Question

How is the acceleration due to the gravity calculated?

Answer 1

Hint: The acceleration due to the gravity by an object due to gravity. Its S.I unit is $m/{{s}^{2}}$ it has both magnitude and direction so it is the quantity of the vector acceleration due to the gravity is expressed by g. at the sea level the normal value of g on the earth’s surface is $9.8m/{{s}^{2}}$.

Formula used:
$\begin{align}
  & F=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}} \\
 & F=mg \\
\end{align}$

Complete step by step solution:
In order to calculate acceleration due to the gravity we will use the newton ‘s law of the universal gravitation which states that every particle attracts every other particle in the universe with a force that is directly proportional to the product of the masses and the inversely proportional to the square of the distance between them.
$F=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$
Where,
F = gravitational force between the two bodies
${{m}_{1}}$and ${{m}_{2}}$is a mass of the two objects
r = distance between two objects
G = universal gravitational constant.

Now if we take earth as one object and any other object which is on the surface on the earth we can rewrite equation as,
$F=\dfrac{GMm}{{{R}^{2}}}....\left( 2 \right)$
Where,
M = mass of the earth
m = mass of the object
R = radius of the earth.

Now we know that force on any object
F = ma

Here we can take acceleration as g which is acceleration due to the gravity
F = mg…… (2)

Now from the equation 1 and the equation 2
$\begin{align}
  & mg=\dfrac{GMm}{{{R}^{2}}} \\
 & \therefore g=\dfrac{GM}{{{R}^{2}}}.....\left( 3 \right) \\
\end{align}$

Here, value of G = $6.67\times {{10}^{-11}}{{m}^{3}}/kg{{s}^{2}}$
\[\begin{align}
  & M=5.972\times {{10}^{24}}kg \\
 & R=6371km \\
\end{align}\]

By substituting all the values in the equation we get value of g
$\therefore g=9.8m/{{s}^{2}}$

Therefore, value of the acceleration due to the gravity (g) is $9.8m/{{s}^{2}}$.

Note:
We know that the value of the earth’s radius is different at some place on the earth therefore the value of the acceleration due to the gravity is varying at some place on the earth but the average value remains same which is $9.8m/{{s}^{2}}$.