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Answer 1 · 2016-12-10T03:40:30

Let the mass of the moon be $Mkg$

Radius of the moon be $R=1.74xx10^6m$

Acceleration due to gravity of moon surface be $g_"moon"=1.67ms^-2$

Gravitational constant be $G=6.674xx 10^-11 m^3 kg^-1s^-2$

considering the force of attraction on a mass of $mkg$ kept on its surface we can write

$mg_"moon"=(GMm)/R^2$

$=>M=(g_"moon"R^2)/G$

$=(1.67ms^-2xx(1.74xx10^6m)^2) /(6.674xx 10^-11 m^3 kg^-1s^-2)=7.576xx10^22kg$