Example Data Table
| Example | Mode | Inputs | Key Output |
|---|---|---|---|
| LEO Satellite | Classical | m=1000 kg, r=7000 km, v=7.5 km/s, θ=90° | |L| ≈ 5.25e13 kg·m²/s |
| Lab Rotor | m r² ω | m=0.25 kg, r=0.12 m, ω=120 rpm | |L| ≈ 4.52e-2 kg·m²/s |
| Electron Orbital | Quantum | ℓ=2, mℓ=1 | |L| ≈ 2.58e-34 J·s |
Formula Used
1) Classical angular momentum
The magnitude of orbital angular momentum for a particle is: |L| = m · r · v · sin(θ) where θ is the angle between the position vector and velocity.
2) Using angular velocity
For a point mass moving with angular velocity: |L| = m · r² · ω. If motion is circular, v = r·ω and both formulas agree.
3) From orbital elements
For a bound Keplerian orbit: |L| = m · √( μ · a · (1 − e²) ), where μ = G·M (or sometimes G·(M+m)).
4) Quantum mechanical orbital momentum
The magnitude is quantized: |L| = √(ℓ(ℓ+1)) · ħ, and the z-component is Lz = mℓ · ħ.
How to Use This Calculator
- Select a calculation mode that matches your known values.
- Enter the inputs and choose units from each dropdown.
- Use θ=90° when velocity is perpendicular to radius.
- For orbital elements, keep eccentricity between 0 and 1.
- Click Calculate to see results above the form.
- Use the CSV/PDF buttons to export the latest calculation.
Orbital Angular Momentum: Practical Notes
Orbital angular momentum measures how strongly a body “circles” a center. In classical mechanics the calculator uses |L| = m r v sin(θ). For circular orbits, θ is 90°, so |L| = m r v and the direction follows the right‑hand rule. The magnitude grows with distance even if speed falls slightly.
1) Low Earth orbit example
For a low‑Earth‑orbit satellite, m=1000 kg, r≈7000 km, v≈7.5 km/s gives |L|≈5.25×10^13 kg·m²/s. That number is large because r is millions of meters. Doubling altitude nearly doubles |L| if speed is similar.
2) Why specific angular momentum matters
Specific angular momentum h = |L|/m is useful for comparing different masses. The same satellite example yields h≈5.25×10^10 m²/s. Engineers often track h because it stays constant for two‑body motion without thrust. In astrodynamics, h also sets the maximum achievable latitude of the orbit plane change.
3) Using angular velocity data
Using angular velocity, the calculator applies |L| = m r² ω. A rotor with m=0.25 kg at r=0.12 m spinning 120 rpm has ω≈12.57 rad/s and |L|≈4.52×10^-2 kg·m²/s. This mode is handy when ω is measured directly.
4) Computing from orbital elements
From orbital elements, h = √(μ a(1−e²)). For Earth’s orbit around the Sun, take a=1 AU, e≈0.0167, μ≈G·M☉. That gives h≈4.46×10^15 m²/s and |L|≈2.66×10^40 kg·m²/s for Earth’s mass.
5) Eccentricity and periapsis behavior
Eccentricity matters: as e increases, (1−e²) decreases, reducing h and |L| for the same a. Highly eccentric comets can have smaller angular momentum yet still reach perihelion speeds. At periapsis the velocity is mostly tangential, so θ stays near 90°.
6) Quantum orbital momentum outputs
In quantum physics, orbital angular momentum is quantized: |L|=√(ℓ(ℓ+1))ħ and Lz=mℓħ. For ℓ=2, |L|≈2.58×10^-34 J·s. The calculator shows both magnitude and z‑component when mℓ is provided. These values connect directly to spectroscopic selection rules.
FAQs
1) What does θ represent in the classical mode?
θ is the angle between the position vector r and velocity v. Use 90° when motion is perpendicular, which is common for circular orbits. Smaller angles reduce |L| by the sin(θ) factor.
2) Why are kg·m²/s and J·s both shown?
They are the same physical unit written differently. Since 1 joule equals 1 kg·m²/s², multiplying by seconds gives J·s = kg·m²/s. Many physics references prefer J·s for momentum.
3) When should I use the orbital elements mode?
Use it when you know semi‑major axis a and eccentricity e, plus the central mass. It is ideal for Keplerian orbits, mission analysis, and comparing different trajectories without needing instantaneous v and r.
4) Why does the elements mode require 0 ≤ e < 1?
The formula h = √(μ a(1−e²)) is for bound elliptical orbits. For parabolic or hyperbolic trajectories, e is ≥ 1 and the relationship between a and energy changes, so this bound‑orbit form is not valid.
5) What is specific angular momentum and why export it?
Specific angular momentum h is |L| divided by mass. It removes the object’s mass from comparisons, stays constant in ideal two‑body motion, and helps check consistency across datasets. Exporting h is useful for reports and audits.
6) How do the quantum results relate to spectroscopy?
Quantum orbital angular momentum depends on ℓ and mℓ. The magnitude √(ℓ(ℓ+1))ħ and component mℓħ determine allowed orientations and selection rules in many atomic transitions, so the values are directly used in quantum models.