitational potential, the tidal tensor is a symmetric matrix, since mixed partial derivatives commute. A spatially uniform gravitational acceleration has zero tidal tensor, however large that acceleration is.

Use local axes radial, tangential in the orbital plane, and perpendicular to that plane around a circular Kepler orbit with ${\Omega }^2=GM/R^3$. The gravitational tidal tensor is ${\Omega }^2\mathrm{diag}(2,-1,-1)$. A uniformly corotating frame adds the gradient of centrifugal acceleration, ${\Omega }^2\mathrm{diag}(1,1,0)$, giving the displayed effective tensor for stationary offsets. Thus the radial differential acceleration is $3GMa/R^3$. These are accelerations, or forces per unit test mass; an actual differential force includes that mass. Moving offsets also experience Coriolis acceleration.

For a source point mass $M$ at the origin, let $\mathbf{R}=R\mathbf{n}$, where $\mathbf{n}$ is an outward unit vector, and let $\mu =GM$ be the source's standard gravitational parameter. Differentiating $g_i=-\mu R_i/R^3$ gives the tidal tensorwhere ${\delta }_{ij}$ is the Kronecker delta and $I$ is the identity matrix. In an orthonormal basis with first axis radial, the tidal tensor isThe radial eigenvalue $2\mu /R^3$ separates nearby freely falling test particles; the two transverse eigenvalues $-\mu /R^3$ bring them together. The farther radial test particle falls inward less rapidly, so its acceleration relative to the nearer one points outward. Transverse gravitational accelerations converge toward the source. Thus an initially resting small cloud stretches radially and squeezes in both transverse directions. These are instantaneous relative accelerations; the subsequent shape also depends on initial velocities and the changing tidal tensor along the cloud's free fall.

Outside the source, the matrix trace is $2\mu /R^3-\mu /R^3-\mu /R^3=0$. This matches $\mathrm{tr}T=-{\nabla }^2\Phi =0$ in a region with zero mass density, by the Poisson equation for Newtonian gravity. Zero matrix trace means no leading change of a small initially resting cloud's volume at order $t^2$, not that each separation stays fixed. Inside matter, the same sign convention gives $\mathrm{tr}T=-4\pi G\rho$ instead.

Subtracting the constant and uniform-acceleration terms from the point-mass relative potential gives ${\psi }_{\mathrm{tid}}=G{m}_p[3(\mathbf{r}\cdot {\hat{\mathbf{r}}}_p{)}^2-r^2]/(2r_p^3)$, with force convention $\nabla \psi$. Its tidal tensor is $G{m}_p(3{\hat{\mathbf{r}}}_p{\hat{\mathbf{r}}}_p^T-I)/r_p^3$. This is the leading term for target size small compared with separation.

Disruption of a self-gravitating body when differential gravitational acceleration across it exceeds its binding acceleration. For a body of radius $R$, mass $m$ at distance $r$ from a much more massive body $M$, the competing scales are $GMR/r^3$ and $Gm/R^2$; their equality gives $r\sim R(M/m{)}^{1/3}$, up to geometry and internal-structure factors. The tidal tensor gives the directional differential acceleration. A tidal disruption event near a black hole is a specific astronomical application.

Differential gravitational acceleration, multiplied by the affected mass, across a spatially extended body. In the linear approximation the acceleration difference is $T\Delta \mathbf{r}$, where $T$ is the tidal tensor. Stellar tidal forces limit the domain of a circumplanetary orbit; sufficiently strong ones can produce tidal disruption.

To isolate a tidal force, subtract the gravitational acceleration shared by the body as a whole. Let a reference test particle follow $\ddot{\mathbf{R}}=\mathbf{g}(\mathbf{R})$, and let another have position $\mathbf{R}+\mathbf{x}$. In a translating accelerated reference frame with parallel, nonrotating axes,for a source point mass, when $|\mathbf{x}|/R\ll 1$. Here $\mu =GM$ is its standard gravitational parameter and the last term uses Big O notation. This Taylor expansion removes the common inward pull and keeps the spatial variation. The near side has an excess pull toward the source; the far side has a deficit of inward pull, which is an outward relative acceleration. Both sides can stretch apart while both still accelerate toward the source.^[ref]

For a material element of mass $m$, the residual external force in this translating accelerated reference frame is $m[\mathbf{g}(\mathbf{R}+\mathbf{x})-\mathbf{g}(\mathbf{R})]$. Actual deformation also depends on elasticity, pressure and self-gravity. A rigid body approximation suppresses deformation but can still have internal forces that resist the tidal force. The point-mass tidal stretching and squeezing pattern describes freely falling, noninteracting test particles, rather than promising the same shape for every solid, star or ocean. A rotating reference frame introduces centrifugal acceleration and Coriolis acceleration, which must be included separately.

A small cohesionless spherical body of radius $s$ and density $\rho$ has surface self-gravitational acceleration $4\pi G\rho s/3$. The radial stretching of the corotating point-mass tidal field is $3GMs/R^3$. Equating them gives the displayed survival-distance estimate. The size $s$ cancels because both accelerations are linear in $s$ at fixed density. This is a local spherical force-balance estimate, not the exact fluid Roche figure: shape deformation, material strength and nonsynchronous spin can change its numerical coefficient.

A tide is a periodic change in ocean level driven mainly by the differential gravitational forces of the Moon and Sun, together with the Earth's rotation. Their tidal forces act across the extended planet; a uniform acceleration shared by the whole Earth does not by itself generate a tide. Ocean-basin resonance and coastal geometry modify the local response.

A spring tide has an enhanced high-to-low tidal range because the Moon and Sun are approximately aligned as seen from the Earth, so their principal tidal contributions reinforce. Such alignments occur twice in a lunar phase cycle.

A neap tide has a reduced high-to-low tidal range when the directions to the Moon and Sun make approximately a right angle at the Earth. Their principal tidal contributions then partly oppose one another. The solar and lunar contributions are not equal, so this does not mean that tides disappear.

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.

For a fixed source point mass, the magnitude $g=\mu /R^2$ of the gravitational acceleration measures the local pull; the radial eigenvalue ${\lambda }_r=2\mu /R^3$ of the tidal tensor measures how that pull changes across space. Doubling $R$ divides $g$ by four but ${\lambda }_r$ by eight. For a fixed small radial separation $L$, the linear relative acceleration is $\Delta a\simeq {\lambda }_{r}L$, soA small body can therefore experience strong common gravitational acceleration with a comparatively small tidal force. A measurement of $g$ alone does not determine the tidal tensor: sources at different distances can have the same $g$ but different spatial variation.

The local stretching time scale is ${\lambda }_r^{-1/2}=\sqrt{R^3/(2\mu )}$: it marks when an initially resting separation could change by order itself if the tidal tensor stayed fixed. During free fall, $R$ changes, so this scale is not a complete trajectory or a universal disruption time. Tidal disruption additionally compares the tidal force with binding by self-gravity and material forces; $g$ by itself supplies no such comparison.

A Love number measures a body's tidal response to an applied potential. The degree-two potential response $k_2$ is sensitive to radial density structure and central concentration, and can complement mass and radius constraints on an exoplanet interior. Inferring it from orbital precession requires separating other contributions and assumptions about rotation and the tidal response.

Gravity gradiometry measures spatial differences of gravitational acceleration. Two simultaneous measurements along a unit vector $\mathbf{e}$, separated by $L\mathbf{e}$, giveThis measures one directional component of the tidal tensor, not its whole matrix. Any identical platform acceleration cancels in the difference. Unequal instrument responses, changing geometry and rotating reference frames can spoil that cancellation. For a large baseline, the quotient is an average of the directional tidal tensor along the baseline, rather than its exact value at one point.

A laboratory gravity gradiometry experiment used two vertically separated clouds of chemical atoms in vacuum, launched into free fall. Shared laser beams applied three pulses ($\pi /2,\pi ,\pi /2$), splitting and recombining matter waves. Balanced photodiodes measured probe-light absorption to infer internal-state populations; scanning the last pulse's wave phase produced an interference fringe.^[ref]

The difference of fringe wave phases obeys $\delta \varphi ={\mathbf{k}}_{\mathrm{eff}}\cdot \Delta \mathbf{g}{\tau }^2$, with effective wave vector ${\mathbf{k}}_{\mathrm{eff}}$ and pulse spacing $\tau$: shared reference motion cancels, leaving differential gravitational acceleration. Eight lead bricks, each approximately $12.5\mathrm{kg}$, were placed symmetrically around the lower chamber, about $0.2\mathrm{m}$ from the fountain apex. The measured signal was $(8.1\pm 2.1)\times 10^{-9}g$, versus a calculated $8.2\times 10^{-9}g$, where $g$ denotes terrestrial gravitational acceleration.^[ref]

Controls deliberately shook the shared reference reflector and compared actively isolated and unisolated operation. Tilting tests exposed sensitivity to Coriolis acceleration and centrifugal acceleration. Thus shared translation is rejectable, while rotation and finite measurement uncertainty remain limitations. The brick signal detects spatially varying gravity; it does not establish a precision value of the gravitational constant.^[ref]

For a point-mass companion at separation $a$, subtract the constant and linear terms of its Newtonian gravitational potential in a frame accelerating with the target's center. The leading residual isIt is degree two in spherical harmonics. In a circular orbit with angular frequency ${\omega }_o$, its time-dependent $m=\pm 2$ components force at $2{\omega }_o$ in a frame with no spin; there is also a static $m=0$ component. This is the leading approximation in $r/a$, and the tidal amplitude may be absorbed into a harmonic-dependent constant.

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.

A collision-free positive-mass configuration satisfies the displayed equation when every gravitational acceleration points proportionally toward a common centre. With $W=-{\sum }_{i<j}G{m}_i{m}_j/|{\mathbf{a}}_i-{\mathbf{a}}_j|$, homogeneity gives $\Lambda I=-W>0$, $I={\sum }_i{m}_i|{\mathbf{a}}_i{|}^2$. Pairwise force cancellation implies ${\sum }_i{m}_i{\mathbf{a}}_i=0$. Such a configuration generates a nonrotating homothetic gravitational motion by multiplying all position vectors by one positive scale factor.

For a Newtonian central configuration, set ${\mathbf{x}}_i(t)=a(t){\mathbf{a}}_i$ with $a>0$. The gravitational force scales as $a^{-2}$, reducing all equations to the displayed scalar ordinary differential equation. Its first integral is ${\dot{a}}^2/2-\Lambda /a=k/2$. The total energy is $kI/2$, so the scalar energy constant exactly tracks the many-body energy for the fixed shape.

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]

An orbit around an isolated source constrains its standard gravitational parameter $\mu =GM$ rather than independently measuring the source's mass and the gravitational constant. In the test particle approximation, a circular orbit of radius $a$ and orbital period $T$ satisfiesThe left equation equates the required centripetal acceleration to the gravitational acceleration. More generally, a Kepler orbit depends on the same $\mu$. In the exact relative two-body problem, replace $GM$ by $G(M+m)$.

For any positive dimensionless number $q$, replacing $G$ by $qG$ and $M$ by $M/q$ leaves $GM$ and every predicted test particle trajectory unchanged. This is the gravitational mass degeneracy: more observations of the same orbit can refine $GM$ without separating its factors. A laboratory source whose mass is calibrated independently supplies the missing information.

Consequently an astronomical mass expressed in kilograms as $M=\mu /G$ inherits measurement uncertainty from $G$, even if $\mu$ is known much more precisely. To first order,When the relative standard uncertainty of $\mu$ is negligible, that of $M$ is approximately $u(G)/G$. Conversely, the ratio of two source masses inferred using the same $G$ is $M_1/M_2={\mu }_1/{\mu }_2$, so this common factor cancels. That distinction explains how excellent orbit determination can coexist with less precise absolute astronomical masses.

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.

In 2014, Gabriele Rosi and colleagues used atom interferometry: two rubidium-87 clouds prepared by laser cooling in vacuum, $328\mathrm{mm}$ apart, received shared three-pulse Raman transitions spaced by $160\mathrm{ms}$. Fluorescence readout and paired-fringe ellipse fitting yielded differential wave phase. Twenty-four tungsten alloy cylinders, about $516\mathrm{kg}$ total, moved between C and F. Cloud subtraction rejected shared vibration; configuration subtraction isolated the source change.^[ref]

Calibration used timed flight and light-sheet crossings; mirror tilt compensated Earth rotation. Wave vector reversal rejected unwanted shifts. Modeling included source Newtonian gravitational potential, atom trajectories, cloud distributions and moving supports. Source positioning, mass density variation and cloud geometry limited the reported combined relative standard uncertainty to $150$ parts per million.^[ref]

To understand the subtraction without reproducing that full analysis, consider an ideal atom interferometer in uniform vertical acceleration during each pulse sequence. Let $k_{\mathrm{eff}}$ be its signed effective vertical wave vector component and $T$ the pulse spacing. Its acceleration-dependent wave phase is $\varphi \simeq k_{\mathrm{eff}}g{T}^2$; labels U and L denote upper and lower clouds. The double difference is thenA background field identical in C and F cancels even if it differs between the two clouds. In a fixed-geometry, linear-response model, write $S=G\mathcal{K}$, where $\mathcal{K}$ is calculated from the source mass density, calibrated pulse sequence and atom trajectories. Thus $G_{\mathrm{inferred}}=S/\mathcal{K}$. Measuring wave phase accurately is one part of the task; calculating $\mathcal{K}$ accurately is another. The real nonuniform field and finite clouds require the trajectory-dependent analysis described above rather than this uniform-field approximation.

For mass density $\rho$, the Newtonian gravitational potential satisfies the Poisson equation ${\nabla }^2\Phi =4\pi G\rho$. The gravitational acceleration is $-\nabla \Phi$. An isolated system also requires the exterior vacuum solution and decay at infinity; solving the interior equation alone does not fix a globally physical field.

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.

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]

For radius $R$, constant mass density $\rho$ and $M=4\pi \rho R^3/3$, normalize the Newtonian gravitational potential to zero at infinity. ThenBy the spherical shell theorem, each spherical shell contributes the potential of a point mass outside it and a constant inside it. Summing inner and outer shells gives $\Phi (r)=-4\pi G\rho [r^2/3+(R^2-r^2)/2]$ inside. Equivalently, angular integration of the free-space Green function of the Laplacian gives ${\int }_{-1}^1(r^2+r^{\prime 2}-2r{r}^{\prime }\mu {)}^{-1/2}d\mu =[r+r^{\prime }-|r-r^{\prime }|]/(r{r}^{\prime })$. The resulting radial integral has the same value.

The potential and its first radial derivative are continuous at $R$. The interior gravitational acceleration is $-\nabla \Phi =-(GM/R^3)\mathbf{r}$, and ${\nabla }^2\Phi =4\pi G\rho$. At the centre $\Phi (0)=-3GM/(2R)$ while its gradient vanishes.

The dimensionless coefficient $J_2$ describes the leading axisymmetric departure of a planet's external Newtonian gravitational potential from a monopole: $\Phi =-G{M}_p/r+[G{M}_p{J}_2{R}_p^2/r^3]P_2(\cos \vartheta )$, where $\vartheta$ is colatitude from the spin axis. An oblate mass distribution has positive $J_2$. The disturbing function is minus the nonspherical potential term. On a circular inclined Kepler orbit, $\cos \vartheta =\sin I\sin M$, so orbit averaging gives ${\overline{\mathcal{R}}}_{J_2}=G{M}_p{J}_2{R}_p^2{a}^{-3}(1/2-3{\sin }^{2}I/4)$. The exact circular nodal rate is $-3n{J}_2(R_p/a{)}^2\cos I/2$; its small-tilt limit is $-3n{J}_2(R_p/a{)}^2/2$.

For $\Phi =-\mu /r+{\alpha }^2\mu (3z^2-r^2)/r^5$, a nearly circular equatorial orbit has $n^2=\mu /R^3$ and azimuthal frequency ${\Omega }_c^2=n^2(1+3{\alpha }^2/R^2)$. The radial epicyclic frequency is ${\kappa }^2=Rd{\Omega }_c^2/dR+4{\Omega }_c^2$, and the vertical epicyclic frequency is ${\nu }^2={\partial }_z^2\Phi {|}_{z=0}$. Radial stability requires $R^2>3{\alpha }^2$. The nodal precession rate ${\Omega }_c-\nu$ is negative and the apsidal precession rate ${\Omega }_c-\kappa$ positive. These sign statements concern small eccentricity and inclination. For an arbitrary inclined orbit, the averaged argument of periapsis instead evolves as $(3/2)n{\alpha }^2[5{\cos }^{2}i-1]/[a^2(1-e^2{)}^2]$ and can regress.

An axisymmetric logarithmic Newtonian gravitational potential has spheroidal equipotential surfaces and a flat galaxy rotation curve: $R{\Phi }_R(R,0)=V^2$. The singular model is defined away from the origin; a fixed reference length makes the logarithm dimensionless. Its central $r^{-2}$ mass density cusp is locally integrable, but its total mass diverges at large radius.

The Poisson equation for Newtonian gravity givesFor $V>0$, nonnegative mass density away from the origin is equivalent to $q^2\ge 1/2$. Strict positivity requires $q^2>1/2$. This condition does not by itself guarantee a nonnegative galactic distribution function.

A radial Newtonian gravitational potential $\Phi =-A{r}^{-\beta }$ produces inward acceleration $A\beta r^{-\beta -1}$. The point-mass case has $\beta =1$ and $A=GM$. Treating the power-law gravitational potential as prescribed does not require it to be sourced by the flowing gas.

For a binary star viewed far from its centre of mass, let $\mathbf{r}$ be the separation and $\mu$ the reduced mass, $M=M_1+M_2$ the total mass, $r=|\mathbf{r}|$ and $x=|\mathbf{x}|$. The Taylor expansion of the point-mass Newtonian gravitational potential has zero dipole because ${\sum }_A{M}_A{\mathbf{x}}_A=0$. Its second moment is ${\sum }_A{M}_A{x}_{Ai}{x}_{Aj}=\mu r_i{r}_j$, soThis holds for an instantaneous separation in any orbit, not only a circular orbit. The mass quadrupole moment convention $Q_{ij}=\mu (r_i{r}_j-r^2{\delta }_{ij}/3)$ gives ${\Phi }_2=-3G{Q}_{ij}{x}_i{x}_j/(2x^5)$.