Celestial mechanics

Celestial mechanics studies motion under gravitational forces.

For a central-force radial energy equation, choose a monotone radius $q=q(r)$ and clock $\tau$ such that $d\tau /dt=(q/r)dq/dr$. Then $q^{\prime }=(r/q)\dot{r}$, and multiplying the old energy equation by $r^2/q^2$ preserves the centrifugal form $L^2/(2q^2)$. A split of the remaining terms into constant energy and a new Newtonian gravitational potential produces a transformed radial orbit.

A transformed radial equation containing $L^2/(2q^2)$ becomes a full planar central-force orbit by choosing $d\bar{\theta }/d\tau =L/q^2$. For $d\tau /dt=(q/r)dq/dr$ and old angular velocity $L/r^2$, this gives $d\bar{\theta }/d\theta =(r/q)dq/dr$. Keeping the old angle generally fails to preserve the stated transformed angular momentum.

With $a=(1-\lambda )GM$, $h=\lambda E>0$ and $\bar{E}>0$, set $\bar{E}{q}^2=h{r}^2+ar$. If $a\ge 0$, its positive-radius root is $r=\sqrt{\bar{E}/h}(\sqrt{q^2+b^2}-b)$, with $b=a/(2\sqrt{h\bar{E}})$. The transformed Newtonian gravitational potential is an additive constant plus $-G\bar{M}/(b+\sqrt{q^2+b^2})$, the spherical isochrone model. The mass sign and branch conditions must be checked.

For bound old Kepler orbits, $E<0$, $\lambda <0$ and $\bar{E}>0$ give a positive radius, positive isochrone scale $b$, and positive transformed mass. The shifted energy is $\bar{E}-k=\bar{E}/\lambda <0$. For unbound old orbits with $E>0$, $1/2<\lambda \le 1$ gives an attractive unbound isochrone branch. The case $\lambda =0$ is handled by the separate Kepler–harmonic radial duality.

Taking $\bar{E}=GMr/q^2>0$ in a radial orbit time transformation gives $r=(\bar{E}/GM)q^2$ and $\bar{\Phi }(q)=-E(\bar{E}/GM{)}^2{q}^2$. Bound Kepler orbits with $E<0$ map to a confining harmonic oscillator; positive energy maps to an inverted oscillator. Full angular reconstruction halves the original orbital angle.

The two-body problem determines the motion of two particles interacting only with one another. For a central force it separates into uniform centre-of-mass motion and one relative-coordinate problem with the reduced mass.

Two equal point masses can move around the same fixed circle of radius $a$ in diametrically opposite positions. Their stationary centre of mass is the circle's center. A separation $2a$ gives gravitational force $G{m}^2/(4a^2)$ on each mass, and Newton's second law equates it to $ma{\omega }^2$. The common angular speed is therefore fixed by the displayed relation. Opposite tangential velocities produce the same sense of angular rotation.

Two point masses released from rest at separation $a$ have negative mechanical energy $-G{m}_1{m}_2/a$ but collide in time $\pi \sqrt{a^3/[8G(m_1+m_2)]}$. Their separation approaches zero along a regular trajectory before collision. The energy sign alone therefore does not supply a positive lower separation bound.

An inverse-square central force produces conic-section trajectories with the force centre at a focus.

For an instantaneous in-plane change of velocity of magnitude $\gamma v_K$ on a circular orbit, with angle $\theta$ measured from the forward tangential direction, the new specific orbital energy gives $a^{\prime }/a=(1-2\gamma \cos \theta -{\gamma }^2{)}^{-1}$. The specific angular momentum gives $e^{\prime 2}=1-(1-2\gamma \cos \theta -{\gamma }^2)(1+\gamma \cos \theta {)}^2$. The signed semi-major axis becomes negative for an unbound hyperbolic Kepler orbit.

Uniform planar kick angle $\theta$ gives $c=\cos \theta$ with probability density $1/(\pi \sqrt{1-c^2})$. This maps onto a one-dimensional curve in orbital eccentricity–semi-major axis space. The two radial-kick signs share that curve but have opposite tangential components of the eccentricity vector. The resulting density is enhanced near tangential-kick endpoints.

For unit mass in $\Phi =-GM/r$, conservation of angular momentum splits the velocity norm into ${\dot{r}}^2+L^2/r^2$. Conservation of energy then gives $\frac{1}{2}{\dot{r}}^2+L^2/(2r^2)=E+GM/r$. It separates radial dynamics from the reconstruction of the orbital angle.

In the centre-of-mass frame of two Newtonian point masses, total orbital energy is reduced mass times specific orbital energy: $E=\mu [|\dot{\mathbf{r}}{|}^2/2-GM/r]$. The factor $\mu$ multiplies both kinetic and potential contributions, while $GM$ is the relative gravitational parameter.

For an attractive inverse-square force, $r=p/(1+e\cos f)$, where $f$ is true anomaly, $e$ is orbital eccentricity, and $p=h^2/\mu$ is the semilatus rectum. The specific angular momentum relation $\dot{f}=h/r^2$ converts this geometric equation into an angular evolution law.

An orbit about a planet. Stellar tides, radiation pressure and other perturbations can limit its stability; a sufficiently small semi-major axis compared with the Hill sphere permits a local approximately planet-centred Kepler orbit.

A circumplanetary orbit whose orbital period equals the central planet's spin period. Its semi-major axis follows from Kepler's third law. Matching periods alone does not fix the satellite above one ground location; that additionally requires a planet-stationary orbit.

A prograde, equatorial circular orbit with orbital period equal to the planet's spin period. In the ideal model its satellite remains above one equatorial ground location. Nonzero orbital inclination or orbital eccentricity destroys exact stationarity even when the periods agree.

The leading uniform stellar radiation pressure is a differential acceleration relative to a planet, of scale $\beta G{M}_{\star }/a_p^2$. Relative to planetocentric gravity at dust radius $a$, its magnitude isThus lying inside a Hill sphere alone does not guarantee small radiative perturbations. Approximately circular tightly bound dust requires this ratio sufficiently small, as well as weak tides; the uniform radiative force can otherwise excite orbital eccentricity or unbind grains.

A planet can block stellar radiation during part of a circumplanetary orbit. Then the full-orbit constant-flux averages for radiation pressure and Poynting–Robertson drag require modification. The detailed shadow duty cycle depends on the orbital inclination, stellar angular extent and dust radius.

An orbit about a star. Dust on such an orbit can experience radiation pressure and Poynting–Robertson drag, in addition to gravitational perturbations.

Binet’s equation rewrites a central-force orbit using $u(\theta )=1/r$. If $\ell =m{r}^2\dot{\theta }$ is angular momentum and the potential is $V(r)$, then

For radial acceleration $-A/r^2-B/r^3$, conserved specific angular momentum $l$ and $u=1/r$ reduce the Binet equation to the displayed linear equation. If $l^2>B$, its angular oscillation frequency is $\sqrt{1-B/l^2}$; attraction proportional to $r^{-3}$ changes the apsidal angle. A zero of $u$ ends the physical positive-radius branch and corresponds to escape, not passage to a negative radius. A spatial self-intersection need not be a periodic return because velocity may differ.

Constant areal velocity gives constant specific angular momentum $l=r^2\dot{\theta }$, hence zero transverse acceleration and a central force. For a focus-based ellipse $u=1/r=(1+\epsilon \cos \theta )/A$, one has $u_{\theta \theta }+u=1/A$. The Binet equation then gives the displayed radial acceleration, proving an attractive inverse-square law. An orbit's shape alone does not specify its acceleration without a time law.

An attractive inverse-square force together with radial drag $-k\dot{r}/r^2$ preserves angular momentum $h=r^2\dot{\theta }$ but modifies the Binet equation for $u=1/r$ by a first derivative. For $k/|h|<2$ and $h\ne 0$, the reciprocal radius is a decaying oscillation about $p/h^2$. Forward angular distance $\mathrm{sgn}(h)(\theta -{\theta }_0)$ makes its damping sign explicit for either orbital orientation.

A bounded trajectory in the damped Binet equation with nonzero angular momentum asymptotically becomes circular. Energy loss is nonnegative in magnitude and vanishes only for zero radial velocity. The centrifugal effective potential energy excludes collision; bounded radius forces indefinitely increasing forward angular distance, so the reciprocal-radius oscillation decays. Escaping trajectories are excluded by the boundedness hypothesis.

ForBinet’s equation isIts circular orbits solve $K=u(1-r_{0}u{)}^2$; the branch $u<1/(3r_0)$ is stable and the branch $u>1/(3r_0)$ is unstable.

Orbital eccentricity distinguishes circular, elliptic, parabolic, and hyperbolic conics.

For an attractive Kepler orbit with gravitational parameter $GM$, velocity $\mathbf{v}$ and specific angular momentum $\mathbf{h}=\mathbf{r}\times \mathbf{v}$, the eccentricity vector isIts magnitude is the orbital eccentricity, and it points toward periapsis for a noncircular orbit. It is a normalized attractive Laplace-Runge-Lenz vector; it remains constant under the pure inverse-square force.

If the relative acceleration is $\ddot{\mathbf{r}}=-GM\mathbf{r}/r^3+\mathbf{f}$, with constant masses, then the specific orbital energy $\epsilon$, specific angular momentum $\mathbf{h}$ and eccentricity vector obeyThe inverse-square-force terms cancel in each derivative. These identities express perturbing work and torque, and track the instantaneous change of the ellipse's shape and orientation.

An isotropic population of bound Kepler orbits has eccentricity probability density $n(e)=2e$ on $0\le e<1$. It follows from the fixed-energy angular-momentum measure $n(L)dL\propto LdL$ and $L=L_c\sqrt{1-e^2}$.

An elliptic orbit is a bound orbit shaped as an ellipse. A Kepler ellipse has negative specific orbital energy and eccentricity $0\le e<1$.

A positive-energy inverse-square orbit is a hyperbola with two scattering asymptotes.

The gravitational scattering angle is the change between incoming and outgoing asymptotic velocity directions.

For a local two-body hyperbolic encounter with central parameter $\mu$ and allowed centre separation $r_{\min }\ge r_0$, conservation of energy gives equal asymptotic speeds. The gravitational scattering angle therefore impliesPut $z=v_{\infty }\sqrt{r_{\min }/\mu }$ and use $2z/(1+z^2)\le 1$. The optimizing grazing orbit has $z=1$, orbital eccentricity two and deflection $\pi /3$. For finite spherical bodies $r_0$ is the sum of their radii; excluding actual contact makes the upper bound a supremum. Without a finite exclusion radius the ideal point-mass problem has no finite bound.

In a gravitational assist, a small body's speed relative to a moving massive body is deflected with nearly unchanged magnitude. Transforming back to the inertial frame can transfer orbital energy between the bodies.

A zero-energy inverse-square orbit is parabolic and has eccentricity one.

Barker’s equation relates time on a parabolic orbit to the tangent of half the true anomaly through a cubic.

A parabolic Kepler orbit obeys $t=\sqrt{2q^3/(GM)}(D+D^3/3)$ with $D=\tan (f/2)$. At the high-eccentricity angular-speed crossover, $q\ll r_x$, so $n_p{t}_x\simeq (\sqrt{2}/3)(r_x/a_p{)}^{3/2}=2^{7/8}(q/a_p{)}^{3/8}/3$. The small time fraction explains why an eccentric orbit turns through almost $\pi$ of true anomaly while a distant perturber barely moves.

The periapsis is the point of an orbit nearest its force centre.

The apoapsis is the point of an orbit farthest from its force centre.

The semi-major axis is one half of an ellipse's longest diameter. For a Kepler ellipse with apsidal distances $r_1,r_2$, it is $a=(r_1+r_2)/2$.

For an ellipse of eccentricity $e$, the semi-minor axis is $b=a\sqrt{1-e^2}$.

An orbital element is one of a set of parameters specifying an orbit at a chosen epoch. For a Kepler orbit, the semi-major axis and eccentricity determine its size and shape, while angular elements determine its orientation and the body's position.

The plane containing a central-force Kepler orbit, perpendicular to its conserved specific angular momentum. Its orientation can be specified by an orbital inclination and an orbital node. Perturbing forces can change that orientation.

An orbital element of the Kepler orbit having the same instantaneous position and velocity as a perturbed trajectory. Its elements vary as the perturbing force acts. Their interpretation depends on the central gravitational parameter used, for example whether static radiation pressure is included in the unperturbed attraction.

The true anomaly is the angle at the focus from periapsis to the current position of an orbiting body, measured in the direction of motion.

Orbital phase specifies where a body lies within its orbital cycle relative to a chosen reference event.

A phase-mixed orbit has a probability distribution uniform in its canonical angle variable. For a Kepler ellipse, this means uniform mean anomaly and therefore probability proportional to time spent in each orbital segment.

Uniform mean anomaly implies a line probability inversely proportional to the orbital speed. For small orbital eccentricity and true anomaly $f$, it is proportional to $1-e\cos f$.

The geometric cross-section passing a fixed point of a steady orbital wire per unit time is $J_{\sigma }={\lambda }_{\sigma }v$, where ${\lambda }_{\sigma }=d\sigma /ds$. Conservation of fragments requires $J_{\sigma }$ constant and ${\lambda }_{\sigma }\propto 1/v$. For total area ${\sigma }_{\mathrm{tot}}$ and orbital period $P$,With the fractional luminosity of a phase-mixed eccentric wire, this becomes $J_{\sigma }=2fh$. The current describes transported area, not the swept area enclosed by the orbit.

For a Kepler orbit with pericentre distance $q$, apocentre distance $Q$, and semi-major axis $a=(q+Q)/2$, a uniform mean anomaly givesThe two radial passages per orbital period are included, and ${\int }_q^Q{p}_r(r)dr=1$.

Mean anomaly is the uniformly increasing phase $M=n(t-\tau )$ of a Kepler orbit, measured from periapsis time $\tau$.

For an elliptic Kepler orbit, Kepler's equation relates mean anomaly $M$ and eccentric anomaly $E$ by

The eccentric anomaly parametrizes a Kepler ellipse through the corresponding point on its auxiliary circle. It gives radius $r=a(1-e\cos E)$.

An orbital node is an intersection of an orbital plane with its reference plane.

The ascending node is the orbital node where a body crosses from the negative to the positive side of the reference plane.

The longitude of ascending node is the reference-plane angle from a fixed direction to the ascending node.

Orbital inclination is the angle between an orbital plane and a reference plane.

In a right-handed reference frame, the rotation matrix from an orbital plane frame whose first axis points along the ascending node is $R_z(\Omega )R_x(I)$. Its columns are the orbital frame's orthonormal basis expressed in reference coordinates. The unit vector normal is $(\sin \Omega \sin I,-\cos \Omega \sin I,\cos I)$; its dot product with another such normal gives the cosine of their mutual inclination.

The mutual inclination is the angle between two orbital planes, equivalently the angle between consistently oriented orbital normal vectors.

For two nearly edge-on orbits with small orbital inclination difference $\Delta I$ and small longitude of ascending node difference $\Delta \Omega$, expanding the dot product of their normals gives $I_m^2=(\Delta I{)}^2+(\Delta \Omega {)}^2+O({\delta }^4)$, with angles in radians and all offsets from edge-on of order $\delta$. Thus equality of observed sky-plane inclinations alone does not imply aligned orbital planes.

The invariable plane is perpendicular to a system's total angular momentum. For an isolated planetary system, it is fixed while individual orbital planes precess.

For two groups carrying angular momenta $L_a,L_b$ with mutual inclination $\Delta I$, resolving total angular momentum parallel and perpendicular to the second group's normal gives its orbital inclination relative to the invariable plane:For a positive denominator, $\tan I_b=\sin \Delta I/(L_b/L_a+\cos \Delta I)$. The larger-angular-momentum group lies closer to the invariable plane.

An orbit is edge-on to an observer when its orbital plane contains the line of sight, so its inclination to the plane of the sky is $90^{\circ }$.

Projected observations that do not identify which node is ascending may determine an orbital plane only up to reversal of its normal or addition of $\pi$ to its nodal longitude.

The longitude of periapsis gives the angular direction of periapsis in the reference plane.

Mean longitude is an angle that advances uniformly at the mean motion in an unperturbed Kepler orbit.

Mean motion is the average orbital angular frequency. A Kepler orbit about gravitational parameter $\mu$ has $n^2{a}^3=\mu$.

The orbital period is the time required to complete one orbit. It is related to mean motion by $T=2\pi /n$.

The synodic period is the interval between successive repetitions of the relative orbital configuration of two bodies. For coplanar prograde mean motions $n_1,n_2$, it is $2\pi /|n_1-n_2|$.

The specific orbital energy is the mechanical energy per unit mass,Its sign distinguishes bound elliptic, marginal parabolic, and unbound hyperbolic Kepler orbits.

A weak perturbing relative acceleration $\mathbf{f}$ changes specific orbital energy at rate $\dot{\epsilon }=\mathbf{v}\cdot \mathbf{f}$. Since $\epsilon =-\mu /(2a)$ for a Kepler orbit,For a circular Kepler orbit under $\mathbf{f}=-\kappa \mathbf{v}+\mathbf{g}$ with constant $\mathbf{g}$, the constant-force term averages to zero over an unperturbed orbit. Thus $⟨\dot{a}⟩=-2\kappa a$. This is a leading-order orbital average; a strong constant force can distort the orbit instead.

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.

The specific angular momentum is angular momentum per unit mass, $\mathbf{h}=\mathbf{r}\times \mathbf{v}$. For a planar Kepler orbit its magnitude is $h=r^2\dot{\theta }=\sqrt{\mu a(1-e^2)}$.

For a nearly radial Kepler orbit with pericentre $q$, specific angular momentum is $h\simeq \sqrt{2GMq}$. Relative to a circular reference orbit of radius $a_p$ and mean motion $n_p$, the instantaneous azimuthal rate changes from faster to slower at $r_x=\sqrt{h/n_p}\simeq a_p(2q/a_p{)}^{1/4}$. In a rotating reference frame the polar angle has a turning point at this radius. The true anomaly there differs from $\pi$ by approximately $2\sqrt{q/r_x}$.

The specific angular momentum of an elliptic Kepler orbit is $h=\sqrt{\mu a(1-e^2)}$. At periapsis, $r=a(1-e)$, soThe mean motion understates instantaneous angular speed near periapsis. A particle in an exterior mean-motion resonance can briefly overtake the planet in angle despite completing full orbits more slowly.

A circular Kepler orbit at radius $r$ has speed $v_K=\sqrt{\mu /r}$ and mean motion $n=\sqrt{\mu /r^3}$.

A Hohmann transfer uses two tangential impulses and one half of an elliptic Kepler orbit to move between two coplanar circular orbits.

The restricted three-body problem follows a negligibly massive particle in the gravitational field of two massive bodies whose motion is unaffected by the particle.

In the circular restricted three-body problem the two massive bodies move on circular orbits about their barycentre. The problem is autonomous in their uniformly rotating frame.

The unforced planar equation of motion in a rotating frame near a circular Kepler orbit has components $\ddot{x}-2n\dot{y}-3n^{2}x=0$ and $\ddot{y}+2n\dot{x}=0$. They describe epicyclic motion and Keplerian shear; mutual gravity adds close-encounter forcing.

The oscillatory integration constants determine the orbital eccentricity vector, the constant radial offset determines the semi-major axis, and the constant along-track offset fixes the mean longitude. The radial and tangential epicycle amplitudes have ratio $1:2$.

The photogravitational restricted three-body problem adds radial radiation pressure from one or both luminous primaries. If the pressure-to-gravity ratio of primary $i$ is ${\beta }_i$, its effective attraction on the test particle is multiplied by $1-{\beta }_i$.

A Lagrange point is an equilibrium of the circular restricted three-body problem in the frame rotating with the two massive bodies.

The rotating-frame effective potential of the circular restricted three-body problem has negative singular wells at both primaries. Its three Collinear Lagrange points are planar saddles, and its two Triangular Lagrange points are planar local maxima for nonzero primary masses. These geometric labels do not replace a dynamical stability analysis including the Coriolis force.

The three collinear Lagrange points $L_1,L_2,L_3$ lie on the line through the two massive bodies.

In barycentric coordinates with unit primary separation, a small secondary mass fraction $\alpha$ gives distances $(\alpha /3{)}^{1/3}$ from that secondary to the two neighboring Lagrange points at leading order. This is the dimensionless Hill radius. The opposite exterior L3 Lagrange point has coordinate $-1-5\alpha /12+O({\alpha }^2)$.

The Lagrange point lying between the components of a circular restricted three-body problem. The critical effective potential through this saddle defines their Roche lobes; gas passing through it can initiate Roche-lobe overflow.

The collinear Lagrange point beyond the smaller primary, away from the larger one, is $L_2$. For a small mass ratio it lies approximately one Hill radius from the secondary and has a growing linear mode.

In unit binary separation and mean motion, displacements obey $\ddot{X}-2\dot{Y}=(1+2B)X$ and $\ddot{Y}+2\dot{X}=(1-B)Y$, where $B={\mu }_1/r_1^3+{\mu }_2/r_2^3$ at $L_2$. The characteristic polynomial is ${\lambda }^4+(2-B){\lambda }^2+(1+2B)(1-B)$. The small-secondary limit has $B\to 4$.

For a small secondary-to-primary mass ratio, the unstable exponent at the outer collinear equilibrium is $n_{\mathrm{bin}}\sqrt{1+2\sqrt{7}}$. Its inverse is approximately $0.0634$ binary periods. Reaching a finite target distance takes this time multiplied by the logarithm of the ratio of final to initial unstable amplitudes.

The $L_3$ Lagrange point lies beyond the more massive primary, on the side opposite the secondary.

The triangular Lagrange points $L_4,L_5$ complete equilateral triangles with the two massive bodies.

The Hill sphere is the neighborhood of a smaller body in which its gravity can dominate tidal perturbations from a larger primary.

For a body of mass $m$ on a circular orbit of radius $a$ about mass $M\gg m$, the Hill radius is approximately

The Jacobi constant is the conserved energy-like integral of the circular restricted three-body problem in the rotating frame.

An orbital resonance occurs when orbital frequencies satisfy an approximately integer relation, allowing perturbations to accumulate coherently.