Why is the weight of an object on the moon 1/6 th its weight on the earth?

Let M_E be the mass of the Earth and m be an object on the surface of the Earth.

Let R_E be the radius of the Earth. According to the universal law of gravitation,

weight W_E of the object on the surface of the Earth is given by,

𝑊_𝐸 ={𝐺𝑀_𝐸𝑚 \over 𝑅_{𝐸^2}}

Let M_M and R_M be the mass and radius of the moon. Then, according to the universal law of gravitation, weight W_M of the object on the surface of the moon is given by:

𝑊_𝑀 = {𝐺𝑀_𝑀𝑚 \over 𝑅_{𝑀^2}}

Now,

⇒ {𝑊_𝑀 \over 𝑊_𝐸}={𝑀_𝑀𝑅_{𝐸^2} \over 𝑀_𝐸𝑅_{𝑀^2}}

Where,

𝑀_𝐸 = 5.98 × 10²⁴ 𝑘𝑔

𝑀_𝑀 = 7.36 × 10²² 𝑘𝑔

𝑅_𝐸 = 6.4 × 10⁶ 𝑚

𝑅_𝑀 = 1.74 × 10⁶ 𝑚

⇒ {𝑊_𝑀 \over 𝑊_𝐸}={{7.36 × 10^{22} × (6.4 × 10^6 )^2} \over {5.98 × 10^{24} × (1.74 × 10^6)^2}}=0.165 ≈ {1\over6}

Therefore, weight of an object on the moon is {1\over6} of its weight on the Earth.