Question

The moon's radius is $\frac{1}{4}$that of earth and its mass is$\frac{1}{80}$ times that of the earth. If g represents the acceleration due to gravity on the surface of the earth, that on the surface of the moon is

• A. $\frac{g}{4}$
• B. $\frac{g}{5}$
• C. $\frac{g}{6}$
• D. $\frac{g}{8}$

Answer 1

The correct option is B

$\frac{g}{5}$

Solution:

Step 1: Given data:

The moon's radius is

$$\begin{gathered} R_m=\frac{R_e}{4} \\ \frac{R_m}{R_e}=\frac{1}{4} \end{gathered}$$

The mass of the moon

$$\begin{gathered} M_m=\frac{M_e}{80} \\ \frac{M_m}{M_e}=\frac{1}{80} \end{gathered}$$.

Step 2: Formula used:

The acceleration is due to gravity on the surface of the earth$g=\frac{G{M}_e}{{R_e}^2}$. Here $M_e$is the mass and $R_e$is the radius of the earth.

The acceleration is due to gravity on the surface of the moon$g=\frac{G{M}_m}{{R_m}^2}$. Here $M_m$is the mass and $R_m$is the radius of the moon.

Step 3: Calculate the surface of the moon:

Note that If $g$ represents the acceleration due to gravity on the surface of the earth, that on the surface of the moon is

$$\begin{gathered} \frac{g_m}{g}=\frac{M_m}{M_e}\times \frac{{R_e}^2}{{R_m}^2} \\ \frac{g_m}{g}=\frac{1}{80}\times \frac{{\left(4\right)}^2}{1^2} \\ \frac{g_m}{g}=\frac{16}{80} \\ \frac{g_m}{g}=\frac{1}{5} \end{gathered}$$

Therefore the surface of the moon is,

$g_m=\frac{g}{5}$.

Hence, the correct option is B.