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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.

Outside a spherical star, its gravitational acceleration is $-G{M}_{\star }\mathbf{r}/r^3$. It provides the dominant central force for Kepler orbits in a star-dominated planetary system.

Half the peak-to-peak line-of-sight orbital velocity range of a star in a single-companion Kepler orbit. It equals $K=2\pi a_{\star }\sin i/[P\sqrt{1-e^2}]$, where $a_{\star }$ is the stellar barycentric semi-major axis, $P$ the orbital period, $i$ the sky-plane orbital inclination, and $e$ the orbital eccentricity. The Doppler spectroscopy formula follows from Kepler's third law for the relative orbit, $a^3=G(M_{\star }+M_p)P^2/(4{\pi }^2)$.

A superintegrable Hamiltonian system with $n$ degrees of freedom has more than $n$ functionally independent integrals of motion, including an integrable set of $n$ in involution. At a regular nonzero Hamiltonian vector field, at most $2n-1$ independent integral differentials can annihilate the flow. A maximally superintegrable system attains that bound. The three-dimensional Kepler orbit has five independent integrals through energy, angular momentum and the Laplace-Runge-Lenz vector. A closed orbit on one isolated resonance is weaker than maximal superintegrability on an open phase-space region.

The same initial distance can lead to strikingly different orbits. Let $s=v_0/\sqrt{\mu /r_0}\ge 1$ and launch tangentially in one orbital plane. Then periapsis is the launch point, specific angular momentum is $h=r_0{v}_0$, andwhere $f$ is true anomaly. These follow by combining the initial specific orbital energy with the orbital eccentricity relation and the Kepler orbit equation. With $\mu =8{\mathrm{m}}^3/{\mathrm{s}}^2$ and $r_0=2\mathrm{m}$, the initial speeds $2$, $\sqrt{6}$, $2\sqrt{2}$, and $2\sqrt{3}m/s$ give $e=0,1/2,1,2$: a circular orbit, an elliptic orbit with apoapsis distance $6\mathrm{m}$, a parabolic trajectory, and a hyperbolic Kepler orbit. The two open paths have asymptotic speeds zero and $2m/s$ respectively. Their different directions at large distance are part of the Kepler orbit, not an extra scalar escape threshold.

A trajectory whose dynamics require gravitational forces from more than one attracting body. Around a synchronously rotating moon, a maintained trajectory involving the planet–moon system is distinct from a simple moon-centred stationary Kepler orbit; the moon's Hill sphere limits the latter approximation.

For bound planet-crossing Kepler orbits, $T=2/(q+Q)+2\sqrt{2qQ/(q+Q)}\cos I$, with the planet radius taken as one. At fixed $q<1$, $T$ is maximized by $I=0$ and $Q=1$. For $2<T<3$ this gives $q\ge (1-\sqrt{3-T}{)}^2/[1+2\sqrt{3-T}-(3-T)]$. This form remains regular at $T=\sqrt{8}$. It is an invariant-based accessibility bound, not a guarantee that a scattering history attains it; for $0<T\le 2$ it gives no positive universal pericentre floor.

For a small planet on a nearly circular Kepler orbit, a stellar chord at normalized transit impact parameter $b$ has length $2R_{\star }\sqrt{1-b^2}$. Dividing by transverse orbital speed $2\pi a/P$ gives the duration. A central transit has the longest duration within this approximation. Eliminating $a$ with Kepler's third law yields $P/D^3\ge {\pi }^{2}G{\rho }_{\star }/3$ for known mean stellar mass density. Finite planet radius, appreciable orbital eccentricity, and $R_{\star }/a$ corrections change the bound.

True longitude is the conventional orbital angle formed from the longitude of ascending node, argument of periapsis and true anomaly. For a coplanar Kepler orbit it is the actual polar longitude of the position vector. For an inclined orbital plane the sum of angles is generally not the projected azimuth. The mean longitude uses mean anomaly $M$ in place of $f$, so ${\ell }_{\mathrm{true}}-\lambda =f-M$. For small orbital eccentricity, Kepler's equation gives $E=M+e\sin M+O(e^2)$, while the eccentric-to-true anomaly relation gives $f=E+e\sin E+O(e^2)$; hence $f-M=2e\sin M+O(e^2)$. This periodic difference matters when separating a slow resonant argument from a rapidly varying geometric angle.

The virial theorem relates the average value of a function along a trajectory to the forces that bind it. For an isolated system of particles interacting through Newtonian gravitational potential energy, it gives $2⟨K⟩+⟨U⟩=0$: the average kinetic energy is half the magnitude of the average negative potential energy. An eccentric Kepler orbit makes the distinction vivid: kinetic energy and potential energy vary throughout the orbit, although their sum is constant. The relation concerns their time averages, rather than their values at every instant. It underlies the virial mass estimate and the negative heat capacity of a virialized gravitational system.

Use constant positive particle masses $m_i$, positions ${\mathbf{r}}_i$, mechanical momenta ${\mathbf{p}}_i=m_i{\dot{\mathbf{r}}}_i$, and total kinetic energy $K={\sum }_i{\mathbf{p}}_i^2/(2m_i)$. The auxiliary quantity $G(t)$ is a sum of dot products, not the gravitational constant, which we denote here by $G_N$:This finite-$T$ identity is exact wherever Newton's second law holds along a regular trajectory; ${\mathbf{F}}_i$ must include every force. The endpoint term is what a finite observation window leaves behind. For the infinite-time theorem, suppose the averages on the left exist and $(G(T)-G(0))/T\to 0$. A sufficient condition is uniformly bounded positions and velocities for finitely many particles. More generally $G(T)=o(T)$ suffices, even without those individual bounds. Bounded positions alone do not supply a bound on momenta, especially near a singular potential energy; nor does a small endpoint term prove thermodynamic equilibrium. Periodic motion with $G(T)=G(0)$ gives the identity without an endpoint term at that finite $T$. For an isolated system, use an inertial centre of mass frame so translational drift does not contribute bulk kinetic energy.^[ref]

For a differentiable potential energy that is a homogeneous function of degree $q$, choose its additive constant so that the scaling law really holds. With conservative forces, the Euler theorem for homogeneous functions givesThus a harmonic oscillator has $q=2$ and $⟨K⟩=⟨U⟩$. An isolated system with Newtonian gravitational potential energy hasEach interacting pair occurs once, and $U=0$ at infinite separation. Here the total energy $E=K+U$ is conserved. An external field or a mixture of different scaling powers requires its own force term; one cannot assign the whole potential energy a single degree $q$. The virial theorem for a homogeneous pair potential states the corresponding general energy relation.

Consider the exact two-body problem with masses $m_1,m_2$. In the inertial centre of mass frame, its internal kinetic energy is $K=m|\dot{\mathbf{r}}{|}^2/2$, where $\mathbf{r}$ is their relative position, $m=m_1{m}_2/(m_1+m_2)$ is the reduced mass, and $U=-\kappa /r$ with $\kappa =G_N{m}_1{m}_2$. Take a noncollision elliptic Kepler orbit with semi-major axis $a$ and orbital eccentricity $0<e<1$. Write $\eta$ for eccentric anomaly, to avoid confusing it with total energy $E$. With mean motion $n=\sqrt{\kappa /(m{a}^3)}$ and orbital period $P=2\pi /n$, Kepler's equation and the orbit geometry giveThe time origin is periapsis. Since $dt=(1-e\cos \eta )d\eta /n$, the time average of $1/r$ over one orbital period is exactly $1/a$. Therefore $⟨K{⟩}_P=\kappa /(2a)$ and $⟨U{⟩}_P=-\kappa /a$, independently of orbital eccentricity. Equal steps in eccentric anomaly are not equal time steps; unweighted averaging of a drawing of the ellipse gives the wrong answer.

The original dimensionless example in both diagrams has $m=a=\kappa =1$ and $e=0.6$. At periapsis, $r=0.4$, $K=2$, $U=-2.5$; at apoapsis, $r=1.6$, $K=0.125$, $U=-0.625$. Both have $E=-0.5$, but $2K+U$ is respectively $1.5$ and $-0.375$. Only the orbital period averages are $⟨K{⟩}_P=0.5$ and $⟨U{⟩}_P=-1$. The instantaneous zero of $2K+U$ occurs at $r=a$, not at every point of the orbit.

The endpoint term can also be evaluated exactly: $G=m\mathbf{r}\cdot \dot{\mathbf{r}}=mn{a}^{2}e\sin \eta$. Starting at periapsis, a window ending at eccentric anomaly $\eta$ hasAt $T=P/4$, Kepler's equation gives $\eta \simeq 2.09133$ and a positive residual $\simeq 0.3314\kappa /a$. At $T=P/2$ or $P$, $G$ returns to zero and the residual vanishes exactly. The lower panel shows the running residual and its endpoint expression coinciding; the shaded quarter-orbital period window explains why a short observation need not satisfy the infinite-time theorem.

For a Kepler orbit with semi-major axis $a$, the vis-viva equation isFor the convention that a hyperbolic Kepler orbit has positive semi-major-axis magnitude $a$, its final term instead has a plus sign.