True Anomaly From Radius⚠ unverified

Understanding the Parameters

• True anomaly $\nu$ — the position angle — measured at the focus (the primary) from the perigee direction to the body's current position. $\nu = 0^\circ$ at perigee (closest), $\nu = 180^\circ$ at apogee (farthest). It is the "true" anomaly because it is the actual geometric angle, as opposed to the mean and eccentric anomalies (mathematical constructs used for timing — see Time Since Perigee).

• The three radii of an ellipse — perigee $r_p = a(1-e)$, apogee $r_a = a(1+e)$, and the semi-latus rectum $p = a(1-e^2)$ at $\nu = 90^\circ$. Any measured $r$ between $r_p$ and $r_a$ corresponds to a specific $\nu$; this formula finds it.

• Why the numerator is $a(1-e^2)/r - 1$ — it is the orbit equation solved for $e\cos\nu$: from $r = a(1-e^2)/(1+e\cos\nu)$, the quantity $a(1-e^2)/r = 1 + e\cos\nu$, so $a(1-e^2)/r - 1 = e\cos\nu$, and dividing by $e$ isolates $\cos\nu$. The whole formula is one algebraic rearrangement of the orbit equation.

• The quadrant ambiguity — a body at radius $r$ on the way out from perigee and one at the same $r$ on the way back in have true anomalies $\nu$ and $360^\circ - \nu$ (or equivalently $\pm\nu$). Radius alone cannot tell them apart, because the orbit is symmetric about the perigee–apogee line. The sign of the radial velocity (climbing vs falling) breaks the tie — information this single-input calculator does not have.

• Circles have no perigee — for $e = 0$ the radius is constant, so no angle can be inferred from it, and true anomaly loses meaning (any reference direction is arbitrary). The calculator returns $0$ in that degenerate case.

Derivation (Approaching a Proof)

Start from the orbit equation (the conic-section solution of the two-body problem), which gives the radius as a function of true anomaly:

$$r = \frac{a(1 - e^2)}{1 + e\cos\nu}.$$

The numerator $a(1-e^2) = p$ is the semi-latus rectum — the radius at $\nu = 90^\circ$. To get $\nu$ from $r$, solve this for $\cos\nu$. Multiply both sides by $(1 + e\cos\nu)$ and divide by $r$:

$$1 + e\cos\nu = \frac{a(1 - e^2)}{r}.$$

Isolate $e\cos\nu$:

$$e\cos\nu = \frac{a(1 - e^2)}{r} - 1.$$

Divide by $e$ (valid for $e > 0$):

$$\cos\nu = \frac{a(1 - e^2)/r - 1}{e}.$$

Finally invert the cosine:

$$\nu = \arccos\!\left(\frac{a(1 - e^2)/r - 1}{e}\right). \qquad\blacksquare$$

The $\arccos$ is the source of the quadrant ambiguity: it returns the principal value in $[0, \pi]$, so it yields $\nu$ but not whether the body is inbound or outbound. Resolving that requires the sign of $\dot r$: $\dot r > 0$ (receding from perigee) means $0 < \nu < 180^\circ$; $\dot r < 0$ (approaching perigee) means $180^\circ < \nu < 360^\circ$.

Dimensional check. $a(1-e^2)/r$ is $\text{m}/\text{m} = $ dimensionless; subtracting 1 and dividing by the dimensionless $e$ leaves a dimensionless argument for $\arccos$, whose output is an angle ✓.

History and Development

• The word "anomaly." The term is a fossil of pre-Keplerian astronomy: ancient and medieval astronomers called a planet's irregular (non-uniform) motion its "anomaly." Kepler's ellipse explained the irregularity, but the name survived for the position angle.

• Kepler's problem. Relating position to time on an ellipse — Kepler's second law made concrete — is the famous Kepler problem, solved through the chain true anomaly ↔ eccentric anomaly ↔ mean anomaly ↔ time (see Time Since Perigee). This page handles the geometric half (radius ↔ true anomaly); the timing half is the transcendental part that has occupied mathematicians from Kepler to today.

• Orbit determination. Recovering orbital angles from range measurements is the everyday work of tracking networks: a radar range, combined with range-rate (which supplies the missing $\dot r$ sign), pins down where a satellite is on its orbit. The quadrant ambiguity here is exactly why range and range-rate are both measured.

Related Concepts: Time Since Perigee, Eccentricity Vector, Orbital Velocity, Orbital Period, Vis Viva Energy, Circular Orbit Velocity

Notes: Inverts the orbit equation $r=a(1-e^2)/(1+e\cos\nu)$ for the angle. $\nu=0$ at perigee, $180^\circ$ at apogee. Output is degrees (mislabelled dimensionless); returns 0 for $e\le0$ (circle has no perigee). Quadrant-ambiguous: $\arccos\in[0,180^\circ]$ can't tell inbound ($-\nu$) from outbound ($+\nu$) — needs the sign of $\dot r$. Geometric half of the Kepler problem; timing half is Time Since Perigee.