Weighing the Earth: You are given the following data: $g = 9.81 \; m/s^2$, $R_E = 6.37 \times 10^6 \; m$, the distance to the moon $R = 3.84 \times 10^8 \; m$, and the time period of the moon's revolution is $27.3$ days. Obtain the mass of the Earth $M_E$ in two different ways.

(N/A) Method $1$: Using the acceleration due to gravity on the surface of the Earth.
$M_E = \frac{g R_E^2}{G}$
$M_E = \frac{9.81 \times (6.37 \times 10^6)^2}{6.67 \times 10^{-11}}$
$M_E \approx 5.97 \times 10^{24} \; kg$
Method $2$: Using the orbital motion of the moon.
From Kepler's third law, $T^2 = \frac{4 \pi^2 R^3}{G M_E}$, so $M_E = \frac{4 \pi^2 R^3}{G T^2}$.
Given $T = 27.3 \; \text{days} = 27.3 \times 24 \times 3600 \; s = 2.3587 \times 10^6 \; s$.
$M_E = \frac{4 \times (3.14)^2 \times (3.84 \times 10^8)^3}{6.67 \times 10^{-11} \times (2.3587 \times 10^6)^2}$
$M_E \approx 6.02 \times 10^{24} \; kg$.
Both methods yield almost the same answer, with a difference of less than $1 \%$.