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 $T=2pisqrt{frac {a^3}{mu_mathrm{Earth}}}$. Neither Newton's gravitational constant nor the mass of the Earth are involved in this expression. This means that, ignoring those details, calculating $mu_mathrm{Earth}$ is merely a matter of calculating a satellite's rotational period and its semimajor axis.
Humanity has lots and lots of artificial satellites in orbit, and the people who model the orbits of those satellites don't ignore those details. A few of those satellites were specially designed to enable the determination of the Earth's non-spherical gravitational field (e.g., GRACE and GOCE), and a few were specially designed to enable extremely precise orbit determination (e.g., LAGEOS). Even with all of those details, the Earth's gravitational parameter is a directly inferable quantity (i.e., knowledge of G is not required). Moreover, the value is known to a very high degree of precision.
The Earth's mass? Not so much. The most precise way to "weigh the Earth" is to divide the high precision Earth's gravitational parameter by the low precision universal gravitational constant G. There's a problem here, which is the notoriously low precision of the gravitational constant when expressed in SI units.
No comments:
Post a Comment