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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 $v=\sqrt{\mu /r}$.

The bulk-parameters table gives the terrestrial standard gravitational parameter and volumetric mean radius $6371\mathrm{km}$. The orbital-parameters table distinguishes the sidereal day, $23.9345\mathrm{hours}$, from the $24$-hour solar day. Use the problem's own radius when it supplies one.

Use the synthetic standard gravitational parameter $\mu =8{\mathrm{m}}^3/{\mathrm{s}}^2$ and launch distance $r_0=2\mathrm{m}$. Then escape speed is $2\sqrt{2}m/s$. Three outward radial launches give exact comparisons: $v_0=2m/s$ has specific orbital energy $-2J/kg$ and stops outward at $r=4\mathrm{m}$; $v_0=2\sqrt{2}m/s$ has zero specific orbital energy; $v_0=2\sqrt{3}m/s$ has specific orbital energy $+2J/kg$ and asymptotic speed $2m/s$. The second case has speed $2m/s$ at $r=4\mathrm{m}$. Doubling $r_0$ divides escape speed by $\sqrt{2}$, whereas doubling $M$ at fixed $r_0$ multiplies it by $\sqrt{2}$. Changing the test particle's mass changes $E$ but cancels from the escape speed.

The escape velocity is the least initial speed that permits a freely coasting body to move arbitrarily far from an isolated attracting body. Despite its conventional name, it is a speed, not a unique velocity vector. For a fixed point mass $M$, or outside a body with spherical symmetry, set Newtonian gravitational potential energy zero at infinity. An exterior test particle of constant mass $m\ll M$ then hasHere $\mu =GM$ is the standard gravitational parameter, and $r$ is distance from the centre, not height above the surface. Set $r=R$ for surface escape velocity. Evaluate velocity in the body's rest inertial frame. Neglect drag, later rocket thrust, other bodies' gravity, and relativistic effects.^[ref]

Conservation of energy gives the threshold by requiring $E\ge 0$: at infinity the potential energy tends to zero and the kinetic energy cannot be negative. If $E=0$, the outward motion loses speed continuously toward zero. If $E>0$, the asymptotic speed is $v_{\infty }=\sqrt{2E/m}=\sqrt{v_0^2-v_{\mathrm{esc}}(r_0{)}^2}$. If $E<0$, infinity is inaccessible. For a radial outward launch, its outer turning point is $r_{\max }=-GMm/E$. For nonradial motion use angular momentum and the escape energy threshold to distinguish zero radial velocity from zero total speed.

Direction still determines the orbit and whether the body hits an obstacle. An initially inward velocity can cause collision even when $E\ge 0$. For an isolated spherical body, an exterior outward launch, or a tangential launch at or above escape velocity, has an outward escape branch without intersecting the body. A parabolic trajectory is the nonradial threshold path; a hyperbolic Kepler orbit has positive energy. The formula specifies their required local speed, not their orientation.^[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.

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]

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.

Here inverse-square potential means the potential energy associated with an attractive inverse-square law for a central force: the potential energy itself varies as inverse distance. For $k>0$ and position $\mathbf{r}$ with $r=|\mathbf{r}|>0$,The gradient makes the distinction explicit: doubling $r$ halves the magnitude of potential energy but quarters the magnitude of force. The neighboring repulsive inverse-square potential instead has potential energy proportional to $r^{-2}$ and an inverse-cube force; the two naming conventions must be distinguished.

For Newton's law of universal gravitation, $k=GMm$, where $G$ is the gravitational constant. Treat the attracting mass $M$ as an isolated fixed point mass, or as a body with spherical symmetry and exterior test particle of constant mass $m\ll M$. The spherical shell theorem gives the same exterior Newtonian gravitational potential energy $U=-GMm/r$ without requiring uniform mass density. Here $r$ is measured from the centre, and the formula applies outside the body; it does not describe its interior. The Newtonian gravitational potential per unit mass is $\Phi =U/m=-GM/r$.^[ref]

The negative sign records binding relative to infinite separation, rather than a negative kinetic energy. Moving from $r_1$ to $r_2>r_1$ increases potential energy byBecause the force is a conservative force, this work depends on the endpoints, including for a curved path. Integrating $dV/dr=k/r^2$ with zero at infinity gives $V=-k/r$; adding a constant leaves the force unchanged but shifts the numerical escape energy threshold by that same constant. This is why the reference at infinity must accompany an escape velocity calculation.

For an exact synthetic comparison, stipulate the standard gravitational parameter $\mu =GM=8{\mathrm{m}}^3/{\mathrm{s}}^2$ and a $1\mathrm{kg}$ test particle. At $r=2\mathrm{m}$, $U=-4\mathrm{J}$ and the inward force has magnitude $2\mathrm{N}$; at $r=4\mathrm{m}$, $U=-2\mathrm{J}$ and the inward force has magnitude $0.5\mathrm{N}$. The increase is $2\mathrm{J}$ of potential energy. These stipulated parameters are an arithmetic example, not observations of a particular body. The escape-energy balance diagram uses the same parameters.

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]

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.

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.

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.