In the previous section we saw the Universal gravitation constant. In this section we will see acceleration due to Gravity.
1. Consider a stone thrown up from the ground. After reaching a height, it will fall down.
2. While falling down, the only force acting on it is the ‘gravitational force of attraction’.
■ That stone is said to be in ‘free fall’.
3. When there is a force, there is an acceleration. We have seen that Force = mass × acceleration. Details here.
4. So the stone which is in free fall is under acceleration. Because of the acceleration, the velocity of the stone increases. This acceleration is due to the gravitational force of the earth.
■ So it is called ‘acceleration due to gravity’. It is denoted by the letter g.
5. The unit of g is same as that of acceleration. That is., ms-2
• d is the distance between the object and the centre of the earth
• G is the universal gravitation constant.
2. Note that the object under consideration is 'on the surface or near the surface of the earth'. So the distance 'd' becomes 'R', the radius of the earth. So we can write:
F = G(m1m2⁄R2)
3. We know that F = m2a. Here, a = g, the acceleration due to gravity. So we can write: F = m2g. That means, the force acting on the object is equal to the product of it’s mass and the acceleration due to gravity.
4. No other force is acting on the object. So we can equate (2) and (3). We get:
m2g = G(m1m2⁄R2)
■ From this we get:
Eq.3.2:
g = G(m1⁄R2)
• Where m1 is the mass of the earth
Let us see how this variation in radius affects the value of g:
At the equator, R is greater. In equation 3.2 above, R is in the denominator. So, the greater value of R will give a lesser value of g at the equator. That means, g is lesser at the equator than at the poles. The values are:
• Value of g at the polar regions = 9.83 m s-2
• Value of g at the equator = 9.78 m s-2
■ The average value of g on the surface of the earth is taken as 9.8 m s-2 for solving numerical problems.
• This is obtained when we use an average value of R as 6.4×106 m. Let us do that calculation ourselves:
From Eq.3.2 above, we have:
g = G(m1⁄R2)
• Where m1 is the mass of the earth = 6×1024 kg
• R is the mass of the earth = 6.4×106 m
• G is the universal constant of gravitation = 6.7×10-11 m
Substituting the values, we get:
■ In our present discussion, we have seen that, the objects under free fall are subjected to a constant acceleration of g.
• So can we use those three equations for freely falling objects, by replacing a with g? Let us find out:
1. When an object falls freely, the force acting on it is mg. So it is clear that, with increase in mass, the force will also increase.
2. Consider two objects shown in the fig.3.4 below. One has a greater mass than the other.
• If we drop both of them from the same height h, which one will reach the ground first?
3. At a first glance, we will be inclined to think that the object 2 which has heavier mass will reach the ground first.
• But the fact is that, both will reach the ground together at the same instant. Let us see the reason:
4. When the two objects are released from rest, their initial velocity u will be zero. Both of them will then begin to gain velocity. That means, both of them will be accelerated.
5. No force other than the gravitational force is acting on them. So the 'acceleration experienced' will be the same 'g' for both of them. It is the acceleration due to gravity.
6. From eq.3.2, we see that, g is a constant. Because on the right side of eq.3.2, G, m1 and R are all constants.
7. Now consider the second equation of motion: s = ut + 1⁄2at2
8. In our present case s = h, u = 0, and a = g. Let us substitute these known values and find the unknown 't'. We get:
h = 0×t + 1⁄2×g×t2 ⇒ h = 1⁄2×g×t2 ⇒
Eq.3.3:
t = √(2h⁄g)
9. We will get this same 't' for both object 1 and object 2. Because, the two objects are distinguished from each other by only one property which is their mass. The final equation 3.3 which gives us the time required to reach the ground does not involve m. That means t is same for both the objects. That means both the objects will reach the ground at the same instant.
A video can be seen here.
■ Several centuries earlier, Galileo had conducted similar experiments. He dropped different objects from the top of the leaning tower of Pisa. He then published his findings. He argued that the feather took more time because, it experienced greater resistance from air. But there were no facilities to create vacuum in a jar at that time.
In the next section, we will see the relation between Mass and Weight.
1. Consider a stone thrown up from the ground. After reaching a height, it will fall down.
2. While falling down, the only force acting on it is the ‘gravitational force of attraction’.
■ That stone is said to be in ‘free fall’.
3. When there is a force, there is an acceleration. We have seen that Force = mass × acceleration. Details here.
4. So the stone which is in free fall is under acceleration. Because of the acceleration, the velocity of the stone increases. This acceleration is due to the gravitational force of the earth.
■ So it is called ‘acceleration due to gravity’. It is denoted by the letter g.
5. The unit of g is same as that of acceleration. That is., ms-2
1. Let the mass of an object on the surface or near the surface of the earth be m2. Then the force F exerted by the earth on that object is given by:
• Where m1 is the mass of the earth. • d is the distance between the object and the centre of the earth
• G is the universal gravitation constant.
2. Note that the object under consideration is 'on the surface or near the surface of the earth'. So the distance 'd' becomes 'R', the radius of the earth. So we can write:
F = G(m1m2⁄R2)
3. We know that F = m2a. Here, a = g, the acceleration due to gravity. So we can write: F = m2g. That means, the force acting on the object is equal to the product of it’s mass and the acceleration due to gravity.
4. No other force is acting on the object. So we can equate (2) and (3). We get:
m2g = G(m1m2⁄R2)
■ From this we get:
Eq.3.2:
g = G(m1⁄R2)
• Where m1 is the mass of the earth
We know that earth is not a perfect sphere. The value of R is greater at the equator than at the poles. This is shown in the fig.3.3 below:
 |
| Fig.3.3 |
• Value of g at the polar regions = 9.83 m s-2
• Value of g at the equator = 9.78 m s-2
■ The average value of g on the surface of the earth is taken as 9.8 m s-2 for solving numerical problems.
• This is obtained when we use an average value of R as 6.4×106 m. Let us do that calculation ourselves:
From Eq.3.2 above, we have:
g = G(m1⁄R2)
• Where m1 is the mass of the earth = 6×1024 kg
• R is the mass of the earth = 6.4×106 m
• G is the universal constant of gravitation = 6.7×10-11 m
Substituting the values, we get:
We have seen 'motion of objects' in chapter 1. We derived the three equations of motion. Details here. Those equations can be used to calculate the following unknown quantities:
• Distance travelled s
• Initial velocity u
• Final velocity v
• Time of travel t
Those three equations are valid when an object moves with a constant acceleration 'a'.
■ In our present discussion, we have seen that, the objects under free fall are subjected to a constant acceleration of g.
• So can we use those three equations for freely falling objects, by replacing a with g? Let us find out:
1. When an object falls freely, the force acting on it is mg. So it is clear that, with increase in mass, the force will also increase.
2. Consider two objects shown in the fig.3.4 below. One has a greater mass than the other.
 |
| Fig.3.4 |
3. At a first glance, we will be inclined to think that the object 2 which has heavier mass will reach the ground first.
• But the fact is that, both will reach the ground together at the same instant. Let us see the reason:
4. When the two objects are released from rest, their initial velocity u will be zero. Both of them will then begin to gain velocity. That means, both of them will be accelerated.
5. No force other than the gravitational force is acting on them. So the 'acceleration experienced' will be the same 'g' for both of them. It is the acceleration due to gravity.
6. From eq.3.2, we see that, g is a constant. Because on the right side of eq.3.2, G, m1 and R are all constants.
7. Now consider the second equation of motion: s = ut + 1⁄2at2
8. In our present case s = h, u = 0, and a = g. Let us substitute these known values and find the unknown 't'. We get:
h = 0×t + 1⁄2×g×t2 ⇒ h = 1⁄2×g×t2 ⇒
Eq.3.3:
t = √(2h⁄g)
9. We will get this same 't' for both object 1 and object 2. Because, the two objects are distinguished from each other by only one property which is their mass. The final equation 3.3 which gives us the time required to reach the ground does not involve m. That means t is same for both the objects. That means both the objects will reach the ground at the same instant.
This was proved experimentally by Newton.
1. He placed a feather and a coin at the top, inside a long glass jar.
Why use a jar made of glass?
Because we want to see the speed at which the objects fall from the top of the jar to the bottom.
2. When the objects were released, the coin reached the bottom first. The feather reached only a little later.
3. He then repeated the experiment. This time, all the air inside the glass jar was sucked out. So there was a perfect vacuum.
4. When the objects were released, both of them reached the bottom at the same instant. So what happened when there was no vacuum?
Ans: When there was no vacuum, the air offered resistance to the fall of the objects. Resistance experienced by the feather will be much greater than that experienced by the coin. So the feather will reach only after some time.A video can be seen here.
■ Several centuries earlier, Galileo had conducted similar experiments. He dropped different objects from the top of the leaning tower of Pisa. He then published his findings. He argued that the feather took more time because, it experienced greater resistance from air. But there were no facilities to create vacuum in a jar at that time.
So now we know that, for free fall, the final velocity and time taken are independent of mass. All the three equations of motion can be applied for objects in free fall. Let us see some solved examples:
Solved example 3.4
A car falls off a ledge and drops to the ground in 0.5 s. Let g = 10 m s-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?
Solution:
1. The car falls off the ledge. So initial velocity u = 0
2. It reaches the ground in time t = 0.5 s
3. We have to find the final velocity v. We can use the first equation of motion:
v = u + at ⇒ v = u + gt ⇒ v = 0 + 10 × 0.5 ⇒ v = 5 m s-1. This is the answer for part (i)
4. Average speed = (initial velocity + final velocity)⁄2 = (0+5)⁄2 = 2.5 m s-1. This is the answer for part (ii)
5. Height of fall is the distance travelled s. We can use the second equation of motion:
s = ut + 1⁄2at2 ⇒ h = 0×t + 1⁄2×g×t2 ⇒ h = 1⁄2×10×0.52 ⇒ h = 1.25 m. This is the answer for part (iii)
Solved example 3.5
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 taken by the object to reach the highest point.
Solution:
1. The object is thrown upwards with an initial velocity u. We have to find this u
2. The distance travelled = s = 10 m
3. This distance travelled is the maximum height reached by the object. When it reaches this maximum height, it's velocity will be zero. So we get: Final velocity v = 0
4. The object is subjected to the 'acceleration due to gravity' g. But g acts down wards. The object is travelling upwards. So acceleration should be taken as negative, since it is opposite to the direction of motion. Thus g = -9.8 m s-2.
5. We can use the third equation of motion:
v2 = u2+ 2as ⇒ v2 = u2+ 2gs ⇒ 02 = u2+ 2 × (-9.8) × 10 ⇒ 02 = u2+ 2 × (-9.8) × 10
⇒ 0 = u2 - 196 ⇒ u2 = 196 ⇒ u = √196 = 14 m s-1. This is the answer for part (i)
6. To find the time, we can use the first equation of motion:
v = u + at ⇒ v = u + gt ⇒ 0 = 14 + (-9.8) × t ⇒ -14 = -9.8t
⇒ 14 = 9.8t ⇒ t = 1.43 s. This is the answer for part (ii)
In the next section, we will see the relation between Mass and Weight.
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