Determine the period of the Moon's orbit around the Earth…

Determine the period of the Moon's orbit around the Earth, given the acceleration due to gravity at the Earth's surface $g_0 = 9.83\ \mathrm{m/s^2}$, the Earth's radius $R_Z = 6.37 \cdot 10^6\ \mathrm{m}$, and the Earth–Moon distance $R = 3.8 \cdot 10^8\ \mathrm{m}$.

Step 1: Idea

This is Newton's famous \"Moon test\". Neither $\gamma$ nor the Earth's mass is given, but we do not need them separately: the surface gravity already contains their product. At the surface, $mg_0 = \gamma Mm/R_Z^2$, so

\[ \gamma M = g_0R_Z^2. \]

Step 2: The Earth's pull at the Moon

Gravity weakens as $1/r^2$. The Moon is $R/R_Z \approx 60$ Earth radii away, so the acceleration there is about $60^2 = 3600$ times smaller than at the surface: $g_M = g_0\frac{R_Z^2}{R^2}$.

\[ g_M = 9.83 \cdot \left(\frac{6.37 \cdot 10^6}{3.8 \cdot 10^8}\right)^2 \approx 2.76 \cdot 10^{-3}\ \mathrm{m/s^2}. \]

Step 3: Period

This acceleration is the centripetal acceleration of the Moon's circular orbit: $\frac{4\pi^2}{T^2}R = \frac{g_0R_Z^2}{R^2}$.

\[ T = 2\pi\sqrt{\frac{R^3}{g_0R_Z^2}}. \]

\[ T = 2\pi\sqrt{\frac{(3.8 \cdot 10^8)^3}{9.83 \cdot (6.37 \cdot 10^6)^2}}, \]

\[ T = 2\pi\sqrt{1.376 \cdot 10^{11}}. \]

\[T \approx 2.33 \cdot 10^6\ \mathrm{s} \approx 27.0\ \mathrm{days}\]

The same force that makes an apple fall keeps the Moon in its orbit: this is exactly the calculation with which Newton confirmed his law of gravitation.

Check

Observed sidereal month: 27.3 days. The small difference comes from the rounded distance; with the mean distance $3.84 \cdot 10^8\ \mathrm{m}$ the formula gives 27.4 days $\checkmark$