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Celestial mechanics predicts the motion of planets, Moons, stars and spacecraft from gravity and other relevant forces. Its simplest useful approximation is the two-body problem, whose Kepler orbits explain the connection between orbital shape, energy and angular momentum.

Johannes Kepler's orbital description and Isaac Newton's inverse-square law meet in this model. Additional bodies introduce orbital perturbations; general relativity changes the prediction when relativistic corrections matter. Comparing a predicted trajectory with observations tests both the inferred masses and the adopted dynamical model.^[ref]

Suppose a rotating reference frame shares its origin with an inertial frame and rotates at constant angular velocity $\mathbf{\Omega }$. Let $\mathbf{x}$ be the relative position, with dots denoting derivatives of its rotating components. The two accelerations, expressed in the same axes, satisfyApply the derivative rule $(d\mathbf{A}/dt{)}_{\mathrm{inertial}}=(d\mathbf{A}/dt{)}_{\mathrm{rotating}}+\mathbf{\Omega }\times \mathbf{A}$ twice to the position vector to derive this relation. Newton's second law therefore gives the rotating-frame equationHere $\mathbf{F}$ contains the actual interaction forces; the extra terms are the Coriolis force and centrifugal force.^[ref]

For example, an object at rest on a rotating table still needs an inward interaction force to maintain its circular motion; in rotating coordinates it balances the outward centrifugal force. If the origin has acceleration ${\mathbf{a}}_O$, add $-m{\mathbf{a}}_O$ to the right side. If the angular velocity varies, also add the Euler force $-m\dot{\mathbf{\Omega }}\times \mathbf{x}$. These terms cannot be omitted merely because the object is instantaneously at rest in the rotating axes.

A gravitational perturbation is a change to a reference motion caused by an additional gravitational force or a correction to the assumed gravitational field. In celestial mechanics, the reference is often a two-body problem with a point-mass central attractor. Another planet or the nonspherical mass distribution of the central body can change the orbital elements. This is a gravitational cause of an orbital perturbation; atmospheric drag is another kind of cause.^[ref]

The third-body acceleration must be relative to the accelerating central body. Let a satellite be at $r$ relative to that body, and let a perturbing mass $m_3$ be at $R$. Subtracting the third body's attraction on the origin givesOmitting the second term would mistake common acceleration of the entire system for a change in the relative orbit. For $\Vert r\Vert \ll R_0=\Vert R\Vert$, expanding yields the tidal approximationThe perturbation stretches separation along $\hat{R}$ and compresses it in perpendicular directions. Its characteristic ratio to a central attraction $GM/\Vert r{\Vert }^2$ is $(m_3/M)(\Vert r\Vert /R_0{)}^3$. This explains why distance as well as mass determines whether a third body strongly disturbs a satellite. Small instantaneous ratios can still produce appreciable accumulated changes over many orbital periods.^[ref]

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.

The Jacobi constant is a conserved quantity for the circular restricted three-body problem. In barycentric coordinates rotating at constant angular frequency $n$, defineThe equation of motion in a rotating frame is ${\dot{\mathbf{v}}}_{\mathrm{rot}}+2\mathbf{\Omega }\times {\mathbf{v}}_{\mathrm{rot}}=\nabla U$. Taking its dot product with ${\mathbf{v}}_{\mathrm{rot}}$ proves $d(|{\mathbf{v}}_{\mathrm{rot}}{|}^2/2-U)/dt=0$: the Coriolis force does no work.

Only positions satisfying $2U-C\ge 0$ are accessible. The boundary gives a zero-velocity surface; openings near Lagrange points allow transit between regions but do not force it. Ordinary inertial-frame energy can change even while $C$ remains constant.^[ref]

A Kepler orbit is a relative trajectory satisfyingFor nonzero specific angular momentum $h$, conservation of angular momentum confines motion to a plane, and the orbit is a conic section:Here $e$ is orbital eccentricity and $\nu$ the true anomaly. Negative specific orbital energy gives a bound ellipse; zero or positive energy gives a parabolic orbit or hyperbolic orbit. Other forces make the exact two-body description an approximation.^[ref]

Orbital eccentricity $e$ measures the shape of a Kepler orbit. A circle has $e=0$, an ellipse $0<e<1$, a parabola $e=1$, and a hyperbola $e>1$ for a nonradial attractive orbit.

For a bound ellipse with semi-major axis $a$, pericentre $r_p=a(1-e)$ and apocentre $r_a=a(1+e)$ giveThus unequal turning distances measure eccentricity directly. In a perturbed orbit, osculating orbital elements can change with time; eccentricity need not remain conserved.^[ref]

The specific orbital energy is the relative mechanical energy per unit reduced mass in the Newtonian two-body problem. For total mass $M=M_1+M_2$, gravitational parameter ${\mu }_g=GM$, separation $r$ and relative speed $v$, it isThe gravitational potential is taken to vanish at infinite separation. The last equality uses a signed semi-major axis $a$: positive for a bound ellipse and negative for a hyperbolic Kepler orbit. A parabolic Kepler orbit has $\epsilon =0$ and no finite semi-major axis. For nonradial attractive Kepler orbits, these signs distinguish the three orbit classes.^[ref]

For a circular orbit, $v^2={\mu }_g/r$ gives $\epsilon =-{\mu }_g/(2r)$. For an escaping hyperbolic Kepler orbit, taking $r\to \infty$ gives the hyperbolic excess speed $v_{\infty }=\sqrt{2\epsilon }$. Multiplication by the reduced mass gives the two-body orbital energy; dividing by a different mass would mix conventions. The units of specific orbital energy are $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}={\mathrm{m}}^2{\mathrm{s}}^{-2}$.

A three-body orbit is a trajectory governed by the gravitational three-body problem. For three point masses at positions ${\mathbf{r}}_i$, Newton's law of universal gravitation givesBoth attracting bodies act simultaneously; switching between separate Kepler orbits is instead a patched conic approximation. In the restricted three-body problem, one mass is negligible, so its trajectory responds to the two primary bodies without appreciably changing their motion. Taking their mutual orbit to be circular gives the circular restricted three-body problem.^[ref]

For example, an asteroid influenced by the Sun and a planet can deviate from any single fixed Kepler orbit. Around a synchronously rotating Moon, a maintained trajectory involving the planet–moon system is likewise distinct from a simple moon-centred stationary Kepler orbit. The Hill sphere estimates where the moon's local gravitational dominance can support that simpler approximation; it is not a sharp boundary beyond which all three-body trajectories are impossible.