Class 9 Gravitation Notes 2024-2025 PDF

M.E.S INDIAN SCHOOL, DOHA– QATAR Notes 2024- 2025 Section: Boys /Girls Date: 13/10/2024 Class & Div.: IX (All Divisions) Sub...

M.E.S INDIAN SCHOOL, DOHA– QATAR Notes 2024- 2025 Section: Boys /Girls Date: 13/10/2024 Class & Div.: IX (All Divisions) Subject : Physics Lesson / Topic: Gravitation xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx Gravitation Universal Law of Free Fall Mass Weight Gravitation Importance of Motion of objects To calculate the 𝑊=𝑚×𝑔 Universal Law of under the influence value of g Gravitation of Gravitation pull 𝐺𝑀 𝑣 = 𝑢 + 𝑔𝑡 Weight of an object 𝑔= on the moon = 1/6 ℎ = 𝑢𝑡 + 𝑔𝑡² weight on earth 𝑣2 = 𝑢2 + 2𝑔ℎ 1. State and derive universal law of gravitation. Universal law of gravitation: Every object in the universe attracts every otherobject with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The force is along the line joining the centers of two objects. Let two objects A and B of masses M and m lie at a distance of d from each other as shown in the figure. F 061, Rev 01, dtd10th March 2020 1  Let F be the force of attraction between two objects. According to the universal law of gravitation G is called a universal constant because its value does not depend on the nature of intervening medium or temperature or any other physical variable. S.I. unit of G = Nm2/kg2 Value of G = 6.673 × 10–11 Nm2/kg2 (Found by Henry Cavendish) 2. Write the importance of universal law of gravitation. A) Universal law of gravitation successfully explained several phenomena like: (i) the force that binds us to the earth. (ii) the motion of moon around the earth. (iii) the motion of planets around the sun. (iv) the tides due to the moon and the sun. 3. Define freefall. Establish the relation between G and g. OR Derive an expression for acceleration due to gravity A) When an object falls down towards the earth under the influence of gravitational forcealone, we say the object is in free fall. This acceleration is called acceleration due to gravity, denoted by ‘g’. Its value is 9.8 and its unit is m/s2. As F = ma (a = g)...(i) F 061, Rev 01, dtd10th March 2020 2  F = mg...(ii) M = Mass of the earth d = Distance between the object and the earth G = Gravitational constant If the object is placed on the earth, then d=R (R = radius of the earth) Note: Earth is not a sphere, it is flattened at poles. 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. Motion of objects under the influence of gravity ‘g’ does not depend on the mass of the body. All objects small, big, heavy, light, hollow or solid fall at same rate. For free fall, value of acceleration a = g = 9.8 ms–2. If an object is just let fall from a height then u = 0 and a = g = 9.8 m/s2. If an object is projected vertically upward with an initial velocity u, then a = –g = – 9.8 ms–2 and the object will go to a maximum height h where its final velocity becomes zero (i.e., v = 0). 4. What are the differences between the mass of an object and its weight? A) mass weight 1. 1. Mass of a body is the measure of its 1. Weight of the body is the force with inertia which it is attracted towards the earth (W = m x g). 2. Its S.I. unit is kg. 2. Its S.I unit is Newton. F 061, Rev 01, dtd10th March 2020 3  3. It remains constant everywhere and it 3. Its value changes from place to place cannot be zero. and it can be zero. 4. it can be measured by beam- balance 4.it can be measured by spring balance 5. It has only magnitude i.e. it is a scalar 5. It has both magnitude and direction i.e. quantity. it is a vector quantity 5. Why will a sheet of paper fall slower than one that is changed into ball form ? a. This is due to the fact that air friction on sheet of paper will be more than on the ball. 6. Difference between Gravitational Constant (G) and Acceleration due to gravity (g) Sl. No. Gravitation Constant (G) Gravitational acceleration (g) 1. Its value is 6.67×10- Its value is 9.8 m/s2. 11 Nm2/kg2. 2. It is a scalar quantity. It is a vector quantity. 3. Its value remains constant Its value varies at various always and everywhere. places. 4. Its unit is Nm2/kg2. Its unit is m/s2. 7. Gravitational force acts on all objects in proportion to their masses. Why then does a heavy object not fall faster than a light object? A) All objects fall on the bottom with constant acceleration called acceleration due to gravity (g). It is constant and therefore the value of ‘g’ doesn’t depend on the mass of the object. So heavy objects don’t fall quicker than light-weight objects provided there’s no air resistance Numericals: 8. What happens to the force between two objects, if (i) The mass of one object is doubled? (ii) The distance between the objects is doubled and tripled? The masses of both objects are doubled? (iii) Solution: (i) If mass of one object is doubled, F = 2F, so force is also doubled. (ii)If the distance between the objects is doubled and tripled If F 061, Rev 01, dtd10th March 2020 4  it’s doubled Hence, F= Gm1m2/(2R)2 F = F/4, Force thus becomes one-fourth of its initial force. If it is tripled F= Gm1m2/(3R)2 F = 9F, Force thus becomes one-ninth of its initial force. (iii)If masses of both the objects are doubled, then F = 4F, Force will therefore be four times greater than its actual value. 9. A ball is thrown vertically upwards with a velocity of 49m/s. Calculate (i) The maximum height to which it rises, (ii) The total time it takes to return to the surface of the earth. Solution: u = 49m/s, v = 0, g = -9.8 m/s2 v2 = u2 – 2gs Substitute all the values in the above equation V = u – gt 0 = 49 – 9.8 x Ta Ta = (49/9.8) = 5s Also, Td = 5s Total time T = Time to ascend (Ta) + Time to descend (Td) Therefore, T = Ta + Td T=5+5 T = 10s 10. A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground u=0 F 061, Rev 01, dtd10th March 2020 5  tower height= distance=19.6m g = 9.8 m/s2 v2 = u2 + 2gs v2 = 0 + 2 × 9.8 × 19.6 v2 = 384.16 v = √(384.16) v = 19.6m/s 11. 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? Ans. u = 40m/s g = 10 m/s2 v=0 v2 = u2 – 2gs [negative as the object goes up] 0 = (40)2 – 2 x 10 x s s = (40 x 40) / 20 Maximum height s = 80m Total Distance = s + s =80 + 80 Total Distance = 160m Total displacement = 0 (The first point is the same as the last point) F 061, Rev 01, dtd10th March 2020 6