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

Explanation

Given :
Weight on earth = 750 N,
Ratio ofmass of the earth `(M_(E))` to mass of the moon `(m_(M))=(M_(E))/(m_(M))=81`
Ratio of radius of the 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 = m g = 750
`=(mGM_(E))/(R_(E)^(2))`
`thereforem=(750R_(E)^(2))/(GM_(E)).........(1)`
weight on the moon `=(mGM_(M))/(R_(M)^(2))........"using(1)"`
`=(750R_(E)^(2))/(GM_(E))xx(Gm_(M))/(R_(M)^(2))`
`=750.(R_(E)^(2))/(R_(E)^(2))xx(m_(M))/(M_(E))`
`=750xx(3.7)^(2)xx(1)/(81)=126.8N`
The weight on the moon is nearly `1//6^(th)` of the weight on the earth. We can write the weight on moon as `mg_(M)(g_(m ` is the acceleration due to gravity on the moon). Thus `g_(M)" is "1//6^(th)` of g on the earth.