Eccentricity Vector⚠ unverified

Understanding the Parameters

• Eccentricity as shape — the orbit's defining shape parameter. $e = 0$: circle (constant radius). $e$ near 0: gentle ellipse. $e$ approaching 1: a long, thin ellipse skimming close at perigee and flung far at apogee (comets, Molniya orbits). $e = 1$: parabola, the escape boundary. $e > 1$: hyperbola, a flyby that never returns. Together with the semi-major axis $a$ (size) and the angles (orientation), $e$ is one of the six classical orbital elements.

• Flight-path angle $\gamma$ — the tilt of the velocity vector relative to the local horizontal (the direction perpendicular to the radius). At perigee and apogee, velocity is purely horizontal so $\gamma = 0$ (and $\cos\gamma = 1$); in between, the orbit is climbing or descending and $\gamma \ne 0$. Only the horizontal component of velocity carries angular momentum — which is why $\cos\gamma$, not $\sin\gamma$, enters the formula (the F22 fix).

• The dimensionless group $rv^2/\mu$ — the ratio of (twice the) kinetic energy to the potential depth, a measure of how energetic the state is. For a circular orbit $rv^2/\mu = 1$; the bracket $(rv^2/\mu)(rv^2/\mu - 2)$ then vanishes at $\gamma = 0$, correctly giving $e = 0$. For $rv^2/\mu = 2$ (escape, $v^2 = 2\mu/r$) the expression gives $e = 1$, a parabola — the two limits the formula must hit.

• What the vector would add — the true eccentricity vector $\mathbf{e}$ points from the primary toward perigee and has magnitude $e$. It fixes not just the shape but where in space the orbit's closest approach lies (the argument of perigee). This calculator drops the direction, keeping only the magnitude — the reason its name overpromises.

Derivation (Approaching a Proof)

The result follows from the two conserved quantities of a two-body orbit: specific energy $\varepsilon$ and specific angular momentum $h$. The key identity relating them to eccentricity is

$$e^2 = 1 + \frac{2\varepsilon h^2}{\mu^2}.$$

This comes from the orbit equation (the conic-section solution of the two-body problem), where the semi-major axis is $a = -\mu/2\varepsilon$ and the semi-latus rectum is $p = h^2/\mu = a(1-e^2)$; eliminating $a$ and $p$ gives the boxed relation. Now express $\varepsilon$ and $h$ from the state $(r, v, \gamma)$.

Specific energy is the vis-viva form (Vis Viva Energy):

$$\varepsilon = \frac{v^2}{2} - \frac{\mu}{r}.$$

Specific angular momentum is the cross product $|\mathbf{r}\times\mathbf{v}|$. Only the component of velocity perpendicular to the radius — the horizontal component $v\cos\gamma$ — contributes:

$$h = r\,v\cos\gamma \quad\Longrightarrow\quad h^2 = r^2 v^2\cos^2\gamma.$$

Substitute both into $e^2 = 1 + 2\varepsilon h^2/\mu^2$:

$$e^2 = 1 + \frac{2\left(\dfrac{v^2}{2} - \dfrac{\mu}{r}\right) r^2 v^2\cos^2\gamma}{\mu^2} = 1 + \frac{\left(v^2 - \dfrac{2\mu}{r}\right) r^2 v^2\cos^2\gamma}{\mu^2}.$$

Factor $r/\mu$ to build the dimensionless group $rv^2/\mu$. Write $v^2 - 2\mu/r = \dfrac{\mu}{r}\!\left(\dfrac{rv^2}{\mu} - 2\right)$:

$$e^2 = 1 + \frac{\dfrac{\mu}{r}\!\left(\dfrac{rv^2}{\mu} - 2\right) r^2 v^2\cos^2\gamma}{\mu^2} = 1 + \frac{r v^2}{\mu}\left(\frac{r v^2}{\mu} - 2\right)\cos^2\gamma.$$

Taking the square root gives the calculator's formula:

$$e = \sqrt{1 + \frac{r v^2}{\mu}\left(\frac{r v^2}{\mu} - 2\right)\cos^2\gamma}. \qquad\blacksquare$$

The $\cos^2\gamma$ (not $\sin^2\gamma$) is now visible as the square of the horizontal velocity fraction that carries angular momentum — exactly the F22 correction.

Dimensional check. $rv^2/\mu = \dfrac{\text{m}\cdot(\text{m}/\text{s})^2}{\text{m}^3/\text{s}^2} = \dfrac{\text{m}^3/\text{s}^2}{\text{m}^3/\text{s}^2} = 1$ (dimensionless), and $\cos^2\gamma$ is dimensionless, so $e$ is dimensionless ✓ — as an eccentricity must be.

History and Development

• Kepler's ellipse (1609). Eccentricity entered astronomy with Kepler's first law: planets move on ellipses with the Sun at a focus. Mars's comparatively large $e \approx 0.093$ was what let Kepler detect the departure from a circle in Tycho's data at all.

• The conic-section solution. Newton (and later the analytical mechanicians) showed that the inverse- square force yields conic-section orbits whose eccentricity is set by energy and angular momentum — the relation $e^2 = 1 + 2\varepsilon h^2/\mu^2$ used above. Eccentricity thus became one of the six orbital elements that completely specify an orbit.

• The eccentricity vector. In modern astrodynamics the eccentricity vector $\mathbf{e} = \dfrac{\mathbf{v}\times\mathbf{h}}{\mu} - \dfrac{\mathbf{r}}{r}$ (the Laplace–Runge–Lenz vector, scaled) is a conserved vector pointing to perigee — a hallmark of the special symmetry of the inverse-square problem, and the reason Keplerian orbits do not precess in the ideal two-body case. Perturbations like Earth's oblateness break that symmetry and make the eccentricity vector slowly rotate.

Related Concepts: Vis Viva Energy, Orbital Velocity, True Anomaly From Radius, Time Since Perigee, Orbital Period, Angular Momentum, Escape Velocity

Notes: Returns the scalar eccentricity magnitude $e$ (shape: $0$ circle, $<1$ ellipse, $1$ parabola, $>1$ hyperbola), not the eccentricity vector (which also gives orientation/perigee direction) despite the name. From $e^2=1+2\varepsilon h^2/\mu^2$ with $h=rv\cos\gamma$ — the $\cos^2\gamma$ (F22 fix; was $\sin^2$) is the horizontal velocity fraction. $\gamma$ in degrees, $\mu$ in m³/s² (both mislabelled dimensionless).