# 03_GRAVITATION_F class 9

Sir Isaac Newton: The Universal Law of Gravitation

There is a popular story that Newton was sitting under an apple tree, an apple fell on his head, and he suddenly thought of the Universal Law of Gravitation. As in all such legends, this is almost certainly not true in its details, but the story contains elements of what actually happened.

Probably the more correct version of the story is that Newton, upon observing an apple fall from a tree, began to think along the following lines: The apple is accelerated, since its velocity changes from zero as it is hanging on the tree and moves toward the ground. Thus, by Newton’s 2nd Law there must be a force that acts on the apple to cause this acceleration. Let’s call this force “gravity”, and the associated acceleration the “acceleration due to gravity”.

Universal Law of Gravitation

Every object in the universe attracts every other object with a force directed along the line of centers of the two objects that is proportional to the product of their masses and inversely proportional to the square of the separation between the two objects.

…..(i)

where,

Fg = gravitational force

m1&m2 are the masses of the two objects,

r = separation between the objects

G = the universal gravitational constant.

The constant of proportionality G is known as the universal gravitational constant. It is termed as “universal constant” because it is thought to be the same at all places and all times, and thus universally characterizes the intrinsic strength of the gravitational force. The SI unit of G is Nm2 kg–2. The accepted value of G is 6.673 × 10–11  Nm2kg–2. If the m1 and m2 of the two bodies are 1 kg each and the distance R between them is 1m, then putting  m1 = 1 kg; m2 = 1 kg and R = 1 m in equation (i) we get

F = G

Thus, the gravitational constant G is numerically equal to the force of gravitation which exists between two bodies of unit masses kept at a unit distance from each other.

Importance of the Universal Law of Gravitation

The universal law of gravitation explained several phenomena such as:

1. i) The rotation of earth around the sun or moon around the earth is explained on the basis of this law.
2. ii) The working of solar and lunar eclipses, made on the basis of this law.

iii)    The formation of tides in the oceans is due the force of attraction between moon and ocean water.

1. iv) The calculations of the orbits and the time period of the satellites is made on the basis of this law.
2. v) Newton’s law of gravitation helps in finding mass and distance of planets from earth.

Gravitation and Gravity

 Gravitation Gravity (i) Gravitation is the force of attraction between any two bodies of the universe. (ii) Gravity is the earth’s gravitational pull on the body lying on or near the surface of the earth. (ii) The gravitational force on a body A of mass m1 due to a body B of mass m2 placed at a distance r is [where G = universal gravitational constant] (ii) The force of gravity on a body of mass m is       F = mg [where g = acceleration due to gravity] (iii) The force of gravitation between two bodies can be zero, if the separation between them becomes infinity (iii) The force of gravity on a body is zero at the centre of the earth.

Free Fall and Acceleration due to Gravity

Our earth attracts all the 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. While falling there is no change in the direction of motion, but due to the earth’s attraction, there is change in the magnitude of the velocity which involves acceleration and this acceleration is due to the force of attraction of the earth on an object. Thus, this acceleration is called the acceleration due to gravity. It is denoted by ‘g’. The unit of g is same as that of acceleration i.e., ms–2.

The uniform acceleration produced in a freely falling body due to the gravitational pull of the earth is known as acceleration due to gravity.

Relation between g and G

Consider the earth to be spherical body of mass M, radius R with centreO. Suppose a body of mass m placed on the surface of the earth, where acceleration due to gravity is g.

According to Newton’s law of gravitation, the force exerted by the earth on the body is given by

…(i)

This force involves acceleration in the body which is equal to the product of the mass of the body and acceleration due to gravity i.e.,

F = m × g                    …..(ii)

From equation (i) and (ii) we get

…..(iii)

Now, to calculate the value of g, we put the above values in equation (iii)

G    = 6.7 × 10–11 N m2kg–2

M    = 6 × 1024 kg

R    = 6.4 × 106 m

= 9.8 ms–2

Variation of g: As we know that the earth is not perfectly spherical. The radius of the earth increases from the poles to the equator. So, the value of g becomes greater at the poles than at the equator. But for most calculations or for the sake of convenience, we take the value of g more or less constant on the surface or near the surface of the earth i.e., (9.8 ms–2 10 ms–2).

But as we go up from the surface of the earth, the distance of the body from the centre of the earth increases, so, the value of g decreases. Also, the value of g decrease as we go inside the surface of the earth. The value of g becomes zero at the centre of the earth.

Motion of Objects Under the Influence of Gravitational Force of the Earth

It was earlier concept that the lighter objects fall slowly and the heavier objects fall more rapidly when dropped from the same height and at the same time. This concept was later found to be wrong by Galileo. Galileo dropped two objects of different masses from the leaning tower of Pisa in Italy and found that they hit the ground at the same time. From this observation, Galileo concluded that (if we neglect the air resistance) all objects big or small, hollow or solid should fall at the same rate i.e., acceleration is uniform and does not depend on the mass of the object.

As g is constant near or on the surface of the earth, all the equations for uniformly accelerated motion of objects become valid with acceleration a replaced by g.

The equations are

v = u + at

v2 = u2 + 2as

Case-I    When we drop the object from a certain height, then u = 0, a = + g (g is along the direction of v)

v = u + at                           or        v = 0 + gtÞ v = gt

s = ut + at2   or                  s = 0 × t + gt2Þ s = gt2

v2 = u2 + 2as            or            v2 = 02 + 2 gsÞ v2 = 2 gs

Case-II   When we throw an object upwards with certain velocity, then final velocity
v = 0 and a = – g.

Therefore,

0 = ugtÞu= gt

s = utgt2

02 = u2 – 2gsÞu2 = 2gs

Mass and Weight

 Mass Weight (i) Mass of a body is the quantity of matter contained in it. (ii) Weight of a body is the force with which it is attracted towards the centre of the earth. (ii) Mass is a scalar quantity. The SI unit of mass is kilogram (kg). (ii) Weight is a vector quantity. The SI unit of weight is newton (N). (iii) Mass of a body is constant and does not change from place to place. (iii) Weight of a body is not constant, it varies from place to place due to the variation of g. (iv) Mass of a body cannot be zero. (iv) Weight of a body can be zero.

Weight of an Object on the Moon

Let the mass of an object be m. Let its weight on the moon be Wm. Also let the mass of the moon is Mm and its radius Rm.

According to Universal law of gravitation, the weight of the object on the moon will be

…..(i)

Let the weight of the same object on earth be We. The mass of the earth is Me and its radius is Re. Then weight of the object on the earth will be

…..(ii)

Now, we know that the mass of the earth is about 100 times that of the moon and the radius of the earth is about 4 times that of the moon.

i.e.,         Me » 100 Mm

Re » 4 Rm

Now dividing equation (i) and (ii) and using these conditions in (i) and (ii). we get

or

or    Wm= We

or    Weight on moon =  th weight on earth.

Centripetal Force: The force which causes the acceleration and keeps the body moving along the circular path and acting towards the centre is called centripetal force. In the absence of this force, the body flies off along a straight line which is tangent to the circular path. For example, the motion of the moon around the earth is due to the centripetal force. This centripetal force is provided by gravitational force of attraction of the earth. If there were no such force, the moon could pursue a uniform straight line motion.

Kepler’s Laws of Planetary Motion: Johannes Kepler formulated three laws, which govern the motion of the planets. These laws are:

1. i) Law of Orbit: Every planet revolves around the sun in an elliptical orbit with the sun situated at one of the focii.
2. ii) Law of Area: The line joining the planet with the sun sweeps out equal areas in equal intervals of time.

iii)    Law of Period: The square of time period of revolution of a planet around the sun is directly proportional to the cube of the mean distance of a planet from the sun i.e.,                                                                   T2µr3

where, T is the time period and r is mean distance of the planet from the sun.

SOLVED EXAMPLES

Example 1:  A planet has three times the mass of Earth with a radius twice as large as our own. What fraction of Earth’s gravitational acceleration would a mass feel on the surface of this planet?

Solution:     The expression for solving for g

Now we plug in 3M for the mass and 2r for the radius to see how it compares to the earth’s g.

The g of this planet would be 3/4 of g on Earth.

Example 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).

(a) What is its speed on striking the ground?

(b) What is its average speed during the 0.5 s?

(c) How high is the ledge from the ground?

Solution:     Time, t = 1/2 second

Initial velocity, u = 0 ms–1

Acceleration due to gravity, g = 10 ms–2

Acceleration of the car, a = + 10 ms–2 (downward)

(a) speed                     v   = at

v   = 10 ms–2 × 0.5 s = 5 m/s

(b) average speed             = = (0 ms–1 + 5 ms–1)/2

= 2.5 ms–1

(c) distance travelled, s   = 1/2 at2 = 1/2 × 10 ms–2 × (0.5)2

= 1/2 × 10 ms–2 × 0.25 s2

= 1.25 m

Thus,

(a) its speed on striking the ground = 5 ms–1

(b) its average speed during the 0.5 s = 2.5 ms–1

(c) height of the ledge from the ground = 1.25 m.

Example 3:  An object is thrown vertically upwards and rises to a height of 10 m. Calculate

(a) the velocity with which the object was thrown upwards

(b) the time taken by the object to reach the highest point.

Solution:     Distance travelled, s = 10 m

Final velocity, v = 0 ms–1

Acceleration due to gravity, g = 9.8 ms–2

Acceleration of the object, a = –9.8 ms–2  (upward motion)

(a) v2 = u2 + 2as

0 = u2 + 2 × (–9.8 ms–2) × 10 m

u2 = –2 × 9.8 × m2 s–2

u = 14 ms–1

(b) v = u + at

0 = 14 ms–1 – 9.8 ms–2 × t

t = 1.43 s.

Thus,

(a) Initial velocity, u = 14 ms–1

(b) time taken, t = 1.43 s.

 EXERCISE

1. a) Does Earth exert a force on the moon?
2. b) What is its direction?
3. c) How would the moon move if this force did not act?
4. a) What would happen to this book’s weight if you managed to double Earth’s

mass?

1. b) What if, instead, you doubled the book’s mass?
2. c) What if you doubled both?
3. a) Does earth’s gravity pull more strongly on a block of wood or on a block of iron having the same size?
4. b) Which one falls faster when dropped (neglect air resistance)?
5. a) Do you exert a gravitational force on Earth?
6. b) If so, how large is it, and in what direction is it?
7. a) If gravity suddenly shut off right now, what would be the shape of Earth’s orbit?
8. b) What about the moon’s orbit?
9. Is there any net force acting on the moon?
10. If earth collapsed to one-tenth of its present radius, keeping its mass some how much would you then weigh?
11. What happens to the force between two objects, if
12. a) the mass of one object is doubled?
13. b) the distance between the objects is doubled and tripled?
14. c) the masses of both objects are doubled?
15. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate
16. a) the maximum height to which it rises,
17. b) the total time it takes to return to the surface of the earth.
18. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking
g = 10 m/s2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
19. A ball thrown up vertically returns to the thrower after 6 s. Find
20. a) the velocity with which it was thrown up,
21. b) the maximum height it reaches, and

c)its position after 4s

WORKSHEET – 1

1. A body is released from rest from a height, find its speed at

(a) t = 2 sec (b) t = 3 sec after the release.

1. A particle is released from rest from a height. Find the distance it falls through in (a) 2 sec (b) 3 sec
2. A person drops a ball from a height of 1.6m.With what velocity does it strike the ground?
3. A body is throw upwards with a speed of 20 m/s. Find its speed after

(a) 1 sec

(b) 2 sec

1. A particle is thrown upwards with a speed of 39.2 m/s. Find

(a) the time for which it moves in the upward direction and

(b)the maximum height reached.

WORKSHEET – 2

1. Find the value of the acceleration due to gravity at a height of 12,800 km from the surface of the earth. Earth’s radius = 6400 km.
2. Find the weight of a boy of mass 10 kg.
3. Find the weight of an object at a height 6400 km above the earth’s surface. The weight of the object at the surface of the earth is 50 N and the radius of the earth is 6400 km.
4. Consider a heavenly body whose mass is thrice that of the earth. What will be the weight of a object on this heavenly body, if its weight m on the earth is 1000N?
5. Consider a heavenly body whose mass is 3×1024 kg (half that of the earth) andradius is 3200 km (half that of earth). What is the acceleration due to gravity at the surface of this heavenly body?

WORKSHEET – 3

1. Two bodies A & B of masses m and 2m respectively are kept a distance a apart.Where should a small particle be placed, so that the net gravitational force on it due to the bodies A and B is Zero?
2. A ball is thrown upwards with some initial speed. It goes up to a height of

19.6m and then returns. Find

(a) The initial speed

(b) The time taken in reaching the highest point

(c) The velocity of the ball one second before and one second after it

reaches the maximum height, and

(d) The time taken by the ball to returns to its original position.

1. A body is dropped from some height. It moves through a distance of
24.5 m in the last second before hitting the ground. Find the height from which it was dropped.
2. A ball is dropped from the edge of a roof. It takes 0.1 s to cross a window ofheight 2m. Find the height of the roof above the top of the window.
3. A ball is dropped from a height of 20m. At the same instant another ball is thrown up from the ground with a speed of 20 m/s. When and where will the ball meet.

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