Question

Find the value of the acceleration due to gravity on the surface of the moon.

Mass of the moon = $7.35\times {10}^{22}kg$

The radius of the moon = $1.74\times {10}^{6}m$

(Given, $G=6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}$)

Answer 1

Step 1: Given data

Mass of the moon $M=7.35\times {10}^{22}kg$

The radius of the moon = $R=1.74\times {10}^{6}m$

Universal gravitational constant $G=6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}$

Step 2: Formula used

The acceleration due to gravity is given as,

$$\begin{gathered} g=G\frac{M}{R^2} \\ where, \\ g\to Accelerationduetogravity \\ G\to Universalgravitationalcons\tan t \\ M\to Massofthemoon \\ R\to Radiusofthemoon \end{gathered}$$

Step 3: Calculation

Putting all the known values in the formula we get,

$$\begin{gathered} g=G\frac{M}{R^2} \\ \Rightarrow g=(6.67\times {10}^{-11}N{m}^{2}k{g}^{-2})\times \frac{(7.35\times {10}^{22}kg)}{{(1.74\times {10}^{6}m)}^2} \\ \Rightarrow g=(6.67\times {10}^{-11})\times \frac{(7.35\times {10}^{22})}{{(1.74\times {10}^6)}^2} \\ \Rightarrow g=1.62m/s^2 \end{gathered}$$

Thus, the value of acceleration due to gravity is $g=1.62m/s^2$.