Acceleration due to Gravity (G)
Let us try to understand the meaning of free fall by performing this activity.
Activity 10.2: ( throwing a stone upward )
* Take a stone.
* Throw it upwards.
* It reaches a certain height and then it starts falling down.
We have learnt that the earth attracts objects towards it. This is due to the gravitational force. Whenever objects fall towards the earth under this force alone, we say that the objects are in free fall. Is there any change in the velocity of falling objects? 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.
We know from the second law of motion that force is the product of mass and acceleration. Let the mass of the stone in activity 10.2 be m. We already know that there is acceleration involved in falling objects due to the gravitational force and is denoted by g. Therefore the magnitude of the gravitational force F will be equal to the product of mass and acceleration due to the gravitational force, that is,
F = m g ________(10.6)
From Eqs. (10.4) and (10.6) we have
\(mg=G\frac{M\times m}{d^2}\\ or \,\,\,\,g=G\frac{M}{R^2}\,\,\,\,\,\,\,\,-----(10.7)\)
where M is the mass of the earth, and d is the distance between the object and the earth. Let an object be on or near the surface of the earth. The distance d in Eq. (10.7) will be equal to R, the radius of the earth. Thus, for objects on or near the surface of the earth,
\(mg=G {\frac{M\times m}{R^2}} \,\,\,\,\,\,\,\, ---(10.8)\\ g=G {\frac{M}{R^2}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, ---(10.9)\)
The earth is not a perfect sphere. As the radius of the earth increases from the poles to the equator, the value of g becomes greater at the poles than at the equator. For most calculations, we can take g to be more or less constant on or near the earth. But for objects far from the earth, the acceleration due to the gravitational force of the earth is given by Eq. (10.7).
Source: This topic is taken from NCERT TEXTBOOK
CALCULATING VALUE OF “G”
To calculate the value of g, we should put the values of G, M, and R in Eq. (10.9), namely, universal gravitational constant, G = 6.7\(\times\) 10–11 Nm2kg-2, the mass of the earth, M = 6 \(\times\)1024 kg, and radius of the earth, R = 6.4 \(\times\) 106 m.
\(g=G\frac{M}{R^2}\\=\frac{ 6.7\times 10^{-11} Nm^2 kg^{-2}\times 6\times 10^{24} kg}{(6.4\times 10^6 m)^2}\)
= 9.8 ms–2.
Thus, the value of acceleration due to gravity of the earth, g = 9.8 ms–2.
Source: This topic is taken from NCERT TEXTBOOK
MOTION OF OBJECTS UNDER THE INFLUENCE OF GRAVITATIONAL FORCE OF THE EARTH
Let us do an activity to understand whether all objects hollow or solid, big or small, will fall from a height at the same rate.
Activity 10.3:
* Take a sheet of paper and a stone. Drop them simultaneously from the first floor of a building. Observe whether both of them reach the ground simultaneously.
* We see that paper reaches the ground a little later than the stone. This happens because of air resistance. The air offers resistance due to friction to the motion of the falling objects. The resistance offered by air to the paper is more than the resistance offered to the stone. If we do the experiment in a glass jar from which air has been sucked out, the paper and the stone would fall at the same rate.
We know that an object experiences acceleration during free fall. From Eq. (10.9), this acceleration experienced by an object is independent of its mass. This means that all objects hollow or solid, big or small, should fall at the same rate. According to a story, Galileo dropped different objects from the top of the Leaning Tower of Pisa in Italy to prove the same.
As 'g' is constant near the earth, all the equations for the uniformly accelerated motion of objects become valid with acceleration 'a' replaced by 'g'. The equations are:
v = u + at ------ (10.10)
s = ut + (1/2)at2 ------ (10.11)
v2 = u2 + 2as ------ (10.12)
where u and v are the initial and final velocities and s is the distance covered in time, t.
In applying these equations, we will take acceleration, 'a' to be positive when it is in the direction of the velocity, that is, in the direction of motion. The acceleration, 'a' will be taken as negative when it opposes the motion.
Illustration 10.2:
A car falls off a ledge and drops to the ground in 0.5 s. Let g = 10 ms–2 (for simplifying the calculations).
i. What is its speed on striking the ground?
ii. What is its average speed during the 0.5 s?
iii. How high is the ledge from the ground?
Sol:
Time, t = ½ second
Initial velocity, u = 0 ms–1
Acceleration due to gravity, g = 10 ms–2
Acceleration of the car, a = + 10 ms–2 (downward)
(i) speed v = at
= 10 ms–2 \(\times\) 0.5 s
v = 5 ms–1
(ii) average speed = \(\frac{u+v}{2}\)
= \((0ms^{-1}+5ms^{-1})/2\)
=2.5 ms-1
(iii) distance travelled, s = ½ a t2
= ½ \(\times\) 10 ms–2 \(\times\) (0.5 s)2
= ½ \(\times\) 10 ms–2 \(\times\) 0.25 s2
= 1.25 m
Thus,
(i) its speed on striking the ground = 5 ms–1
(i) its average speed during the 0.5 s = 2.5 ms–1
(iii) height of the ledge from the ground= 1.25 m.
Illustration 10.3:
An object is thrown vertically upwards and rises to a height of 10 m. Calculate
(i) the velocity with which the object was thrown upwards and
(ii) the time is taken by the object to reach the highest point.
Sol:
Distance travelled, s = 10 m Final velocity, v = 0 m s–1
Acceleration due to gravity, g = 9.8 m s–2
Acceleration of the object, a = –9.8 m s–2
(upward motion)
(i) v2 = u2 + 2a s
0 = u2 + 2 \(\times\) (–9.8 m s–2) \(\times\) 10 m
–u2 = –2 \(\times\) 9.8 \(\times\)10 m2s–2
u = \(\sqrt{196}\) ms-1
u = 14 ms-1
(ii) v = u + a t
0 = 14 ms–1 – 9.8 ms–2 \(\times\) t
t = 1.43 s.
Thus,
Initial velocity, u = 14 ms–1, and
Time taken, t = 1.43 s.
Questions
1. What do you mean by free fall?
2. What do you mean by the acceleration due to gravity?
Source: This topic is taken from NCERT TEXTBOOK