If a person weighs 750 N on Earth,how much would be his weight on the Moon given that Moon's mass is `frac(1)(81)` of that Earth and its radius is `frac(1)(3.7)` of that of Earth

Explanation

Weight on earth = 750 N,
Ratio of mass of the earth `(M_(E))` to mass of the moon `(M_(M)) =(M_(E))/(M_(M))=81`
Ratio of radius of earth `(R_(E))` to radius of moon `(R_(M)) =(R_(E))/(R_(M)) =3.7`
Let the mass of the person be m kg
Weight on the earth `=mg =750 =(m G M_(E))/(R_(E)^(2))" "therefore m =(750 R_(E)^(2))/((GM_(E))......(i)`
Weight on Moon `=(m G M_(M))/(R_(M)^(2))` using (i)
`=(750 R_(E)^(2))/((GM_(E)))xx(GM_(M))/(R_(M)^(2))=750 (R_(E)^(2))/(R_(M)^(2))xx(M_(M))/(M_(E))=750xx(3.7)^(2)xx(1)/(81) =126.8 N`
The weight on the moon is nearly `1//6^(th)` of the weight on the earth. We can write the weight on moon as mgm (`g_(m)` is the acceleration due to gravity on the moon). Thus `g_(m)" is 1/6"^(th)` of the g on the earth.