Question

The radius of the moon is one-fourth, and its mass is one eighty-first that of the earth. If the acceleration due to gravity on the surface of the earth is 9.8 m/s 2, what is its value on the moon’s surface.

Answer 1

it is given that radius of the moon is 1/4th of radius of the earth. i.e., $r_{moon}=\frac{1}{4}r_{earth}$....(1)

and mass of the moon is one eighty - first that of the earth.i.e., $M_{moon}=\frac{1}{81}M_{earth}$....(2)

we know, acceleration due to gravity on the earth's surface is given by, $g=\frac{GM_{earth}}{r_{earth}^2}$......(3)

similarly, acceleration due to gravity on the moon's surface is given by, $g'=\frac{GM_{moon}}{r_{moon}^2}$

from equations (1) and (2),

$g'=\frac{16}{81}\frac{GM_{earth}}{r_{earth}^2}$

from equation (3),

g' = $\frac{16}{81}g$

= 16 × 9.8/81 [ given g = 9.8 m/s² ]

= 1.93 m/s²