Answer
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Hint
Let, the mass of the body is m and the weight of the body is mg. Also, this is the acceleration produced in a freely falling body. Also, this is inversely proportional to the distance between the body and the center of the earth.
Complete step by step answer
The ratio of the weight of a body to its mass gives the acceleration due to gravity at the given place.
Let, the mass of the body is $ m $ .
then, the weight of the body will be $ mg $ .
So, the ratio of the weight of a body to its mass gives the $mg/m = g$ (acceleration due to gravity).
So, we can say that acceleration produced in a freely falling body under the action of gravity is called the acceleration due to gravity.
Additional Information
So, in case of a body of mass $ m $ , using Newton’s second law of motion, we have, $ F = mg $ .
Again, the force of attraction on a body due to earth is $ F = \dfrac{{GMm}}{{{r^2}}} $ (Where, $ G $ is the gravitational constant, $ M $ is the mass of the earth, $ m $ is the mass of the body, and $ r $ is the distance of the body from the center of the earth).
Then, we can write, $ mg = \dfrac{{GMm}}{{{r^2}}} $ or, $ g = \dfrac{{GM}}{{{r^2}}} $
As both the gravitational constant and mass of the earth are constant terms, hence, $ g \propto \dfrac{1}{{{r^2}}} $
So, in other words, we can say that the acceleration due to gravity at a point near the surface of the earth is inversely proportional to the distance between the body and the center of the earth.
Note
As we know, $ g = \dfrac{{GM}}{{{r^2}}} $ , this is the relation between acceleration due to gravity and the gravitational constant. The gravitational constant is a mutually operative force of attraction between two particles of unit mass kept unit distance apart.
Let, the mass of the body is m and the weight of the body is mg. Also, this is the acceleration produced in a freely falling body. Also, this is inversely proportional to the distance between the body and the center of the earth.
Complete step by step answer
The ratio of the weight of a body to its mass gives the acceleration due to gravity at the given place.
Let, the mass of the body is $ m $ .
then, the weight of the body will be $ mg $ .
So, the ratio of the weight of a body to its mass gives the $mg/m = g$ (acceleration due to gravity).
So, we can say that acceleration produced in a freely falling body under the action of gravity is called the acceleration due to gravity.
Additional Information
So, in case of a body of mass $ m $ , using Newton’s second law of motion, we have, $ F = mg $ .
Again, the force of attraction on a body due to earth is $ F = \dfrac{{GMm}}{{{r^2}}} $ (Where, $ G $ is the gravitational constant, $ M $ is the mass of the earth, $ m $ is the mass of the body, and $ r $ is the distance of the body from the center of the earth).
Then, we can write, $ mg = \dfrac{{GMm}}{{{r^2}}} $ or, $ g = \dfrac{{GM}}{{{r^2}}} $
As both the gravitational constant and mass of the earth are constant terms, hence, $ g \propto \dfrac{1}{{{r^2}}} $
So, in other words, we can say that the acceleration due to gravity at a point near the surface of the earth is inversely proportional to the distance between the body and the center of the earth.
Note
As we know, $ g = \dfrac{{GM}}{{{r^2}}} $ , this is the relation between acceleration due to gravity and the gravitational constant. The gravitational constant is a mutually operative force of attraction between two particles of unit mass kept unit distance apart.
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