An artificial satellite moves in the Earth's equatorial…

An artificial satellite moves in the Earth's equatorial plane at a distance $R = 2 \cdot 10^7\ \mathrm{m}$ from its centre, from west to east (the same direction as the Earth's rotation). It passes over the same point on the equator every $T_f = 11.6\ \mathrm{h}$. From this, determine the mass of the Earth.

Step 1: Idea

The 11.6 h is not the satellite's orbital period. It is the time between two passes over a point that itself turns with the Earth. The key is to convert it into the satellite's true angular velocity, and then use gravity as the centripetal force.

Step 2: Angular velocity of the satellite

Both the satellite and the ground point turn eastward. The satellite turns faster (angular velocity $\omega_s$) than the Earth ($\omega_Z = 2\pi/T_Z$, $T_Z = 24\ \mathrm{h}$). Seen from the ground, the satellite drifts eastward with the relative angular velocity $\omega_s - \omega_Z$, and it is above the same point again once it has gained a full turn, $2\pi$:

\[ (\omega_s - \omega_Z)\,T_f = 2\pi, \]

\[ \omega_s = 2\pi\left(\frac{1}{T_f} + \frac{1}{T_Z}\right). \]

(This is the same reasoning as for the two hands of a clock.)

Step 3: Mass of the Earth

\[ \omega_s^2R = \gamma\frac{M}{R^2}, \]

\[ M = \frac{\omega_s^2R^3}{\gamma} = \frac{4\pi^2R^3}{\gamma}\left(\frac{1}{T_f} + \frac{1}{T_Z}\right)^2. \]

In seconds, $T_f = 41\,760\ \mathrm{s}$ and $T_Z = 86\,400\ \mathrm{s}$: $\frac{1}{41\,760} + \frac{1}{86\,400} = 3.552 \cdot 10^{-5}\ \mathrm{s^{-1}}$.

\[ M = \frac{4\pi^2 \cdot 8 \cdot 10^{21}}{6.67 \cdot 10^{-11}} \cdot (3.552 \cdot 10^{-5})^2. \]

\[M \approx 5.97 \cdot 10^{24}\ \mathrm{kg}\]

Check

The satellite's own period is $2\pi/\omega_s $ $= 28\,150\ \mathrm{s} $ $= 7.8\ \mathrm{h}$, shorter than a day, so it is indeed faster than the Earth turns $\checkmark$

Kepler check: $T $ $= 2\pi\sqrt{R^3/\gamma M} $ $= 28\,150\ \mathrm{s}\ \checkmark$