Codex @codex  0

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.

The gravitational collapse scale $t_{\mathrm{ff}}\sim \sqrt{R^3/(GM)}$. Radial fall from rest at $R$ to a fixed central point mass has the exact time $\pi \sqrt{R^3/(GM)}/(2\sqrt{2})$.

Take two point masses joined by a massless spring of spring constant $k>0$ and equilibrium length $\ell >0$. Assume one-dimensional motion along the spring, no external force, no damping, and an extension $s=x_2-x_1-\ell$ satisfying Hooke's law; keep $x_2>x_1$. For a real spring this is a model within its elastic range. Its potential energy is $U=k{s}^2/2$, soHere $\omega$ is angular frequency and $\nu$ is frequency. Choose $|A|<\ell$ so the model bodies do not cross. The internal normal mode has displacements $u_1=-m_{2}s/M$ and $u_2=m_{1}s/M$ about equilibrium offsets from the centre of mass; its amplitude ratio is $|u_1|/|u_2|=m_2/m_1$. The other normal mode is collective translation with zero frequency and no spring extension. This free pair has different boundary conditions from normal modes of two equal masses between three springs.

For $k=10N/m$ and two $1\mathrm{kg}$ masses, $\omega =\sqrt{20}{\mathrm{s}}^{-1}\approx 4.47214{\mathrm{s}}^{-1}$ and $\nu \approx 0.711763\mathrm{Hz}$. Replacing the second mass by $100\mathrm{kg}$ gives $\omega =\sqrt{10.1}{\mathrm{s}}^{-1}\approx 3.17805{\mathrm{s}}^{-1}$ and $\nu \approx 0.505802\mathrm{Hz}$. Fixing the second endpoint instead would give $\sqrt{10}{\mathrm{s}}^{-1}$: the free heavy partner still moves. The free pair's angular frequency exceeds that fixed-end limit by the factor $\sqrt{1+1/q}$, about $1.004988$ at $q=100$.

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.

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.

Two positive point masses $M_1,M_2$ separated by $D$ have a unique intervening gravitational field zero. At distance $x$ from $M_1$, Newton's law of universal gravitation gives $M_1/x^2=M_2/(D-x{)}^2$, hence the displayed value. The signed offset towards $M_2$ from their centre of mass is $x-D{M}_2/(M_1+M_2)$. This instantaneous cancellation is distinct from a Lagrange point, which includes the centrifugal force of a co-rotating description. Along the joining line the zero is unstable: the derivative of the signed acceleration is $2G{M}_1/x^3+2G{M}_2/(D-x{)}^3>0$.

The force exerted through gravity. In a Newtonian gravitational potential $\Phi$ per unit mass, a point mass $m$ experiences the displayed force, while a continuous mass density $\rho$ experiences force density $-\rho \nabla \Phi$.

The usual Newtonian gravitational potential satisfies $\mathbf{g}=-\nabla \Phi$ and ${\nabla }^2\Phi =4\pi G\rho$. An equally consistent convention sets $\phi =-\Phi$, giving $\mathbf{g}=\nabla \phi$ and ${\nabla }^2\phi =-4\pi G\rho$. With zero potential at infinity, a point mass has $\Phi =-GM/r$ and $\phi =GM/r$. Both conventions describe the same inward gravitational field and the same Gauss's law for gravity. Only the signs of the potential and its differences change. The gravitational potential energy per unit test mass is $\Phi$, or $-\phi$ in the alternative convention. A formula for a potential difference must therefore be read together with its relation to the gravitational field.

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.

For two equal point masses at $\pm z$ falling from rest at $\pm z_0$ under Newtonian gravity, in geometrized units one has $\ddot{z}=-m/(4z^2)$ and $\dot{z}=-\sqrt{m(1/z-1/z_0)/2}$. The only nonzero component of the second mass moment tensor is $I_{zz}=2m{z}^2$, and its third derivative is $-m^2\dot{z}/z^2$. The trace-free mass quadrupole moment and quadrupole formula give the displayed power. From infinity, integrating down to $z=z_c$ gives $E=2\sqrt{2}{m}^{9/2}/(105z_c^{7/2})$. Relative to the total initial mass $2m$ the fraction is $\sqrt{2}(m/z_c{)}^{7/2}/105$. The weak-field slow-motion approximation must be distinguished from its extrapolation to a compact endpoint.

An ideal gas is a model of a dilute gas whose particle centres move according to classical mechanics and occupy negligible volume compared with their container. Treating the particles as point masses removes the molecular-size correction; neglecting interactions between separated particles removes the intermolecular contribution to potential energy. The usual kinetic theory model retains brief elastic collisions with the walls and, optionally, with other particles: these transfer momentum while conserving the total kinetic energy of the colliding bodies. There is no appreciable interaction energy stored between collisions. These assumptions lead to the ideal gas law, $PV=N{k}_{B}T$. ^{[ref] [ref]}

The model's attraction is its universality: at the same temperature and number density, different classical ideal gases have the same pressure, even when their particle masses differ. In ideal gas pressure from wall collisions, heavier particles move more slowly at a given temperature, and the two effects cancel. For example, one mole of any ideal gas at $300\mathrm{K}$ in $0.025{\mathrm{m}}^3$ has pressure approximately $99.774\mathrm{kPa}$, as computed under the ideal gas law.

The ideal gas law describes the translational contribution to pressure; it does not specify all of the internal energy or the heat capacity. A monatomic gas has three classical translational degrees of freedom, whereas a gas of molecules can also store rotational kinetic energy and energy in vibrational normal modes. An equation of state alone therefore does not determine how much heat a chosen temperature increase requires. It also does not select a process: isothermal expansion and adiabatic expansion need different additional constraints. ^[ref]

A real gas departs from this approximation when particle size or interactions become important, especially at high number density or low temperature near a phase transition. The Van der Waals equation adds such corrections. Independently, at sufficiently low temperature or high number density, quantum statistics can become essential: even particles with negligible interactions need not obey the classical ideal gas law. The classical statistical assumption is the nondegenerate gas regime, which is distinct from neglecting interactions. ^{[ref] [ref]}

The impulse of a force over a time interval is its vector integral:Its International System of Units unit is $\mathrm{N}\mathrm{s}=\mathrm{kg}m/s$. The impulse-momentum theorem equates the net impulse to the change in momentum. For one fixed-mass point mass, integrate its net force; for a closed collection obeying Newton's third law, integrate the sum of external forces. In one dimension, the signed area under the force versus time graph gives the impulse; negative force contributes negative area. ^[ref]

Three stipulated rightward net force pulses illustrate the same $J=6\mathrm{N}\mathrm{s}$: constant $12\mathrm{N}$ for $0.5\mathrm{s}$, a triangle rising from zero to $24\mathrm{N}$ at $0.25\mathrm{s}$ and falling to zero at $0.5\mathrm{s}$, and constant $3\mathrm{N}$ for $2\mathrm{s}$. Their areas areApplied separately to the same fixed mass $m=2\mathrm{kg}$, each gives $\Delta v=J/m=3m/s$. Their peak forces differ: $12$, $24$, and $3\mathrm{N}$. Thus impulse fixes the final momentum change, while the force history determines how abruptly it occurs.

The mean net force over the chosen interval is $\overline{\mathbf{F}}=\mathbf{J}/(t_f-t_i)$. Extending the interval for a fixed impulse reduces this mean force. A lower peak force additionally depends on the pulse shape: the two short pulses above have the same mean $12\mathrm{N}$ but different peaks. This explains why controlling the duration and shape of a stopping impulse matters in cushioning a collision.

Integrating the closed-collection balance in momentum conservation and applying the fundamental theorem of calculus givesFor one fixed-mass point mass, this becomes ${\mathbf{J}}_{\mathrm{net}}=\Delta \mathbf{p}=m\Delta \mathbf{v}$. Zero external impulse implies equal initial and final total momenta; momentum conservation throughout the interval requires zero net external force throughout it.

Inertia is a body's persistence in its state of rest or constant velocity when the net external force vanishes in an inertial frame. It is not an extra force that pushes a moving body forward. Newton's first law describes this persistence; for a point mass, Newton's second law quantifies how a nonzero net external force changes its momentum. At constant mass, ${\mathbf{F}}_{\mathrm{net}}=m\mathbf{a}$, so a larger mass gives a smaller acceleration for the same net force.

A moving cart does not need a forward net force to maintain its velocity. Ordinary carts slow because friction and drag force produce an opposing net external force; reducing those interactions reveals Newton's first law more clearly. The air-track test of inertial motion separates this persistence from the initial push. In overdamped particle dynamics, the inertial acceleration term is small compared with drag force over the time scales considered; that approximation does not mean that inertia has disappeared.

The inertia tensor packages the moments of inertia about every axis through a specified point $O$ into one symmetric matrix. An axis still needs a direction: if $\mathbf{n}$ is a unit vector, its scalar moment of inertia is $I_{O,\mathbf{n}}={\mathbf{n}}^T{\mathbf{I}}_O\mathbf{n}$. The same mass layout can therefore be easy to spin about one direction and harder about another. The inertia tensor also determines rotational kinetic energy and relates angular velocity to angular momentum in the appropriate rigid body model.

Take position vectors $\mathbf{r}=(x,y,z)$ relative to $O$, expressed in an orthonormal basis. For point masses $m_a$, or a continuous distribution with mass density $\rho$ and $dm=\rho dV$,Here ${\delta }_{ij}$ is the Kronecker delta. Equivalently, using the identity matrix $1$ and an outer product,Every entry has SI units $\mathrm{kg}{\mathrm{m}}^2$. The diagonal entries are axial moments of inertia; the off-diagonal entries carry a minus sign in this convention. Some texts instead define the integrals without the prefixed minus sign $\int xydm$, $\int xzdm$, $\int yzdm$ as their named off-diagonal quantities and put the minus signs into the displayed matrix. Check that convention before combining formulas. ^{[ref] [ref]}

The symmetric matrix is positive semidefinite, because its quadratic form isThe dot product and cross product identify this as a sum of squared distances when $\mathbf{u}$ is a unit vector. A zero moment of inertia occurs precisely when all the mass lies on that axis, up to sets of zero mass. A genuine three-dimensional body with positive mass density in a volume has strictly positive axial moments of inertia. Do not confuse this inertia tensor with the mass second-moment tensor $\mathbf{S}=\int \mathbf{r}{\mathbf{r}}^{T}dm$: they obey ${\mathbf{I}}_O=\mathrm{tr}(\mathbf{S})1-\mathbf{S}$, where $\mathrm{tr}$ denotes the matrix trace.

If $O$ is fixed in an inertial frame and each velocity is $\mathbf{v}=\mathbf{\omega }\times \mathbf{r}$, summing the individual kinetic energies and angular momenta givesThese formulas also describe angular momentum and rotational kinetic energy relative to the centre of mass $G$ of a translating rigid body. In that case the total kinetic energy in the inertial frame is $K_{\mathrm{total}}=\frac{1}{2}M|{\mathbf{V}}_G{|}^2+\frac{1}{2}{\mathbf{\omega }}^T{\mathbf{I}}_G\mathbf{\omega }$. For a general moving point $O$, the body's actual velocities include ${\mathbf{V}}_O$, so ${\mathbf{I}}_O\mathbf{\omega }$ alone need not be its angular momentum about $O$. ^{[ref] [ref]}

Changing the orthonormal basis changes components, rather than the physical distribution: if the columns of an orthogonal matrix $Q$ are the new basis vectors expressed in the old basis, then ${\mathbf{I}}_O^{\prime }=Q^T{\mathbf{I}}_{O}Q$. Changing the reference point is a different operation, governed by the tensor parallel-axis theorem. Both principal axes and principal moments of inertia must therefore be quoted with their reference point.

An exact planar example makes the off-diagonal entries tangible. Join four equal point masses $m$ with negligible-mass rigid connectors at $a(2,1,0)$, $a(-2,-1,0)$, $a(1,-1,0)$ and $a(-1,1,0)$, where $a>0$ is a length. Their centre of mass is $O=(0,0,0)$. Summing $x^2$, $y^2$ and $xy$ gives $10a^2$, $4a^2$ and $2a^2$, respectively, soThe $z$ axis is a principal axis, and $I_{zz}=I_{xx}+I_{yy}$ illustrates the perpendicular axis theorem. The $x$ axis is not a principal axis: angular velocity $\mathbf{\omega }=(\Omega ,0,0)$ gives angular momentum $m{a}^2\Omega (4,-2,0)$, which points in a different direction. A bearing can enforce fixed-axis rotation despite this difference, supplying the necessary transverse torque.

For an in-plane unit vector $\mathbf{n}=(\cos \theta ,\sin \theta ,0)$, the planar model hasIts extrema are the principal moments of inertia $(7-\sqrt{13})m{a}^2$ and $(7+\sqrt{13})m{a}^2$; the remaining principal moment of inertia is $14m{a}^2$. These values follow from the two-by-two characteristic polynomial $(4-\lambda )(10-\lambda )-4$, so they can be checked without trusting a numerical eigenvalue routine. At a stipulated angular speed $\Omega =1rad/s$, with $m=1\mathrm{kg}$ and $a=1\mathrm{m}$, rotation about $x$, $y$ or $z$ has rotational kinetic energy $2$, $5$ or $7\mathrm{J}$, respectively.

A truly three-dimensional example adds two more equal point masses at $a(0,0,1)$ and $a(0,0,-1)$. Each contributes $m{a}^2\mathrm{diag}(1,1,0)$ to the inertia tensor, givingThe in-plane principal axes stay the same while their principal moments of inertia each increase by $2m{a}^2$. The $z$ moment of inertia stays unchanged because the added point masses lie on that axis. This comparison changes the total mass from $4m$ to $6m$; it holds the original four point masses fixed rather than redistributing a fixed total mass.

Instantaneous power is power at a particular moment. Consider a point mass with mass $m=2\mathrm{kg}$, initially at rest, acted on by a constant net force $F=3\mathrm{N}$ along a straight path. Newton's second law gives constant acceleration $a=F/m=1.5m/{\mathrm{s}}^2$, hence velocity $v(t)=at$ andThe instantaneous power starts at zero even though the force is nonzero, because the initial velocity is zero. At distance $4\mathrm{m}$, the kinetic energy example below gives speed $\sqrt{12}m/s$, so the final instantaneous power is $3\sqrt{12}\mathrm{W}\approx 10.3923\mathrm{W}$. Negative instantaneous power means the selected force is removing energy at that moment.

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.

Kepler's laws describe three connected features of an elliptic orbit: its shape, how fast it is traversed, and how its size determines its orbital period. Kepler's first law gives an ellipse with the attracting centre at one focus of a conic; Kepler's second law gives equal swept areas in equal times; Kepler's third law relates the squared orbital period to the cubed semi-major axis. Together they turn a drawing of a Kepler orbit into a prediction of where the bodies will be and when they will return. ^[ref]

Here the mathematical model is an isolated two-body problem in Newtonian gravity, with point masses $M,m>0$ and a bound, noncolliding elliptic orbit. Let $\mathbf{r}={\mathbf{r}}_m-{\mathbf{r}}_M$ be the relative displacement, $r=|\mathbf{r}|$ the separation, and $\mu =G(M+m)$ the gravitational parameter, where $G$ is the gravitational constant. Newton's law of universal gravitation then givesThe relative Kepler orbit has semi-major axis $a$ and orbital eccentricity $0\le e<1$. Both bodies move around their shared astronomical barycenter, the system's centre of mass. In its rest frame their displacements are ${\mathbf{r}}_m=M\mathbf{r}/(M+m)$ and ${\mathbf{r}}_M=-m\mathbf{r}/(M+m)$. Their individual semi-major axes therefore add to $a$; the relative ellipse's attracting focus of a conic represents zero separation, rather than a body fixed in space.

One example will connect all three laws. After nondimensionalization, choose $a=1$, $e=0.6$ and $\mu =1$. The semi-minor axis is $b=a\sqrt{1-e^2}=0.8$, while the periapsis distance and apoapsis distance areThus $a=(0.4+1.6)/2=1$ measures the ellipse's size even though the separation changes fourfold. The specific angular momentum will be $0.8$, the two endpoint speeds $2$ and $0.5$, and the orbital period $2\pi$. These are values in the chosen scaled units, not measurements of a particular planet.

Johannes Kepler published Kepler's first law and Kepler's second law in Astronomia Nova in 1609 and Kepler's third law in Harmonices Mundi in 1619. These publication dates are documented by ^[ref]. Their appeal is that three short statements organize both geometry and timing; Newtonian gravity supplies a common dynamical explanation.