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

#### Solution

Let *M*_{E} be the mass of the Earth and *m* be an object on the surface of the Earth. Let *R*_{E} be the radius of the Earth. According to the universal law of gravitation, weight *W*_{E} of the object on the surface of the Earth is given by,

`W_E=(GM_Em)/R_E^2`

Let M_{M} and R_{M }be the mass and radius of the moon. Then, according to the universal law of gravitation, weight W_{M} of the object on the surface of the moon is given by:-

`W_E=(GM_Mm)/R_M^2`

`W_M/W_E=(M_MR_E^2)/(M_ER_M^2)`

`"Where, "M_E=5.98xx10^24kg," "M_M=7.36xx10^22kg`

`R_E=6.4xx10^6m," "R_M=1.74xx10^6m`

`thereforeW_M/W_E=(7.36xx10^22xx(6.37xx10^6)^2)/(5.98xx10^24xx(1.74xx10^6)^2)=0.165approx1/6`

Therefore, weight of an object on the moon is `1/6` of its weight on the Earth.