Question #db42e

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Question #db42e

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Answer 1 · 2016-12-10T03:40:30

Let the mass of the moon be $M k g$ $Mkg$

Radius of the moon be $R = 1.74 \times 10^{6} m$ $R=1.74xx10^6m$

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

Gravitational constant be $G = 6.674 \times 10^{- 11} m^{3} k g^{- 1} s^{- 2}$ $G=6.674xx 10^-11 m^3 kg^-1s^-2$

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

$m g_{moon} = \frac{G M m}{R^{2}}$ $mg_"moon"=(GMm)/R^2$

$\Rightarrow M = \frac{g_{moon} R^{2}}{G}$ $=>M=(g_"moon"R^2)/G$

$= \frac{1.67 m s^{- 2} \times {(1.74 \times 10^{6} m)}^{2}}{6.674 \times 10^{- 11} m^{3} k g^{- 1} s^{- 2}} = 7.576 \times 10^{22} k g$ $=(1.67ms^-2xx(1.74xx10^6m)^2) /(6.674xx 10^-11 m^3 kg^-1s^-2)=7.576xx10^22kg$

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Question #db42e

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