Reduced Mass Calculator
Enter two masses to compute reduced mass μ=m₁m₂/(m₁+m₂), the mass ratio and total mass. m₁=1, m₂=35 → μ≈0.972; electron-proton μ≈m_e (planet-star approximation).
Input Data
Results
At a glance:Reduced mass: in the two-body problem, the mass that reduces the relative motion of two particles to an equivalent one-body problem, μ=m₁m₂/(m₁+m₂). Derivation: two masses m₁, m₂ under a central force F(r) — after separating variables the relative motion equation is μ·d²r/dt²=F(r), with μ the reduced mass. Physical meaning: (1) equal masses → μ=m/2 (halved); (2) m₁≪m₂ → μ≈m₁ (the light one dominates, e.g. electron-atom, planet-star); (3) μ is always less than min(m₁,m₂); (4) total mass M=m₁+m₂ relates as μ=M·q/(1+q)² (q=m₁/m₂); (5) centre-of-mass motion (M) and relative motion (μ) decouple, so the two-body problem splits into two one-body problems. History: Newton used reduced mass to derive binary-star orbits under gravity; quantum mechanics uses μ to correct the Bohr model of hydrogen. Applications: (1) binary stars and planet-star orbits (Kepler's third law μ correction); (2) molecular vibration spectra (oscillator frequency ω=√(k/μ)); (3) hydrogen Rydberg-constant correction (μ replaces m_e); (4) rotational spectra moment of inertia I=μr²; (5) two-body quantum mechanics and scattering; (6) isotope effects (different μ → different vibration frequency).
Formula
Reduced mass: μ = m₁·m₂/(m₁ + m₂)
Total mass: M = m₁ + m₂
Mass ratio: q = m₁/m₂
μ = M·q/(1+q)²
Limits: m₁≪m₂ → μ≈m₁; m₁=m₂ → μ=m/2
$$\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{M q}{(1+q)^2}, \quad M = m_1 + m_2, \quad q = \frac{m_1}{m_2}$$How to Use
- Enter mass m₁ (kg).
- Enter mass m₂ (kg).
- The tool computes reduced mass μ, mass ratio and total mass.
Case Studies
Molecular vibration and isotope effect
Molecular vibration is a harmonic oscillator, frequency ω=√(k/μ), k the bond force constant, μ the reduced mass. H-Cl μ≈0.972 amu, D-Cl μ≈1.91 amu → ω(D-Cl)/ω(H-Cl)=√(0.972/1.91)≈0.714 — deuterated compounds vibrate ~30% lower.
IR spectrometers measure molecular vibration; the isotope effect validates the μ correction.
Rotational spectra moment of inertia I=μr² — μ correction makes isotopologues have different rotational constants.
Hydrogen Rydberg and binary orbits
Hydrogen Rydberg constant R∞ uses μ instead of m_e: R_H=R∞·μ/m_e≈109677 cm⁻¹ (vs R∞ 109737).
The ~0.05% shift is the reduced-mass correction of the electron against the proton.
Binary stars: Kepler's third law a³/T²=G(m₁+m₂)/(4π²) uses total mass; the relative orbit uses μ for the effective one-body picture.
Content review: Calculatorism Science Team. Results are for reference only; please refer to the relevant authorities for the official figures.