Question #db42e

1 Answer
P dilip_k
Dec 10, 2016

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#

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