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

Explanation

The weight of an object of mass m is W=mg. g is the acceleration due to gravity. `g=(GM)/R^2`
Now, the weight of an object on the earth,
`W_(e)=mg_(e)=(GM_(e))/R_(e)^(2) ……..(1)`
The weight of the same object on the moon, `W_(m)=(GM_(m))/(R_(m)^(2)) ...........(2)`
From (2) and (1),
`W_(m)/(W_(e))=(M_(m))/R_(m)^(2)) divide M_(e)/(R_(e)^(2))=M_(m)/M_(e) xx ((R_(e))/(R_(m))^(2)`
Now the mass `(M_(e))` of the earth is 100 times than that of the moon `(M_(m))` and the radius `(R_(e))` of the earth is 4 times than that of the moon `(R_(m))`,
i.e, `M_(e)=100 M_(m) and R_(e)=4 R_(m)`
`W_(m)/W_(e)=1/(100) xx (4/1)^(2)=(16)/(100) appro 1/6`
the weight of an object on the moon is `1/6th` of its weight on the earth.