In Newtonian gravity, two bodies do not usually have one fixed object and one moving object.
Both bodies move around their common center of mass, called the barycenter.
The reduced-mass method turns the two-body problem into an equivalent one-body problem.
Instead of tracking both masses separately, we track one equivalent particle of reduced mass \(\mu\).
1. Total mass and reduced mass
For two masses \(m_1\) and \(m_2\), the total mass is
\[
\boxed{
M=m_1+m_2
}.
\]
The reduced mass is
\[
\boxed{
\mu=\frac{m_1m_2}{m_1+m_2}
}.
\]
The reduced mass is always less than or equal to the smaller mass.
If one body is much more massive than the other, then \(\mu\) is approximately equal to the smaller mass.
2. Barycenter location
Let \(a\) be the separation between the two bodies.
The barycenter divides the line between them so that
\[
m_1r_1=m_2r_2,
\]
where \(r_1\) is the distance from body 1 to the barycenter, and \(r_2\) is the distance from body 2 to the barycenter.
Since
\[
r_1+r_2=a,
\]
the barycenter distances are
\[
\boxed{
r_1=\frac{m_2}{m_1+m_2}a
},
\qquad
\boxed{
r_2=\frac{m_1}{m_1+m_2}a
}.
\]
The more massive body is closer to the barycenter.
If \(m_1\gg m_2\), then \(r_1\) is small and \(r_2\) is nearly equal to \(a\).
3. Circular two-body orbit
For a circular two-body orbit, both bodies have the same angular speed \(\omega\) and the same orbital period \(T\).
The relative separation \(a\) behaves like an orbit around the total mass \(m_1+m_2\).
\[
\boxed{
\omega=\sqrt{\frac{G(m_1+m_2)}{a^3}}
}.
\]
The period is
\[
\boxed{
T=2\pi\sqrt{\frac{a^3}{G(m_1+m_2)}}
}.
\]
This is the two-body version of Kepler’s third law.
4. Solving separation from period
If the period is known, rearrange
\[
T=2\pi\sqrt{\frac{a^3}{G(m_1+m_2)}}.
\]
Square both sides:
\[
T^2=4\pi^2\frac{a^3}{G(m_1+m_2)}.
\]
Solve for \(a\):
\[
a^3=\frac{G(m_1+m_2)T^2}{4\pi^2}.
\]
Therefore,
\[
\boxed{
a=\sqrt[3]{\frac{G(m_1+m_2)T^2}{4\pi^2}}
}.
\]
5. Orbital speeds
The speeds of the two bodies around the barycenter are
\[
\boxed{
v_1=\omega r_1
},
\qquad
\boxed{
v_2=\omega r_2
}.
\]
The relative speed is
\[
\boxed{
v_{\mathrm{rel}}=\omega a
}.
\]
The smaller body usually has the larger speed around the barycenter because it is farther from the barycenter.
6. Force between the bodies
The gravitational force magnitude is
\[
\boxed{
F=\frac{Gm_1m_2}{a^2}
}.
\]
This force pulls each body toward the barycenter and provides the centripetal acceleration needed for circular motion.
7. Energy in the circular two-body problem
The gravitational potential energy is
\[
\boxed{
U=-\frac{Gm_1m_2}{a}
}.
\]
The total kinetic energy of both bodies is
\[
K=\frac12m_1v_1^2+\frac12m_2v_2^2.
\]
In the reduced-mass view, this is equivalent to
\[
K=\frac12\mu v_{\mathrm{rel}}^2.
\]
For a circular two-body orbit,
\[
\boxed{
K=\frac{Gm_1m_2}{2a}
}.
\]
Therefore, the total mechanical energy is
\[
E=K+U.
\]
\[
E=
\frac{Gm_1m_2}{2a}
-
\frac{Gm_1m_2}{a}.
\]
\[
\boxed{
E=-\frac{Gm_1m_2}{2a}
}.
\]
8. Why reduced mass is useful
The reduced-mass method allows the relative motion to be treated as if one particle of mass \(\mu\) moves in a central potential:
\[
U(r)=-\frac{Gm_1m_2}{r}.
\]
The relative coordinate is
\[
\mathbf r=\mathbf r_2-\mathbf r_1.
\]
Then the relative equation of motion can be written in the form
\[
\mu \ddot{\mathbf r}
=
-\frac{Gm_1m_2}{r^2}\hat{\mathbf r}.
\]
This is mathematically similar to a one-body central-force problem.
9. Worked example: Earth-Moon system
Use approximately
\[
m_{\oplus}=5.9722\times10^{24}\ \mathrm{kg},
\qquad
m_{\mathrm{Moon}}=7.3477\times10^{22}\ \mathrm{kg}.
\]
The average separation is
\[
a=384400\ \mathrm{km}=3.84400\times10^8\ \mathrm{m}.
\]
The reduced mass is
\[
\mu=
\frac{
(5.9722\times10^{24})(7.3477\times10^{22})
}
{5.9722\times10^{24}+7.3477\times10^{22}}.
\]
\[
\mu\approx7.26\times10^{22}\ \mathrm{kg}.
\]
The distance from Earth’s center to the barycenter is
\[
r_{\oplus}
=
\frac{m_{\mathrm{Moon}}}{m_{\oplus}+m_{\mathrm{Moon}}}a.
\]
\[
r_{\oplus}
\approx
4.67\times10^6\ \mathrm{m}
=
4670\ \mathrm{km}.
\]
Since Earth’s radius is about \(6371\ \mathrm{km}\), the Earth-Moon barycenter lies inside Earth.
10. Planet-moon versus binary systems
| System type |
Mass relation |
Barycenter behavior |
| Planet-moon system |
One mass is much larger |
Barycenter may lie inside the larger body |
| Binary planet or dwarf planet system |
Masses are more comparable |
Barycenter may lie outside both bodies |
| Equal-mass binary stars |
\(m_1=m_2\) |
Barycenter lies halfway between them |
11. Summary formulas
| Goal |
Formula |
Use |
| Total mass |
\(M=m_1+m_2\) |
Controls the relative orbit |
| Reduced mass |
\(\mu=m_1m_2/(m_1+m_2)\) |
Equivalent mass for the one-body reduced problem |
| Barycenter from body 1 |
\(r_1=m_2a/(m_1+m_2)\) |
Orbit radius of body 1 around the barycenter |
| Barycenter from body 2 |
\(r_2=m_1a/(m_1+m_2)\) |
Orbit radius of body 2 around the barycenter |
| Angular speed |
\(\omega=\sqrt{G(m_1+m_2)/a^3}\) |
Common angular speed of both bodies |
| Period |
\(T=2\pi\sqrt{a^3/[G(m_1+m_2)]}\) |
Common orbital period |
| Force |
\(F=Gm_1m_2/a^2\) |
Gravitational attraction between the bodies |
| Total energy |
\(E=-Gm_1m_2/(2a)\) |
Circular two-body orbital energy |
12. Assumptions and limits
- The calculator uses Newtonian gravity.
- The orbit preview assumes circular motion.
- The separation is treated as the center-to-center distance.
- Atmospheric drag, tides, non-spherical bodies, and third-body perturbations are ignored.
- Real binary systems may have eccentric orbits, so the separation and speed can vary over time.
Key idea: the two-body problem is not “one object fixed and one object moving”; both bodies orbit the barycenter, and the reduced mass lets the relative motion be solved like a one-body problem.
