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SCIENCE · INVESTIGATE AND EXPLAIN

Reconstruct an orbit from two turning radii

At an apsis of a noncircular orbit, radial velocity vanishes and speed is tangential. With specific angular momentum h, that speed is h/r. The same total energy must therefore equal h²/(2rp²) − μ/rp at pericenter and h²/(2ra²) − μ/ra at apocenter.

Learning outcome: Derive energy and angular momentum from both apsides, check both energy substitutions, and identify the circular limiting case.

Starting recap

Integrate mass moments about named axes, derive torque and angular-momentum constraints, and test orbital energy, angular and numerical consistency against independent evidence.

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Try

For rp = ra = 3 m and μ = 1, the orbit is circular: eccentricity zero, speed 1/sqrt(3) m/s, energy −1/6 m²/s² and period 2π sqrt(27) s.

Open the experiment

Predict

For rp = 1 and ra = 3 in the supplied μ = 1 orbit, pericenter/apocenter speed ratio is:

Experiment

Derive energy and angular momentum from both apsides, check both energy substitutions, and identify the circular limiting case.

Reconstruct an ellipse from two turning radii

Worked teaching model · synthetic values · Ideal test-particle Newtonian orbit with fixed mu=1. Selected radii satisfy rp<=ra, including a circular equality. Radii are from the force center, not ellipse center. No real mission design or body surfaces.

Reset keeps your prediction, captures and practice history.

Ellipse with force center at origin

north coordinate (m)east coordinate (m)-311.732051-1.732051

Spatial axes use the same pixels per physical unit, preserving geometry. 1: geometric orbit, not equally timed markers (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Ellipse with force center at origin

geometric orbit, not equally timed markers

east coordinate (m)north coordinate (m)
10
0.99383470.1358951
0.97537670.2709524
0.94473980.4043392
0.9021130.5352331
0.84775910.6628271
0.7820130.7863346
0.70528030.9049941
0.6180341.018074
0.52081191.124877
0.41421361.224745
0.29889611.317062
0.17557051.401259
0.044997131.476816
-0.0920191.543269
-0.23463311.600206
-0.3819661.647278
-0.53310931.684194
-0.68713111.710726
-0.84308181.726711
-11.732051
-1.1569181.726711
-1.3128691.710726
-1.4668911.684194
-1.6180341.647278
-1.7653671.600206
-1.9079811.543269
-2.0449971.476816
-2.1755711.401259
-2.2988961.317062
-2.4142141.224745
-2.5208121.124877
-2.6180341.018074
-2.705280.9049941
-2.7820130.7863346
-2.8477590.6628271
-2.9021130.5352331
-2.944740.4043392
-2.9753770.2709524
-2.9938350.1358951
-32.12115e-16
-2.993835-0.1358951
-2.975377-0.2709524
-2.94474-0.4043392
-2.902113-0.5352331
-2.847759-0.6628271
-2.782013-0.7863346
-2.70528-0.9049941
-2.618034-1.018074
-2.520812-1.124877
-2.414214-1.224745
-2.298896-1.317062
-2.175571-1.401259
-2.044997-1.476816
-1.907981-1.543269
-1.765367-1.600206
-1.618034-1.647278
-1.466891-1.684194
-1.312869-1.710726
-1.156918-1.726711
-1-1.732051
-0.8430818-1.726711
-0.6871311-1.710726
-0.5331093-1.684194
-0.381966-1.647278
-0.2346331-1.600206
-0.092019-1.543269
0.04499713-1.476816
0.1755705-1.401259
0.2988961-1.317062
0.4142136-1.224745
0.5208119-1.124877
0.618034-1.018074
0.7052803-0.9049941
0.782013-0.7863346
0.8477591-0.6628271
0.902113-0.5352331
0.9447398-0.4043392
0.9753767-0.2709524
0.9938347-0.1358951
1-4.24230e-16

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
QuantityValueUnit
Semimajor axis2m
Eccentricity0.5ratio
Specific angular momentum1.224745m²/s
Specific energy-0.25m²/s²
Pericenter speed1.224745m/s
Apocenter speed0.4082483m/s
Orbital period17.77153s

The same energy and angular momentum must hold at both turning radii.

At apsides radial speed is zero and tangential speed h/r. Equate h²/(2rp²)−1/rp and h²/(2ra²)−1/ra to obtain h²=2rp ra/(rp+ra), with the circular case obtained by continuity. Energy is−1/(rp+ra), semimajor axis is their half sum, and period2pi a^(3/2). The ellipse parameter in the plot is not elapsed time.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

The same energy and angular momentum must hold at both turning radii.

At apsides radial speed is zero and tangential speed h/r. Equate h²/(2rp²)−1/rp and h²/(2ra²)−1/ra to obtain h²=2rp ra/(rp+ra), with the circular case obtained by continuity. Energy is−1/(rp+ra), semimajor axis is their half sum, and period2pi a^(3/2). The ellipse parameter in the plot is not elapsed time.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Reconstruct an orbit from two turning radii

At an apsis of a noncircular orbit, radial velocity vanishes and speed is tangential. With specific angular momentum h, that speed is h/r. The same total energy must therefore equal h²/(2rp²) − μ/rp at pericenter and h²/(2ra²) − μ/ra at apocenter.

Equating these expressions and solving gives h² = 2μrpra/(rp + ra). Substitution gives specific energy −μ/(rp + ra). The semimajor axis is (rp + ra)/2 and eccentricity is (ra − rp)/(ra + rp). These relations assume an ideal Newtonian test particle around a fixed central source, not arbitrary attractive motion.

At μ = 1, rp = 1 m and ra = 3 m, h = sqrt(1.5) m²/s, energy is −0.25 m²/s² and eccentricity is 0.5. Speeds are sqrt(1.5) at pericenter and sqrt(1.5)/3 at apocenter. The period is 2π(2)^(3/2) ≈ 17.7715 s.

The diagram places the force center at a focus, not at the ellipse center. Its angle parameter draws geometry and is not equally spaced elapsed time. Conservation of angular momentum means equal areas are swept in equal times, so motion is faster near pericenter. At rp = ra the circular formulas follow by a limit rather than dividing by a vanished radius difference.

Changed-case check

Predict before reading the comparison. For rp = ra = 3 m and μ = 1, the orbit is circular: eccentricity zero, speed 1/sqrt(3) m/s, energy −1/6 m²/s² and period 2π sqrt(27) s.

Retained investigation

Derive energy and angular momentum from both apsides, check both energy substitutions, and identify the circular limiting case. Why can equal spacing around an ellipse fail to represent equal time steps?

Source and boundary

Primary conceptual reference checked 30 October 2026: OpenStax University Physics Volume 1. All models, packets and teaching prose are original. These are bounded mathematical investigations, not measured apparatus, physical procedures, practical certification or institutional credit.

Apply

For rp = 1 and ra = 3 in the supplied μ = 1 orbit, pericenter/apocenter speed ratio is:

Public practice checks a specific claim. It does not establish independent mastery.

Why can equal spacing around an ellipse fail to represent equal time steps?

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