Calculate reduced mass μ = m₁m₂/(m₁+m₂) for two or three bodies. Supports kg, amu, electron masses with reference table of common systems.
The reduced mass μ = m₁m₂/(m₁+m₂) transforms a two-body problem into an equivalent one-body problem. It appears throughout physics: orbital mechanics (planet-star systems), molecular spectroscopy (vibrational frequencies), atomic physics (hydrogen energy levels), and nuclear scattering.
This calculator handles masses from electrons (9.1 × 10⁻³¹ kg) to stars (2 × 10³⁰ kg) with six unit options: kg, grams, atomic mass units (amu), pounds, electron masses, and proton masses. For three-body systems, it performs sequential reduction: μ₁₂₃ = (μ₁₂ × m₃)/(μ₁₂ + m₃).
Results include the reduced mass, its percentage of each input mass, total mass, mass ratio, and a visual comparison bar chart. The reference table lists five classic systems—from proton-electron to Sun-Earth—showing how μ approaches the lighter mass when one body is much heavier. Check the example with realistic values before reporting. Use the steps shown to verify rounding and units. Cross-check this output using a known reference case. Use the example pattern when troubleshooting unexpected results.
Reduced mass is a fundamental concept that students and professionals encounter across physics and chemistry. This calculator eliminates unit conversion headaches (amu ↔ kg ↔ electron masses) and provides immediate context through the mass-ratio analysis and reference table.
The 3-body extension and the visual mass comparison make this more than a simple formula evaluator—it builds physical intuition about the two-body problem.
Reduced mass: μ = m₁m₂/(m₁+m₂) = m₁/(1 + m₁/m₂). For equal masses: μ = m/2. For m₂ >> m₁: μ ≈ m₁. Three-body reduction: μ₁₂₃ = (μ₁₂ × m₃)/(μ₁₂ + m₃).
Result: μ = 7.253 × 10²² kg
μ = (5.972e24 × 7.342e22) / (5.972e24 + 7.342e22) = 7.253 × 10²² kg ≈ 98.8% of Moon's mass, since Earth is ~81× heavier.
| Field | Application | Typical μ | |---|---|---| | Atomic physics | Hydrogen energy levels | 0.9995 × m_e | | Molecular spectroscopy | IR vibrational frequencies | 1–20 amu | | Nuclear physics | Scattering cross-sections | ~1 amu | | Orbital mechanics | Two-body orbit calculations | ≈ lighter body | | Gravitational waves | Inspiral chirp mass | 1–50 M_☉ |
The reduced mass connects to the center of mass (CM) frame: - Total mass: M = m₁ + m₂ - CM velocity: v_CM = (m₁v₁ + m₂v₂)/M - Relative velocity: v_rel = v₁ − v₂ - Kinetic energy in CM frame: KE_rel = ½μv_rel² - Angular momentum in CM frame: L = μ × r × v_rel
The chirp mass in gravitational wave astronomy is: M_chirp = (m₁m₂)^(3/5) / (m₁+m₂)^(1/5) = μ^(3/5) × M^(2/5).
It simplifies two-body problems: instead of tracking two particles, we solve for one particle of mass μ in the relative coordinate. The mathematics is identical to a one-body problem.
μ approaches the smaller mass. For Earth-Sun: μ ≈ 5.97 × 10²⁴ kg ≈ Earth's mass, because the Sun is 333,000× heavier. This is why we can approximately treat Earth as orbiting a fixed Sun.
The vibrational frequency of a diatomic molecule is ω = √(k/μ), where k is the bond force constant and μ is the reduced mass of the two atoms. Heavier isotopes have lower frequencies.
The Rydberg constant includes a reduced mass correction: R = R_∞ × μ/(m_e). For hydrogen, μ/(m_e) = 0.99946, giving a 0.054% correction to energy levels.
Only if one mass is zero (a massless particle). In practice, reduced mass is always positive and less than or equal to either individual mass.
First reduce bodies 1 and 2 to get μ₁₂, then reduce μ₁₂ with body 3. The result depends on the order of reduction and is an approximation for 3-body systems.