# Free Fall

When an object falls from any height under the influence of gravitational force only, it is known as free fall. In the case of free fall no change of direction takes place but the magnitude of velocity changes because of acceleration.

This acceleration acts because of the force of gravitation and is denoted by ‘g’. This is called acceleration due to gravity.

## Acceleration Due to Gravitation

Let mass of the object put under free fall = m.

And acceleration due to gravity = g.

Therefore, according to Newton’s Second Law of Motion which states that Force is the product of mass and acceleration,

F = m × g -----------------(i)

Now, according to Universal Law of gravitation,

`F= G*(M*m)/d^2` ---(ii)

Thus, from above two expressions, we get

`mg = G*(M*m)/d^2`

`=>g = (GMm)/(d^2*m)`

`=>g=(GM)/d^2` ---(iii)

Where, g is acceleration due to gravity,

G is the Universal Gravitational Constant.

M is the mass of earth.

And d is the distance between object and centre of earth.

### When object is near the surface of earth

When an object is near the surface of earth, the distance between object and centre of the earth will be equal to the radius of earth because the distance of object is negligible in comparison of the radius of earth.

Let the radius of earth is equal to R.

Therefore, after substituting ‘R’ at the place of ‘d’ we get,

`g = (GM)/R^2` ----(iv)

Since, earth is not a perfect sphere rather it has oblique shape. Therefore, radius at the equator is greater than at the poles.

Since, value of ‘g’ is reciprocal of the square of radius of earth, thus, the value of ‘g’ will be greater at the poles and less at the equator.

And the value of ‘g’ will decrease with increase of distance of object from earth.

## Calculation of value of `g`

We know that,

The accepted value of `G` is `6.673xx10^(-11)Nm^2kg^(-2)`

The mass of earth, `M=6xx10^(24)kg`

The radius of earth, `R = 6.4xx10^6m`

Therefore, by using expression `g = (GM)/R^2`, the value of `g` can be calculated.

Therefore, after substituting the value of G, M and R in the expression for ‘g’ we get.

`g=(6.673xx10^(-11)Nm^2kg^(-2)xx6xx10^(24)kg)/(6.4xx10^6m)^2`

`=>g= 9.8\ m\ s^(-2)`

Motion of an object under the influence of gravitational force of earth

The expression for ‘g’ is written as `g=(GM)/R^2`

Since, the value of ‘g’ does not depend upon the mass or distance of an object, therefore, all objects fall over the earth with the same rate.

The equations for motion are as follows:

`v=u+at` ---(i)

`s=ut+1/2at^2` ---(ii)

`v^2=u^2+2as` ---(iii)

Therefore, the equations of motion are also applied to calculate the velocity, distance, etc by replacing ‘a’ by ‘g’. After substituting ‘g’ at the place of ‘a’ we get above equations as follows:

`v=u+g\t` ---(i)

`s=ut+1/2g\t^2` ---(ii)

`v^2=u^2+2gs` ---(iii)

In the calculation; initial velocity (u), final velocity (v), time taken (t), or distance covered (s), the value of ‘g’ is taken as positive in the case of object moving towards earth and taken as negative in the case of object is thrown in opposite direction of earth.