Reduced Mass Calculator: μ = m₁m₂/(m₁+m₂)
Reduced Mass Calculator
The reduced mass is the trick that turns a two-body problem into a one-body problem. Give it two masses and it returns μ, the single mass that one imaginary particle would need in order to move exactly as the real pair does — together with the two orbit radii, the moment of inertia, and the proof that μ is always smaller than either mass you put in.
Reduced mass, orbit radii about the centre of mass, moment of inertia
A hydrogen chloride molecule is modelled as hydrogen (1.0 u) bonded to chlorine (35.4 u). Find the reduced mass, and find how far each atom sits from the centre of mass as a fraction of the bond length.
Two forms of the same statement, and the one that explains the name
- m₁, m₂
- the two masses, in any one consistent unit.
- μ
- the reduced mass, in that same unit. It is the mass a single particle would need so that its motion under the same force reproduces the real pair’s relative motion.
- 1/μ = 1/m₁ + 1/m₂
- the reciprocal form. This is the one to remember, because it makes the “always smaller” result obvious and it is easier to use.
- r₁ = d m₂/M, r₂ = d m₁/M
- the radii of the two circles about the common centre of mass. Note the cross-over: m₁’s orbit radius carries m₂’s mass.
- I = μd²
- the moment of inertia of the pair about their centre of mass. The two-mass sum m₁r₁² + m₂r₂² collapses into this one term, which is the cleanest demonstration of what μ is for.
Worked example
A hydrogen chloride molecule is modelled as hydrogen (1.0 u) bonded to chlorine (35.4 u). Find the reduced mass, and find how far each atom sits from the centre of mass as a fraction of the bond length.
Total mass. M = 1.0 + 35.4 = 36.4 u.
Reduced mass. μ = (1.0 × 35.4) ÷ 36.4 = 35.4 ÷ 36.4 = 0.9725 u, which rounds to the 0.97 u OpenStax quotes for this molecule. Compare it with the hydrogen mass it came from: 0.9725 against 1.0, so μ is 2.7% below the lighter atom and 97% below the heavier one.
Check it the other way. 1/μ = 1/1.0 + 1/35.4 = 1 + 0.02825 = 1.02825, so μ = 1 ÷ 1.02825 = 0.9725 u. Same answer, and this is the version worth doing in your head: adding a small reciprocal to a big one barely changes it, so μ barely differs from the lighter mass.
Orbit radii. r₁ = d m₂/M = d × 35.4/36.4 = 0.9725 d for the hydrogen, and r₂ = d m₁/M = d × 1.0/36.4 = 0.0275 d for the chlorine. The hydrogen does almost all the moving. Notice r₁ and μ share the same number here, which is a coincidence of this particular pair: r₁/d and μ/m₁ are both m₂/M.
Moment of inertia. I = μd² = 0.9725 d². Doing it the long way, m₁r₁² + m₂r₂² = 1.0(0.9725d)² + 35.4(0.0275d)² = 0.9458d² + 0.0268d² = 0.9726d², which agrees to rounding. The two-term sum always collapses to μd², and that collapse is the reason reduced mass exists.
Where this goes wrong. Writing μ = (m₁ + m₂)/(m₁m₂) upside down, and giving the masses in different units — one in grams and one in atomic mass units — are the two reliable ways to lose the marks. The second produces a number that is not absurd, which is what makes it dangerous.
What μ does as the mass ratio changes
| m₁ | m₂ | μ | μ ÷ the lighter mass | What it means |
|---|---|---|---|---|
| 1 | 1 | 0.5000 | 0.5000 | Equal masses: μ is half of each, the maximum possible |
| 1 | 2 | 0.6667 | 0.6667 | Already two thirds of the lighter mass |
| 1 | 4 | 0.8000 | 0.8000 | Four fifths |
| 1 | 10 | 0.9091 | 0.9091 | Within 10% of the lighter mass |
| 1 | 100 | 0.9901 | 0.9901 | Within 1% |
| 1 | 1 000 000 | 1.000000 | 0.999999 | Indistinguishable from the lighter mass |
| 1.0 | 35.4 | 0.9725 | 0.9725 | Hydrogen and chlorine — this page’s default |
Reduced mass in four real systems
| System | m₁ | m₂ | μ ÷ the lighter mass | Why it matters |
|---|---|---|---|---|
| Electron and proton | electron | proton (1836 × heavier) | 0.999 4557 | Shifts every hydrogen spectral line by about 0.054%, and the identical sum for deuterium is how deuterium was discovered |
| Earth and Moon | 5.97 × 10²⁴ kg | 0.073 × 10²⁴ kg | 0.98792 | Both bodies orbit a barycentre about 4 640 km from Earth’s centre, which is inside the Earth |
| Hydrogen chloride | 1.0 u | 35.4 u | 0.9725 | Sets the rotational and vibrational line spacings in the infrared spectrum |
| Two identical stars | m | m | 0.5000 | The extreme opposite case: μ = m/2, and neither star can be treated as fixed |
Why two bodies collapse into one, and why μ can never be large
Two bodies pulling on each other is a harder problem than one body moving in a fixed field, and the reduced mass is the device that removes the difference. Earth and Sun both move; so do the two stars of a binary; so do the electron and the proton in a hydrogen atom. Write out Newton’s second law for each body separately and you have two coupled equations. Subtract them, and something remarkable happens: the separation vector obeys a single equation of exactly the familiar form, with the mass in it replaced by m₁m₂/(m₁ + m₂). One imaginary particle of that mass, moving in a fixed field, reproduces the real relative motion exactly.
The reciprocal form is the one to carry around. 1/μ = 1/m₁ + 1/m₂ says that you add the reciprocals, the way you add resistors in parallel or focal lengths of thin lenses in contact. That similarity is not a coincidence of notation — all three are cases where two things each offer some resistance and the combination is easier to shift than either alone — and it gives you the result for free. Adding a positive number to 1/m₁ makes 1/μ bigger than 1/m₁, so μ is smaller than m₁. The same argument against m₂. There is no arrangement of two positive masses for which μ exceeds either.
How much smaller? That depends entirely on the ratio. When the masses are equal, μ is exactly half of each — the extreme case, and the largest μ can be for a given total mass. As one mass grows relative to the other, μ creeps towards the lighter mass without ever reaching it. By a ratio of 100 to 1, μ is within one per cent of the lighter mass; by 1836 to 1, the proton-to-electron ratio, it is within 0.055%. So μ always lies between half the lighter mass and the lighter mass, and that is a useful check on any answer.
That near-collapse is exactly why the simple pictures work. Treating the Sun as fixed while the Earth orbits is not quite right — the Sun moves too — but the error is a part in 330 000. Treating the proton as fixed in the Bohr model is wrong by about one part in 1836, which sounds negligible and is not: it shifts every line in the hydrogen spectrum by 0.054%, enough to measure easily. Doing the same sum for deuterium, whose nucleus is about twice as heavy, gives a slightly different shift, and the gap between the hydrogen and deuterium lines is how deuterium was first identified. A correction arithmetic calls negligible can be the whole experiment.
The other half of the picture is the pair of orbit radii, and they carry a cross-over that catches people out. The body of mass m₁ orbits at radius r₁ = d m₂/M — its own radius is set by the other body’s mass. That is the same inverse relationship as a see-saw: the heavy child sits close to the pivot. In a binary star system with a 10:1 mass ratio, the light star swings round a circle ten times the radius of the heavy star’s, and both complete one orbit in the same time. Watching a star wobble on a small circle is how a great many unseen companions have been found.
Then the moment of inertia tidies itself up. The honest sum for the pair about their centre of mass is m₁r₁² + m₂r₂², which looks like two terms to keep track of. Substituting the radii and simplifying gives exactly μd² — one term, one mass, one distance. Rotational and vibrational spectroscopy of diatomic molecules runs entirely on that expression, which is why a chemistry textbook will quote a molecule’s reduced mass as a property worth tabulating.
Where this goes wrong is mixed units and an upside-down fraction. Both masses must be in the same unit; the formula contains a ratio, and a ratio of a gram to an atomic mass unit is meaningless. And μ = (m₁ + m₂)/(m₁m₂) is the reciprocal of the right answer, which for masses near 1 gives something near the right size and is therefore easy to miss. Use the window: if your μ is not between half the lighter mass and the lighter mass, you have made one of those two mistakes.
A short derivation, because it is three lines and it makes the formula stop being arbitrary. Call the separation vector r = r₁ − r₂. Each body feels the same size of force F from the other, in opposite directions, so a₁ = F/m₁ and a₂ = −F/m₂. Subtract to get the acceleration of the separation: a = a₁ − a₂ = F(1/m₁ + 1/m₂) = F/μ, with 1/μ defined as that bracket. Rearranged, F = μa — which is Newton’s second law for a single particle of mass μ. Nothing was assumed about the kind of force, so the result holds for gravity, for electrostatic attraction and for the spring that models a chemical bond alike.
It is worth noticing what the derivation does not say. It does not say either body has mass μ; neither does. It does not describe where the pair is, only how the two bodies move relative to each other — the position of the centre of mass is a separate, much easier problem, and with no external force it simply drifts at constant velocity. The full motion is the sum of the two: a centre of mass going in a straight line, and a reduced-mass particle orbiting the origin. Splitting a problem into those two halves is one of the most reusable moves in mechanics, and this page is the arithmetic for the second half of it.
Frequently asked questions
Why is the reduced mass always smaller than both masses?
Because 1/μ = 1/m₁ + 1/m₂, and you are adding two positive quantities. That makes 1/μ larger than 1/m₁ on its own, so μ must be smaller than m₁; the same argument gives μ smaller than m₂. The tightest statement is that μ lies between half the lighter mass (when the two are equal) and the lighter mass itself (approached but never reached as the ratio grows).
What is the reduced mass actually for?
It turns a two-body problem into a one-body problem. Once you use μ, the relative motion of the two bodies obeys the same equation as a single particle of mass μ moving in a fixed field — so every result you already know for one body orbiting a fixed centre carries straight over. It also collapses the pair’s moment of inertia from m₁r₁² + m₂r₂² to the single term μd².
Does the separation change μ?
No. μ depends only on the two masses. The separation box on this page exists for the orbit radii and the moment of inertia, which do depend on it. Leave it at 1 and the radii come out as plain fractions of the separation.
How is this different from the total mass?
They are opposite ends of the same pair of quantities. The total mass M = m₁ + m₂ governs how the centre of mass moves — which, with no external force, is in a straight line at constant speed. The reduced mass governs the relative motion about that centre. Between them the two numbers describe the whole system, and a full treatment of a binary uses both.
Why does the electron–proton reduced mass matter if it is only 0.05% off?
Because spectroscopy measures wavelengths to far better than 0.05%. Putting μ in place of the electron mass in the Bohr energy expression shifts every hydrogen line by that amount, and the shift is different for deuterium because the nucleus is roughly twice as heavy. Urey found deuterium in 1931 by looking for precisely those displaced lines.
Where will I actually meet reduced mass in an exam?
Most often in three places. In a two-body gravitational or binary-star problem, where neither body can be held fixed. In the rotational and vibrational spectra of diatomic molecules, where the moment of inertia is μd² and the vibration frequency is √(k/μ). And as a correction to the Bohr model, where replacing the electron mass with the electron–proton reduced mass shifts every hydrogen line by 0.054%. A collision problem will sometimes use it too, because the kinetic energy available to be lost in a collision is ½μv² with v the relative speed.
Can μ ever equal one of the masses?
Only in the limit, never exactly. As m₂ grows without bound, μ tends to m₁ from below, but for any finite m₂ there is still a gap. A calculator displaying only a few decimal places will show them as equal long before they are — this page gives the ratio to six places so you can see the gap is still there.
Related calculators
References
- OpenStax, University Physics Volume 3, §9.2 “Molecular Spectra” (CC BY 4.0). Equation 9.8 is I = μr0², and Example 9.2 works the hydrogen-chloride reduced mass as (1.0 u)(35.4 u)/(36.4 u) = 0.97 u — the pair this page opens with.
- NIST, CODATA Internationally Recommended 2022 Values of the Fundamental Physical Constants, ASCII table read 11 November 2026. Standard acceleration of gravity 9.806 65 m s−2, marked (exact) rather than carrying an uncertainty; Newtonian constant of gravitation 6.674 30(15)×10−11 m3 kg−1 s−2.
physics.nist.gov/cuu/Constants/Table/allascii.txt - NASA, Planetary Fact Sheet — Metric, NASA Space Science Data Coordinated Archive, read 11 November 2026. The “Gravity (m/s²)” and “Diameter (km)” rows supply every surface gravity and radius on this page.
nssdc.gsfc.nasa.gov/planetary/factsheet/ - BIPM, The International System of Units (SI), 9th edition (2019, updated 2026). The newton, the pascal and the kilogram: the kilogram is a unit of mass and the newton a unit of force, and the brochure keeps them in separate tables for the reason this page keeps repeating.