# Tag Info

Ignoring details such as the oblateness of the Earth, atmospheric drag, third body influences such as the Moon and the Sun, relativity, ..., the period of a satellite of negligible mass (even the International Space Station qualifies as a "satellite of negligible mass") is $T=2\pi\sqrt{\frac {a^3}{\mu_\mathrm{Earth}}}$. Neither Newton's gravitational ...
$\mu$ is a quantity that can be easily observed (semi-) directly. It can be easily derived from orbital period of orbiting bodies or acceleration of falling bodies, even if their mass is significantly smaller than the mass of the body you're measuring $\mu$ for. Now the only way to really obtain G is to divide $\mu$ by mass of your celestial body. And ...