Explain why weight of an object on moon is only `(1)/(6)th` of the weight of the object on earth.

Explanation

(i) Let , m= mass of an object,
`M_m` = Mass of the moon
`W_m` = Weight of the object on the moon
`R_m` = Radius of moon
`M_e` = Mass of the earth
`R_e` = Radius of the earth
`W_e` = Weight of the object on the earth
G = Universal gravitational constant
(ii) Acceleration due to gravity of an object is given as ,
`g= (GM)/(R^2)`
(iii) Weight of the object on the moon will be ,
`W_m = m(g_m) = (GM_(m)m)/(R_m^2) " "....(i)`
(iv) Weight of the same object on earth will be ,
`W_e = m(g_e) = (GM_(e) m)/(R_2^2) " ".....(ii)`
(v) Dividing equation (i) and (ii)
`W_m/W_e = (GM_(m)m)/(R_m^2) xx(R_e^2)/(GM_(e)m)`
`therefore W_m/W_e = (M_(m) R_e^2)/(R_m^2 M_e) `
For earth and moon ,
`R_E=6.37xx10^6 m` ,
`R_m = 1.76xx10^6 m`
`M_e = 5.98xx10^24` kg
`M_m = 7.36xx10^22 kg`
Substituting these values
`(W_m)/(W_e) = (7.36xx10^22 xx(6.37xx10^6)^2)/((1.76xx10^6)^2 xx 5.98 xx10^24)`
`therefore W_m/W_e ~~ 1/6`
Hence weight of an object on the moon is nearly `(1/6)^(th)` its weight on the earth .