= Gravitational mass degeneracy
{title2=$\mu=GM$}
An <orbit> around an isolated source constrains its <standard gravitational parameter> $\mu=GM$ rather than independently measuring the source's <mass> and the <gravitational constant>. In the <test particle> approximation, a <circular orbit> of radius $a$ and <orbital period> $T$ satisfies
$$
\frac{4\pi^2a}{T^2}=\frac{GM}{a^2},\qquad \mu=\frac{4\pi^2a^3}{T^2}.
$$
The left equation equates the required <centripetal acceleration> to the <gravitational acceleration>. More generally, a <Kepler orbit> depends on the same $\mu$. In the exact relative <two-body problem>, replace $GM$ by $G(M+m)$.
For any positive dimensionless number $q$, replacing $G$ by $qG$ and $M$ by $M/q$ leaves $GM$ and every predicted <test particle> <trajectory> unchanged. This is the <gravitational mass degeneracy>: more observations of the same <orbit> can refine $GM$ without separating its factors. A laboratory source whose <mass> is calibrated independently supplies the missing information.
Consequently an astronomical <mass> expressed in kilograms as $M=\mu/G$ inherits <measurement uncertainty> from $G$, even if $\mu$ is known much more precisely. To first order,
$$
\frac{\delta M}{M}=\frac{\delta\mu}{\mu}-\frac{\delta G}{G}.
$$
When the relative <standard uncertainty> of $\mu$ is negligible, that of $M$ is approximately $u(G)/G$. Conversely, the ratio of two source <masses> inferred using the same $G$ is $M_1/M_2=\mu_1/\mu_2$, so this common factor cancels. That distinction explains how excellent <orbit determination> can coexist with less precise absolute astronomical <masses>.
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