# Question #db42e

Dec 10, 2016

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

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

Acceleration due to gravity of moon surface be ${g}_{\text{moon}} = 1.67 m {s}^{-} 2$

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

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

$m {g}_{\text{moon}} = \frac{G M m}{R} ^ 2$

$\implies M = \frac{{g}_{\text{moon}} {R}^{2}}{G}$

$= \frac{1.67 m {s}^{-} 2 \times {\left(1.74 \times {10}^{6} m\right)}^{2}}{6.674 \times {10}^{-} 11 {m}^{3} k {g}^{-} 1 {s}^{-} 2} = 7.576 \times {10}^{22} k g$