Why is the weight of an object on the moon $\frac{1}{6}^{th}$ its weight on the Earth?

(N/A) Let $M_E$ be the mass of the Earth and $m$ be the mass of an object on the surface of the Earth. Let $R_E$ be the radius of the Earth. According to the universal law of gravitation,the weight $W_E$ of the object on the surface of the Earth is given by:
$W_E = \frac{G M_E m}{R_E^2}$
Let $M_M$ and $R_M$ be the mass and radius of the moon,respectively. Then,according to the universal law of gravitation,the weight $W_M$ of the object on the surface of the moon is given by:
$W_M = \frac{G M_M m}{R_M^2}$
Now,taking the ratio of the two weights:
$\frac{W_M}{W_E} = \frac{M_M R_E^2}{M_E R_M^2}$
Given values:
$M_E = 5.98 \times 10^{24} \text{ kg}$
$M_M = 7.36 \times 10^{22} \text{ kg}$
$R_E = 6.4 \times 10^6 \text{ m}$
$R_M = 1.74 \times 10^6 \text{ m}$
Substituting these values:
$\frac{W_M}{W_E} = \frac{7.36 \times 10^{22} \times (6.4 \times 10^6)^2}{5.98 \times 10^{24} \times (1.74 \times 10^6)^2} \approx 0.165 \approx \frac{1}{6}$
Therefore,the weight of an object on the moon is $\frac{1}{6}^{th}$ of its weight on the Earth.