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In the isolated model, conservation of energy and angular momentum conservation reduce the exterior test particle's motion toHere $\epsilon$ is specific orbital energy and $h$ is specific angular momentum. The effective potential is ${\Phi }_{\mathrm{eff}}=h^2/(2r^2)-\mu /r$. A zero radial velocity is a radial turning point, and need not mean zero total speed: at the periapsis or apoapsis of a nonradial Kepler orbit, tangential velocity remains. Thus the vertical gap between $\epsilon$ and $-\mu /r$ represents total kinetic energy per unit mass, whereas the gap above ${\Phi }_{\mathrm{eff}}$ represents its radial part.

For $h>0$, the orbital eccentricity obeys $e^2=1+2\epsilon h^2/{\mu }^2$. Negative $\epsilon$ gives an elliptic orbit, including a circular orbit at its minimum allowed energy; zero $\epsilon$ gives a parabolic trajectory; positive $\epsilon$ gives a hyperbolic Kepler orbit. The effective potential tends to zero from below at infinity, so its outer region creates no extra positive escape barrier. This explains why escape velocity depends on energy even though angular momentum changes the path. The $h=0$ case is radial motion, rather than a nondegenerate parabola or hyperbola; inward radial motion can end in collision.

The apocentre distance is the greatest distance from the attracting centre on a bound Kepler orbit. For an elliptic orbit,where $q$ is pericentre distance. Thus the two turning distances determine both semi-major axis and orbital eccentricity.

A particle moves more slowly near apoapsis and spends more time there, an important effect when translating an orbital population into observed spatial counts. A parabolic or hyperbolic escape trajectory has no finite outer turning point, so assigning it an elliptic apocentre distance would misdescribe the motion.

The apoapsis of an Earth-centred orbit. For an elliptic orbit its distance from the Earth's centre is $a(1+e)$, in terms of semimajor axis and orbital eccentricity.

A heliocentric model places Earth and the other planets in orbits around the Sun. The apparent annual journey of the Sun through the celestial sphere then reflects Earth's own orbital motion. Apparent backward loops in a planet's path can arise from viewing its motion from another moving planet.

The geometric arrangement alone does not specify the shape or governing force of an orbit. Nicolaus Copernicus used combinations of circular motion; Johannes Kepler subsequently described elliptic orbits. Distinguishing a reference arrangement, a mathematical trajectory and a physical explanation prevents these separate achievements from being conflated.

A heliocentric orbit describes a body's trajectory relative to the centre of the Sun. For a low-mass body whose motion is dominated by solar gravity, a useful first approximation is a Kepler orbit withwhere $\mathbf{r}$ is the body's position relative to the Sun, $r=\Vert \mathbf{r}\Vert$, $G$ the gravitational constant and $M_{\odot }$ the solar mass. A heliocentric description specifies an origin; the two-body equation additionally assumes that other gravitational forces can be neglected.^[ref]

For a bound elliptic orbit with semi-major axis $a$ and orbital eccentricity $e$, the specific orbital energy and orbital period obeyFor example, a hypothetical solar orbit with $a=2\mathrm{AU}$ and $e=1/2$ ranges from one to three astronomical units from the Sun. Its orbital period is $2^{3/2}=2\sqrt{2}$ times that of a circular solar orbit of radius one astronomical unit, independent of its orbital eccentricity in this approximation.^[ref]

The Sun itself accelerates as other bodies attract it. Accurate Solar system calculations must therefore include the relevant orbital perturbations and specify the reference frame. A spacecraft orbiting Earth also travels around the Sun, but solar-centred coordinates alone do not make a solar two-body problem an adequate model of that spacecraft's local motion.

A Hohmann transfer orbit between coplanar circular Kepler orbits of radii $r_1,r_2$ traverses half an elliptic orbit of semi-major axis $a=(r_1+r_2)/2$. Kepler's third law therefore gives $t=\pi \sqrt{a^3/(GM)}$. Relative to the initial orbital period $T_1$, $t=(T_1/2)[(r_1+r_2)/(2r_1){]}^{3/2}$. Radii must be measured from the attracting centre, rather than from its surface.

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.

For an isolated two-body problem in Newtonian gravity, the relative elliptic orbit's orbital period $T$ obeysHere $a$ is the relative semi-major axis, and $M+m$ is the sum of the two masses. The separation $r$ varies during an elliptic orbit, so replacing $a$ by $r$ works only for a circular orbit, where $r=a$. Likewise, either body's individual semi-major axis about the centre of mass must first be converted to the relative one. When $m$ is negligible compared with $M$, the denominator can be approximated by $GM$.

For fixed gravitational parameter, the orbital period depends on the semi-major axis and is independent of orbital eccentricity. Our $a=1$, $\mu =1$ example gives $T=2\pi$ both for $e=0.6$ and for a circular orbit with $e=0$. The eccentric body's rapidly changing speed compensates for its different path. A compact way to see the cancellation uses Kepler's second law: divide the ellipse's area $\pi ab$ by its areal velocity $h/2$. With $b=a\sqrt{1-e^2}$ and specific angular momentum $h=\sqrt{\mu a(1-e^2)}$, the result is $2\pi \sqrt{a^3/\mu }$.

Compare two separate Kepler orbits with the same gravitational parameter, $a_1=1$ and $a_2=4$:With $\mu =1$, their orbital periods are $2\pi$ and $16\pi$; measured in units of $T_1$, these are $1$ and $8$. Quadrupling the size therefore takes eight times as long to complete a circuit. This power law makes comparisons possible without first choosing physical units.

The same relation can be inverted to weigh a system: $M+m=4{\pi }^2{a}^3/(G{T}^2)$. An observed relative semi-major axis and orbital period determine the total mass, while additional information is needed to separate the two masses. This is useful for binary stars, as discussed in NASA, Orbits and Kepler's Laws: measuring masses. Geometry and a clock together become a weighing instrument.

Nicolaus Copernicus reorganized planetary astronomy by placing Earth among the planets moving around the Sun. His 1543 De revolutionibus orbium coelestium by Nicolaus Copernicus treated terrestrial motion as part of the explanation of observed celestial appearances.

This changes the geometry of apparent backward planetary motion. The observer is moving too: when Earth overtakes an outer planet, the changing relative position can reverse the observed direction of its progress against distant stars. Copernicus's system still used combinations of circular motion; the heliocentric model did not already contain Johannes Kepler's elliptic orbits.^[ref]

Orbital energy is the mechanical energy associated with relative motion and mutual attraction. In the isolated Newtonian two-body problem, let $\mu =m_1{m}_2/(m_1+m_2)$ be the reduced mass, $r$ the separation and $v$ the relative speed. ThenDividing by $\mu$ gives specific orbital energy. With zero potential energy at infinity, negative $E$ gives a bound Kepler orbit, zero gives the parabolic limit and positive $E$ gives an unbound encounter.

For an elliptic orbit, $E=-G{m}_1{m}_2/(2a)$ with semi-major axis $a$. Energy loss can therefore shrink a bound orbit, whereas a sufficiently large gain can unbind it. External forces or interactions with additional bodies invalidate conservation of the isolated pair's orbital energy.

The instantaneous distance from an attracting centre to an orbiting body. It is constant on a circular orbit but varies along an elliptic orbit. For Kepler's third law the relevant size is the semi-major axis, which equals the orbital radius only for a circular orbit.

For unit mass, the inverse-square force has potential energy $-k/r$, with zero potential at infinity. Conservation of angular momentum gives $r^2\dot{\theta }=h$. Differentiate the polar equation of the Kepler orbit:Consequently the total mechanical energy iswhere $k=h^2/\ell$. This also explains the sign distinction between an elliptic orbit, a parabolic Kepler orbit, and a hyperbolic Kepler orbit.

At the original periapsis, the parabolic Kepler orbit has $e=1$, $r=d=\ell /2$, and purely tangential speed $V$. ThusThe outward radial impulse has zero moment about the Sun, so it preserves angular momentum and hence $\ell$. Its additional velocity is perpendicular to the original velocity, so the new mechanical energy is $E^{\prime }={\alpha }^2{V}^2/2$. Applying the energy formula givesFor $\alpha >0$, the new orbit is a hyperbolic Kepler orbit; $\alpha =0$ leaves the parabola unchanged.

To fix the orientation of the sketch, put the Sun at the origin, the impulse point at $(d,0)$, and take the original motion there upwards. The new velocity is $(\alpha V,V)$. The eccentricity vector isIt points towards the new periapsis, which is rotated clockwise through $\beta =\arctan (2\alpha )$ from the original periapsis direction. Thus, measured using the original polar angle $\theta$, the new orbit isIn particular, the impulse point is already on the outgoing part of the new orbit and is not its new periapsis. The old parabola satisfies $x=d-y^2/(4d)$. Both orbits pass through $(d,0)$ and have the same focus, but the hyperbola's new periapsis axis is tilted.

The transverse velocity is $v_{\theta }=h/r$. Subtracting its square from the vis-viva equation, and putting $u=a/r$, givesConsequentlyThe positive branch is outward motion and the negative branch is inward motion. The PDF's unsigned expression gives the radial speed, or the outward branch, rather than the signed radial velocity everywhere. For example, on the inward part of any noncircular elliptic orbit, $\dot{r}<0$.

The same result follows from $\dot{r}=(\mu e/h)\sin f$, with $f$ the true anomaly. The radicand is nonnegative precisely on $1/(1+e)\le u\le 1/(1-e)$; it vanishes at periapsis and apoapsis. For a circular Kepler orbit, $e=0$, $u=1$ and $\dot{r}=0$.

Use the separation vector $\mathbf{r}$ from the star to the comet, and let $m_{\mathrm{red}}=m{M}_{\star }/(m+M_{\star })$. The two-body problem separates into uniform centre of mass motion andTaking a cross product with $\mathbf{r}$ proves conservation of angular momentum; taking the scalar product with $\dot{\mathbf{r}}$ proves conservation of energy. The relative constants areThe total relative angular momentum and energy are $m_{\mathrm{red}}\mathbf{h}$ and $m_{\mathrm{red}}\epsilon$, respectively. The conserved eccentricity vector is $\mathbf{e}=\dot{\mathbf{r}}\times \mathbf{h}/\mu -\hat{\mathbf{r}}$. Its scalar product with $\hat{\mathbf{r}}$ gives $r=h^2/[\mu (1+e\cos f)]$, with $f$ the true anomaly. The periapsis and apoapsis distances are therefore $h^2/[\mu (1+e)]$ and $h^2/[\mu (1-e)]$. Their sum is twice the semi-major axis, soAt either apsis $v=h/r$; substituting an apsidal radius gives $\epsilon =-\mu /(2a)$. Thus the specific orbital energy implies the vis-viva equationThese are exact relative-coordinate formulas for an elliptic orbit, $0\le e<1$. In the barycentric frame the comet's speed is $M_{\star }/(M_{\star }+m)$ times the relative speed; the printed identification with the comet's stellar orbital speed uses $M_{\star }\gg m$.

Put $A=\beta \mu /c$. The Poynting–Robertson drag acceleration has $\bar{R}=-2A\dot{r}/r^2$ and $\bar{T}=-A{v}_{\theta }/r^2$. Its work is strictly dissipative:For an unperturbed elliptic orbit, use $dt=r^{2}df/h$, $\dot{r}=(\mu /h)e\sin f$ and $v_{\theta }=(\mu /h)(1+e\cos f)$. Averaging over its orbital period $P$ givesSince $\dot{a}=2a^2\dot{\epsilon }/\mu$, $P=2\pi a^{3/2}/\sqrt{\mu }$ and $h^3={\mu }^{3/2}{a}^{3/2}(1-e^2{)}^{3/2}$, the Poynting–Robertson decay of a circumstellar orbit isFor a circular Kepler orbit this becomes $-2A/a$. This is a secular first-order average, not an exact instantaneous decay law. If static radiation pressure is appreciable, use a bound conservative orbit with central parameter ${\mu }_{\mathrm{eff}}=(1-\beta )\mu$, requiring $0\le \beta <1$, and define its orbital elements accordingly; repeating the work calculation cancels ${\mu }_{\mathrm{eff}}$ and leaves the same coefficient $A=\beta G{M}_{\star }/c$.

Let $h=r^2\dot{f}$ be the specific angular momentum. The radial Kepler orbit equation has semi-latus rectum $p=h^2/\mu$, so comparison withgives $h^2=\mu a(1-e^2)$ for an elliptic orbit and $h^2=\mu a(e^2-1)$ for a hyperbolic Kepler orbit. At either apsis, $\dot{r}=0$ and $v=h/r$. Substituting the apsidal radius and angular momentum into part (i), or equivalently using the vis-viva equation, givesThus the signs are $C=\mp \mu /(2a)$ in the convention of the question.

For a differentiable central potential, the force is $\mathbf{F}=-V^{\prime }(r)\hat{\mathbf{r}}$. The angular momentum is $\mathbf{J}=m\mathbf{r}\times \dot{\mathbf{r}}$, soThis proves conservation of angular momentum. For $J\ne 0$, the trajectory lies in the fixed plane perpendicular to $\mathbf{J}$. In plane polar coordinates, $J=m{r}^2\dot{\varphi }$. Substituting this into the kinetic energy gives the radial energy equationThe second term is the centrifugal contribution to the effective potential. Its radial derivative reproduces $m\ddot{r}=-V^{\prime }(r)+J^2/(m{r}^3)$. Purely radial motion is the $J=0$ case.

For the satellite, write $\mu =GM$, with $M$ Earth's mass and $m$ the satellite's mass. At apogee and perigee, the radial velocity is zero, so angular momentum conservation and conservation of energy giveEliminating $v_{\mathrm{P}}$ yieldsFor an elliptic orbit, $r_{\mathrm{A}}+r_{\mathrm{P}}=2a$. Therefore the vis-viva equation at apogee isSubstitution into the orbital energy gives $E=-GMm/(2a)$. The circular case follows by continuity, or directly from the radial force balance.

An instantaneous tangential retrograde burn at apogee leaves the position unchanged. For the speed change $v\mapsto (1-ϵ)v$, the orbital energy change isDifferentiating $E=-\mu m/(2a)$ gives $\Delta E=\mu m\Delta a/(2a^2)$ to first order. Since $r_{\mathrm{A}}=a(1+e)$, the vis-viva equation implies $v^2=\mu (1-e)/[a(1+e)]$. HenceThe factor of two distinguishes the major axis from the semimajor axis. As a finite-change check, the new semimajor axis is exactlywithin the instantaneous-burn, unchanged-mass model. Expanding this expression gives the stated result.

For a satellite of mass $m$ moving in the Newtonian gravitational field of a much larger, effectively fixed mass $M$, put $\mu =GM$. Its conserved mechanical energy, with gravitational potential energy zero at infinity, isHere $r$ is the instantaneous centre-to-centre distance, $v$ the satellite's speed, $G$ the gravitational constant, and $a$ the semi-major axis of its bound elliptic orbit. The kinetic energy and potential energy vary around the orbit, while their sum remains constant. Dividing by $m$ and solving for $v$ gives the vis-viva equation, $v^2=\mu (2/r-1/a)$.

For this Hohmann transfer orbit, the periapsis and apoapsis distances are $r_1=1\mathrm{AU}$ and $r_2=1.52\mathrm{AU}$, soThe vis-viva equation gives the endpoint speedsAs a check, conservation of angular momentum at the two endpoints gives $r_1{v}_{\mathrm{p}}=r_2{v}_{\mathrm{a}}$ because each velocity is perpendicular to its radius. Using $1\mathrm{AU}=1.49598\times 10^{11}\mathrm{m}$ and $\mu =1.32712\times 10^{20}{\mathrm{m}}^3{\mathrm{s}}^{-2}$ gives $v_1=29.78\mathrm{km}{\mathrm{s}}^{-1}$ andThe departure impulse must increase the speed in the direction of the initial circular orbit:These are Sun-centred speeds; leaving the Earth's own gravitational field would require an additional calculation beyond this idealized solar orbit transfer.

At $r_2$, the desired circular orbit speed is $v_2=\sqrt{\mu /r_2}=24.16\mathrm{km}{\mathrm{s}}^{-1}$. The arrival impulse is thereforeThe required thrust on the spacecraft is forward, tangent to its orbit, to accelerate it. The rocket ejects exhaust backwards, opposite to the spacecraft's velocity; this distinguishes the direction of thrust from the direction in which the exhaust is fired.

The Hohmann transfer time is half the orbital period of the transfer ellipse. From Kepler's third law, relative to the Earth's one-year orbital period,Meanwhile Mars's orbital period is $T_M=1.52^{3/2}\mathrm{yr}=1.8740\mathrm{yr}$. The spacecraft arrives $\pi$ radians from its departure position. For a rendezvous, Mars must reach that same position after advancing through $2\pi t_H/T_M$. Thus its initial lead angle over the departing spacecraft isMars must be ahead by this angle in the common direction of motion around their orbits. At departure the spacecraft and the Earth share the inner orbital position; arrival occurs on the opposite side of the Sun.

Assume the station's mass remains fixed through an instantaneous kick, with subsequent motion governed only by the Earth's Newtonian gravitational field. Put $\kappa =G{M}_E$, and work per unit station mass. The initial circular orbit has specific angular momentum $j$ and specific orbital energy ${\epsilon }_0$ given byLet $\delta$ denote the fractional increase of kinetic energy. The position and potential energy are unchanged during the kick, soUnchanged angular momentum at unchanged position fixes the tangential velocity; the added kinetic energy must be radial. Thus this is a radial kick, not a purely tangential speed increase. The new orbital eccentricity follows fromAt either radial turning point, the Kepler radial energy equation gives $\epsilon =j^2/(2r^2)-\kappa /r$. Its two roots, equivalently the periapsis and apoapsis of the elliptic orbit, areFor $\delta =0.01$, $e=0.1$ andAn inward or outward kick gives the same pair of radial turning points, although it places the station at a different phase of the orbit. These two finite positive radii require $0\le \delta <1$.

Classification of attractive inverse-square orbits. Write the potential energy as $U=-\kappa /r$, with $\kappa =GMm>0$ and zero at infinity. For nonzero angular momentum $J$, the Kepler orbit is a conic,Negative total orbital energy gives an elliptic orbit, $0\le e<1$. Its special case $e=0$ is the circular Kepler orbit, at the minimum energy permitted for that $J$, $E=-m{\kappa }^2/(2J^2)$. Zero energy gives a parabolic Kepler orbit, $e=1$, and positive energy a hyperbolic Kepler orbit, $e>1$. For $J=0$, the motion is radial: these are degenerate limiting trajectories and may strike the centre or the finite planetary surface.

The scattering geometry. The impact parameter $b$ is the perpendicular offset of the undeflected incoming asymptote from the centre, not the actual periapsis radius $r_{\min }$. Gravity bends the incoming path toward the planet.

The grazing threshold. At infinity, conservation of energy and angular momentum give $E=m{v}_{\infty }^2/2$ and $J=mb{v}_{\infty }$. At closest approach the radial speed is zero, so the velocity is tangential andEliminating $v_p$ givesThis increases monotonically with $r_{\min }>0$. Setting $r_{\min }=R$ gives the grazing impact parameter in gravitational focusing,Here $v_{\mathrm{esc}}=\sqrt{2GM/R}$ is the surface escape speed. The expression has units of length. Equality is the tangency limit; a strictly noncolliding trajectory requires $b>b_{\min }$. At high speed, $b_{\min }\to R$, while at low speed $b_{\min }\sim \sqrt{2GMR}/v_{\infty }$. Thus gravitational focusing enlarges the collision cross-section to $\pi b_{\min }^2$.

Deflection angle. For this hyperbola,The asymptotes occur at $\cos {\varphi }_{\infty }=-1/e$. The change of velocity direction is $\chi =2{\varphi }_{\infty }-\pi$, giving $\sin (\chi /2)=1/e$. Hence the gravitational scattering angle satisfiesFor the grazing trajectory this isIt is the attractive counterpart of the Rutherford impact-parameter relation; the sign of the transverse bending changes, but the positive deflection magnitude has this same form.

Range of angles. Along the grazing family, varying $v_{\infty }>0$ givesThe limits are $\chi \to 0$ as $v_{\infty }\to \infty$ and $\chi \to \pi$ as $v_{\infty }\to 0^{+}$. Neither endpoint is reached at finite positive speed. The low-speed limiting orbit still has a finite periapsis, but an increasingly distant incoming asymptote.