Contents
Physics Topics can help us understand the behavior of the natural world around us.
Newton’s third law of motion says that: If an object exerts a force on another object, then the second object exerts an equal and opposite force on the first object. The Newton’s third law of motion also holds good for the force of gravitation. This means that when earth exerts a force of attraction on an object, then the object also exerts an equal force on the earth, in the opposite direction. Thus, even a falling object attracts the earth towards itself. When an object, say a stone, is dropped from a height, it gets accelerated and falls towards the earth and we say that the stone comes down due to the gravitational force, of attraction exerted by the earth. Now, if the stone also exerts an equal and opposite force on the earth, then why don’t we see the earth rising up towards the stone ? We will try to answer this question now.
From Newton’s second law of motion, we know that :
Force = Mass × Acceleration
So, Acceleration = \(\frac{\text { Force }}{\text { Mass }}\)
or a = \(\frac{F}{m}\)
It is clear from this formula that the acceleration produced in a body is inversely proportional to the mass of the body. Now, the mass of a stone is very small, due to which the gravitational force produces a large acceleration in it. Due to large acceleration of stone, we can see the stone falling towards the earth. The mass of earth is, however, very, very large. Due to the very large mass of the earth, the same gravitational force produces very, very small acceleration in the earth. Actually, the acceleration produced in the earth is so small that it cannot be observed. And hence we do not see the earth rising up towards the stone.
It can be shown by calculations that the gravitational force between the earth and a 1 kilogram stone produces an acceleration of 9.8 m/s2 in the stone. Since the acceleration produced in the stone is quite large, we can see the stone falling towards the earth. Now, the same gravitational force produces an acceleration of 1.63 × 10-24m/s2 in the earth. Since the acceleration produced in the earth is extremely small, we cannot see the motion of the earth towards the falling stone.
Free Fall
The falling of a body (or object) from a height towards the earth under the gravitational force of earth (with no other forces acting on it) is called free fall. And such a body is called ‘freely falling body’ (or ‘freely falling object’). So, whenever a body (or object) falls towards the earth on its own, we say that it is under free fall or that it is a freely falling body (or freely falling object). Let us discuss this in a little more detail.
Our earth attracts all the bodies (or objects) which are near its surface so that when a body (say, a stone) is dropped from the roof of a house, it falls down to the earth. Earlier it was thought that the lighter objects fall slowly and the heavier objects fall more rapidly when dropped from the same height and at the same time, because when a feather and metal coin are dropped simultaneously from a roof, the metal coin reaches the ground first and feather takes more time to reach the ground. This was later on found to be wrong by Galileo.
Galileo dropped two stones of different masses from the top of the leaning tower of Pisa, and found that they hit the ground at the same time. From this Galileo concluded that the acceleration of an object falling freely towards the earth does not depend on the mass of the object. During free fall, the heavier objects as well as the lighter objects accelerate at the same rate. It was suggested by Galileo that the slow speed of feather, while falling, is due to the fact that its surface area is very large as compared to its
mass, so the feather experiences much more resistance from air and its speed is slowed down. The metal coin, being small and heavy, does not get so much resistance from air and falls to the ground at a faster rate. If there were no air, the feather and the coin would fall at the same rate. This has been shown to be so by means of experiments done in vacuum by Robert Boyle. The feather and the coin, both, were put in a tall glass jar and the air from the jar was removed by using a vacuum pump. When this jar was inverted, both, feather and the coin fell to the bottom of the jar at the same time showing that in the absence of air resistance, that is, in vacuum, all the objects fall at the same rate. In other words, the acceleration produced in the freely falling bodies is the same for all the bodies and it does not depend on the mass of the falling body.
Acceleration due to Gravity (g)
When an object is dropped from some height, its velocity increases at a constant rate. In other words, when an object is dropped from some height, a uniform acceleration is produced in it by the gravitational pull of the earth and this acceleration does not depend on the mass of the falling object. The uniform acceleration produced in a freely falling body due to the gravitational force of the earth is known as acceleration due to gravity and it is denoted by the letter g. The value of g does not depend on the mass of the body.
The value of g changes slightly from place to place but for most of the purposes it is taken as 9.8 m/s2. Thus, the acceleration due to gravity, g = 9.8 m/s2. In other words, when an object falls to the ground under the action of earth’s gravity, its velocity increases at the constant rate of 9.8 metres per second for every second of time it is falling. When a body is dropped freely, it falls with an acceleration of 9.8 m/s2 and when a body is thrown vertically upwards, it undergoes a retardation of 9.8 m/s2. So, the velocity of a body thrown vertically upwards will decrease at the rate of 9.8 m/s2. The velocity decreases until it reaches zero. The body then falls back to the earth like any other body dropped from that height.
Before we go further and derive a formula to calculate the acceleration due to gravity of earth, please remember that when a body (or object) is ‘on the surface of earth’ or ‘near the surface of earth’ then the distance of the body from the centre of earth is taken as equal to radius of earth (R). So, in this case the distance r in the formula for gravitational force becomes ‘radius of earth’ R.
Calculation of the Acceleration Due to Gravity (g)
If we drop a body (say, a stone) of mass m from a distance R from the centre of the earth of mass M, then the force exerted by the earth on the body is given by universal law of gravitation as :
F = G × \(\frac{M \times m}{R^2}\) (G = Gravitational constant) …. (1)
This force exerted by the earth produces acceleration in the stone due to which the stone moves downwards. We also know that:
Force = Mass × Acceleration
or F = m × a
So, Acceleration of stone, a = \(\frac{F}{m}\) ……. (2)
Putting the value of force F from equation (1) in the above relation, we get:
Acceleration, a = \(\frac{G \times M \times m}{R^2 \times m}\)
or a = G × \(\frac{M}{R^2}\)
The acceleration produced by the earth is known as acceleration due to gravity and represented by the symbol g. So, by writing ‘g in place of ‘a in the above equation, we get:
Acceleration due to gravity, g = G × \(\frac{M}{R^2}\)
where G = gravitational constant
M = mass of the earth
and R = radius of the earth
This is the formula for calculating the acceleration due to gravity on or near the surface of the earth. To calculate the value of g, we should put the values of G, M and R in the above formula.
Now, Gravitational constant, G = 6.7 × 10-11 Nm2/kg2
Mass of the earth, M = 6 × 1024 kg
And, Radius of the earth, R = 6.4 × 106 m
Putting these values of G, M and R in the above formula, we get:
g = \(\frac{6.7 \times 10^{-11} \times 6 \times 10^{24}}{\left(6.4 \times 10^6\right)^2}\)
or g = 9.8 m/s2
Thus, the value of acceleration due to gravity is 9.8 m/s2. But sometimes, to make the calculations easy, the value of g is taken as a round figure of 10 m/s2. This is done just for the sake of convenience. Please note that this acceleration due to gravity acts in the direction of the line joining the body to the centre of the earth.
Variation of Acceleration Due to Gravity (g)
The value of acceleration due to gravity of earth (g) depends on the values of gravitational constant (G), mass of the earth (M), and radius of the earth (R). As gravitational constant (G) and mass of earth (M) are always constant , the value of acceleration due to gravity (g) is.constant as long as the radius of earth R remains constant. Hence the value of g is constant at a given place on the surface of the earth. Please note that the value of acceleration due to gravity, g, is not constant at all the places on the surface of the earth.
This is due to the fact that the earth is not a perfect sphere, so the value of its radius R is not the same at all the places on its surface. In other words, due to the flattening of the earth at the poles, all the places on its surface are not at the same distance from its centre and so the value of g varies with latitude. Since the radius of the earth at the poles is minimum, the value of g is maximum at the poles. Again, the radius of earth is maximum at the equator, so the value of g is minimum at the equator (because radius occurs in the denominator of the formula for g).
We have just seen that:
g = G × \(\frac{M}{R^2}\)
We find that the value of g is inversely proportional to the square of distance from the centre of the earth. Now, as we go up from the surface of the earth, the distance from the centre of the earth increases, and hence the value of g decreases (because R increases in this case). The value of acceleration due to gravity, g, at an altitude of 200 km above the surface of the earth is 9.23 m/s2; at an altitude of 1000 km, g is 7.34 m/s2; at 5,000 km above earth g is 3.08 m/s2; at 10,000 km g is 1.49 m/s2; at 20,000 km, g is 0.57 m/s2 whereas at a height of 30,000 km above the surface of earth, the value of g is only 0.30 m/s2.
The above formula suggests that the value of g should increase on going down inside the earth because then the value of R decreases. This, however, is not true. Actually this formula for g is not applicable at any point inside the surface of earth. At the moment, it will be sufficient for us to know that the value of g also decreases as we go down inside the earth, and it becomes zero at the centre of the earth. Please note that the value of acceleration due to gravity, g, is maximum on the surface of the earth, it decreases on going above the surface of earth or on going inside the surface of the earth.
Acceleration Due to Gravity (g) Does Not Depend on the Mass of a Body
Let us write down the formula for the acceleration due to gravity, g, once again :
g = G × \(\frac{M}{R^2}\)
where G = gravitational constant
M = mass of earth and
R = radius of earth
We find that this formula for the acceleration due to gravity involves only the mass of earth. It does not involve the mass of the body on which the force of gravity of earth acts. Since the acceleration due to gravity does not depend on the mass of the body, all the bodies (whether heavy or light) fall with the same acceleration towards the surface of the earth. Thus, a big stone will fall with the same acceleration as a small stone because the acceleration due to gravity of earth (which acts on them during their free fall) does not depend on their mass. Both the stones are acted upon by the same acceleration of 9.8 m/s2. In other words, if a big stone and a small stone are dropped from the roof of a house simultaneously, they will reach the ground at the same time. Let us solve some problems now.
Example Problem 1.
Calculate the value of acceleration due to gravity on the surface of the moon. (Given : Mass of the moon = 7.4 × 1022 kg; Radius of moon = 1740 km; G = 6.7 × 10-11 Nm2/kg2)
Solution.
The formula for calculating the acceleration due to gravity is :
g = G × \(\frac{M}{R^2}\)
Here, Gravitational constant, G = 6.7 × 10-11 Nm2/kg2
Mass of the moon, M = 7.4 × 1022 kg
And, Radius of the moon, R = 1740 km
= 1740 × 1000 m
= 1.74 × 106 m
Now, putting these values of G, M and R in the above formula, we get:
g = \(\frac{6.7 \times 10^{-11} \times 7.4 \times 10^{22}}{\left(1.74 \times 10^6\right)^2}\)
or g = 1.63 m/s2
Please note that the value of g on the moon is about one-sixth \(\left(\frac{1}{6}\right)\) of the value of g on the earth (the value of acceleration due to gravity on the earth being 9.8 m/s2).
Example Problem 2.
The earth’s gravitational force causes an acceleration of 5 m/s2 in a 1 kg mass somewhere in space. How much will the acceleration of a 3 kg mass be at the same place ?
Solution.
The acceleration produced by the gravitational force of earth does not depend on the mass of the object. So, the acceleration produced in the 3 kg mass will be the same as that produced in 1 kg mass. That is, the acceleration produced will be 5 m/s2.