Free Fall & Gravity
Whenever objects fall towards the earth under this force alone, we say that the objects are in free fall. While falling, there is no change in the direction of motion of the objects. But due to the earth’s attraction, there will be a change in the magnitude of the velocity. Any change in velocity involves acceleration.
Whenever an object falls towards the earth, an acceleration is involved. This acceleration is due to the earth’s gravitational force. Therefore, this acceleration is called the acceleration due to the gravitational force of the earth (or acceleration due to gravity). It is denoted by g. The unit of g is the same as that of acceleration, that is, ms^{-2}.
Calculation of Acceleration due to Gravity
Let the mass of the object put under free fall is m & acceleration due to gravity is g.
Therefore, according to Newton’s second law of motion which states that force is the product of mass and acceleration,
F = m x g ------------ (i)
Now, according to the universal law of gravitation,
F = G⋅M⋅m/ d^{2} ------------ (ii)
Thus, from the above two expressions, we get
mg = G⋅M⋅m/ d^{2}
⇒ g = G.M.m/ d^{2}⋅m
⇒ g = G.M/ 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 the earth.
We know that an object experiences acceleration during free fall. From Eq. (iii), this acceleration experienced by the object is independent of its mass. This means that all objects hollow or solid, big or small, should fall at the same rate.
When the object is near the surface of the earth
When the object is near the surface of the earth, the distance between object and centre of the earth will be equal to the radius of the earth because the distance between the object & the earth is negligible in comparison of the radius of the earth. Let the radius of the earth is equal to R. Therefore, after substituting ‘R’ at the place of ‘d’ we get,
g = G.M/ R^{2 }------------- (iv)
Since, the earth is not a perfect sphere rather it has oblique shape. Therefore, the radius at the equator is greater than at the poles. Since, the value of ‘g’ is reciprocal of the square of the radius of the 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 the increase of distance of the object from earth.
Value of g on the Earth' Surface
We know that, G = 6.673×10−11 Nm^{2}/kg^{−2}
Mass of the earth, M = 6 × 10^{24} kg
Radius of the earth, R = 6.4 × 10^{6} m
Therefore, the value of g can be calculated after substituting the value of G, M and R in the expression for ‘g’ we get.
g = G.M/R^{2},
⇒ g = 6.673×10^{−11} Nm^{2}/kg^{−2} × 6 × 10^{24} kg / (6.4 × 10^{6} m)^{2}
⇒ g = 9.8 ms^{−2}
Difference between Gravitation & Gravity