Physics, asked by aartirai923, 1 year ago

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

Answers

Answered by abhi178
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

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