The moon's radius is 1/4 that of the earth and its mass is 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

The correct answer is g/5.

To find the acceleration due to gravity on the surface of the moon, we can use the formula for acceleration due to gravity on the surface of a spherical celestial body:

$g=\frac{GM}{R^2}$

where:
• $G$ is the universal gravitational constant,
• $M$ is the mass of the body,
• $R$ is the radius of the body.

Let the mass of the Earth be $M_e$ and its radius be $R_e$. The acceleration due to gravity on the Earth's surface is:

$g=\frac{G{M}_e}{{R_e}^2}$

According to the problem statement, the Moon's mass ($M_m$) and radius ($R_m$) in terms of the Earth's parameters are:
• $M_m=\frac{M_e}{80}$
• $R_m=\frac{R_e}{4}$

Now, let us write the expression for the acceleration due to gravity on the surface of the Moon ($g_m$):

$g_m=\frac{G{M}_m}{{R_m}^2}$

Substitute the values of $M_m$ and $R_m$ into the equation:

$g_m=\frac{G\left(\frac{M_e}{80}\right)}{{\left(\frac{R_e}{4}\right)}^2}$

Simplify the denominator:

$g_m=\frac{G{M}_e/80}{{R_e}^2/16}$

Rearranging the fraction gives:

$g_m=\left(\frac{16}{80}\right)\times \frac{G{M}_e}{{R_e}^2}$

Simplify the numerical coefficient ($\frac{16}{80}=\frac{1}{5}$):

$g_m=\frac{1}{5}\times g$

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

Thus, the acceleration due to gravity on the surface of the moon is $\frac{g}{5}$.