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Reduced mass
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Two-body problems simplify using an effective mass.
μ = m₁m₂/(m₁+m₂)For two interacting bodies it is convenient to use μ = m₁m₂/(m₁ + m₂): the relative motion behaves like a single body of mass μ.
Earth–Moon: μ ≈ 0.99 of the Moon's mass.
Used in orbits and atomic spectra.
In more detail
The reduced mass of two bodies is μ = m1m2 / (m1 + m2). It simplifies the two-body problem, since the relative motion behaves as if one body of mass μ moved around a fixed centre.
Real-world uses
- Astronomers use reduced mass in the two-body problem for binary stars.
- Atomic physics uses it for the energy levels of hydrogen-like atoms.
- Two stars orbiting each other can be treated as one body of reduced mass moving around a fixed centre.
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Frequently asked questions
Why do we need reduced mass?
It turns the two-body problem into an effective one-body problem, which is much simpler.
What happens when one mass is very large?
The reduced mass approaches the smaller mass.
What is the reduced mass for a heavy and a light body?
It is close to the mass of the lighter body, because the heavier body barely moves.
Related topics in Momentum & Collisions
Light hits heavy object · Heavy hits light object · Energy lost in inelastic collision · Collisions in two dimensions · Billiard ball collisions · Ballistic pendulum