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

Explanation

Let the mass of a given object be M. It weight on earth i.e. the force with which the earth attracts this mass is given by
`W_(e)=(G.M_(e)xxm)/(R_(2))`…………1
Where Me is the mass of earth Re is its radius and G is the universal gravitational constant. Its vlue of `6.67xx10^(-11)Nm^(2)//kg^(2)`.
When that object is taken on the moon its weight `(W_(m))` will be given by
`W_(m)=(G.M_(m)xxm)/(R_(m)^(2))`
Where `M_(m)` is the mass and `R_(m)` is the radius of the moon
Dividing 2 by 1 we get
`(W_(m))/(W_(e))=(M_(m))/(M_(e))xx((R_(e))/(R_(m)))^(2)`
Now the mass of the earth is about 100 times the mass of the moon and its radius is 4 times that of the moon.
`:.(W_(m))/(W_(e))=1/100xx(4)^(2)=16/100=1/6`
or `W_(m)=(W_(e))/6`
Hence the weight of the object on the moon is about one sixth of its weight on earth.
It should be noted that mass of the object on the moon is the same as on the earth.