Free Fall

Key Concepts

• Free Fall
• Acceleration Due To Gravity
• Mass & Weight
• Weight of an object on the Moon

Recall: 

• A body falling towards the Earth also pulls the Earth towards itself. 
• The acceleration gained by the Earth is negligibly small for its movement to be noticeable. 
• Yes, there is a force of attraction between any two bodies in this universe. 
• The attractive force varies directly as the product of the masses and inversely as the square of the distance between them. 

Introduction: 

The bodies falling towards the Earth experience two forces, namely the downward gravitational force and the upward air resistance (generally small). If the upward air resistance is neglected, the body moves only under the action of gravity. Such a motion in which the body moves under the action of gravity alone is called free fall. An example of a freely falling body without air resistance is the revolution of the Moon around the Earth, as outer space has no air. 

Explanation: 

Free fall- A uniformly accelerated motion 

Consider two points A and B, in the path of a ball falling from the top of a building, as shown in  
figure 3.1 below. A is the initial point of the motion, and B is located somewhere in the path. The ball was dropped at point A, which means it had zero velocity at that point. And, at point B, it is moving, which means it has some velocity. Therefore, there has been an increase in velocity from the initial point A and another point B in the motion of the ball. This means that the ball has been accelerating (increasing its velocity with time) throughout its motion. 

The gravitational force of pull exerted by the Earth on a falling body is constant. Moreover, from Newton’s second law of motion, it is known that the acceleration of a moving body is directly proportional to the force which acts on it. From the two statements above, it can be concluded that the acceleration of a freely falling body is constant as the gravitational force that pulls it down is constant throughout. 

Acceleration due to gravity 

The acceleration experienced by a freely falling body on the Earth is called the acceleration due to gravity. It is denoted by the letter g. It is measured in the SI unit of m/s², i.e., the same as the acceleration.  

Consider a body of mass m freely falling under the action of Earth’s gravity. Let the Earth’s mass be M, its radius R, and the height at which the object is located from the Earth’s surface be r, as shown in figure 3.2 below. 

From Newton’s second law, 

The force experienced by the body = F = ma = mg 

From the universal law of gravitation, 

The force experienced by the body = F = G (mM/r²) 

On equating the previous two equations, we get, 

F = mg = G (mM/r²) 

g = G (mM/r²)/m 

g = (GM/r²) 

Acceleration due to the gravity of the Earth: 

The acceleration due to gravity on the Earth is given by, 

g = (GM/r²) 

The universal gravitational constant = G = 6.67 × 10^–11 N m²/kg²  

Mass of the Earth, M = 6 × 10²⁴ kg 

The radius of the Earth, R = 6.4 × 10⁶ m 

The acceleration due to the gravity of the Earth is, 

g = (6.67 × 10^–11 × 6 × 10²⁴/(6.4 × 10⁶)²) 

g = 9.8 m/s² 

Therefore, the value of the acceleration due to gravity on the Earth is 9.8 m/s². 

The acceleration due to gravity is the same for various objects falling towards the Earth. It does not depend on the mass of an object. It means, that no matter how heavy or light an object is, it falls with the same acceleration toward the Earth. To check this, two objects need to be dropped at the same time from the same height above the Earth. If they land at the same time, the acceleration due to gravity is the same for both, otherwise not.  

Consider two objects, namely a feather and a stone. The stone is heavier as compared to the feather. If they are dropped at the same time from the same height above the Earth, the stone lands first, and the feather lands later. This is because the air resistance acting on the feather is greater than that of the stone, which reduces its speed. If the same experiment is conducted inside an evacuated glass tube, the two will land at the same time, as the air resistance is eliminated. 

It was believed earlier that the acceleration of a falling body depends upon its mass. In the late 16th century, Galileo Galilei proved this to be wrong by dropping two objects of different masses from the top of the leaning tower of Pisa. Thus, the acceleration of a freely falling body is independent of its mass. 

Mass and weight 

The mass of a body is a measure of the amount of matter contained in a body or a measure of its inertia. It is measured in the SI unit of kg. Its value remains the same everywhere. It is a scalar quantity, as it only has size (magnitude). 

The weight of an object is the force with which it is attracted toward the Earth. 

F = m × a 

F = m × g, as the acceleration due to gravity is g 

Therefore, weight, W = mg 

The weight of an object is measured in the SI units of Newton (N). 

It varies from place to place with variation in the value of the acceleration due to gravity (g). 

Weight of an object on the moon 

The weight of an object of mass m on the Moon is given by, 

W_m = G m × M_m/ R_m² 

Where G = 6.67 × 10^-11 N m²/kg² = the universal gravitational constant 

            M_m = 7.36 × 1022 kg = mass of the Moon 

             R_m = 1.74 × 106 m = radius of the Moon 

Similarly, the weight of an object of mass m on the Earth is given by, 

W_e = G m × M_e/ R_e² 

Where G = 6.67 × 10^-11 N m²/kg²  

             M_e = 6 × 10²⁴ kg = mass of the Earth 

              R_e = 6.4 × 10⁶ m = radius of the Earth 

On substituting all the values in the respective equations, we get, 

W_m / W_e = (G m x M_m/ R_m²)/ (G m x Me/ R_e²) 

W_m / W_e = 0.165, which is nearly equal to 1/6 

Therefore, the weight of an object on the Moon = (1/6) of its weight on the Earth. 

Question: 

Calculate the weight of a person of mass 55 kg on the Earth and the Moon. 

Solution: 

The mass of the person = 55 kg 

The acceleration due to gravity on the Earth = 9.8 m/s² 

His weight on the Earth, We = m × g 

 = 55 × 9.8 

 = 539 N 

His weight on the Moon, W_m = (1/6) × 539 N 

  = 89.83 N