# Why is acceleration due to gravity constant?

Jul 25, 2018

Mass of the falling body cancels out because the force of gravity is directly proportional to the mass and the acceleration caused by that force is inversely proportional to the mass.

#### Explanation:

The force of gravity on a body of mass $m$ is what we call its weight, $W$, and is given by

$W = m \cdot g$

The only data used in the calculation of $g$ is 3 constants.

$g = \frac{G \cdot {M}_{e}}{R} _ {e}^{2}$

where

$G$ is the universal constant of gravitation, value: $6.673 \cdot {10}^{-} 11 \frac{N \cdot {m}^{2}}{\text{kg}} ^ 2$
${M}_{e}$ is the mass of the Earth: value $5.983 \cdot {10}^{24} k g$
${R}_{e}$ is the radius of the Earth: value $6.37 \cdot {10}^{6} m$

Newton"s 2nd Law tells us that trhe acceleration of that body of mass $m$ is given by

$a = \frac{W}{m} = \frac{\cancel{m} \cdot g}{\cancel{m}} = g$

Since $g$ is calculated using only constants, $g$ is a constant.

I hope this helps,
Steve

Jul 26, 2018

As per Law of Universal Gravitation the force of attraction ${F}_{G}$ between two bodies of masses is directly proportional to the product of masses of the two bodies. it is also inversely proportional to the square of the distance $r$ between the two. In case one of the body is the earth of mass ${M}_{e}$ and other body has mass $m$ we have

${\vec{F}}_{G} = G \frac{{M}_{e} \cdot m}{r} ^ 2 \hat{r}$ .......(1)

Comparing modulii with the equation for Newton's Second Law of Motion

$\vec{F} = m \vec{a}$ ......(2)

we get acceleration due to gravity, which is also called gravity, as

$a = G \frac{{M}_{e}}{r} ^ 2$ ......(3)

from above we see that when the body is located on the surface of the earth gravity is

$a = G \frac{{M}_{e}}{R} _ {e}^{2}$ ......(4)
where ${R}_{e}$ is radius of the earth.

As earth is not a perfect sphere, value of gravity $g$ is not a constant at all the locations on the surface of the earth.

We also know that average radius of earth is $6 , 371 \setminus k m$. This gives us average value of $g$ as $9.81 \setminus m {s}^{-} 2$ on the surface of earth. Also that value of $g$ varies insignificantly if distance between the body and the surface of earth is very small as compared to average radius of the earth.

Therefore, for practical purposes we take acceleration due to gravity as a constant.