Class : 9 – Science (English) : Lesson 9. Gravitation

EXPLANATION & SUMMARY

🔵 Introduction
Gravitation is a natural phenomenon where every object in the universe attracts every other object with a certain force.
This universal attraction explains why objects fall on Earth, why planets revolve around the Sun, and why the Moon orbits the Earth.

🟢 Universal Law of Gravitation
Stated by Sir Isaac Newton:
“Every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.”

Formula:
F = G (m₁ m₂) / r²
where:
F = force of attraction
m₁, m₂ = masses of objects
r = distance between them
G = universal gravitational constant (6.67 × 10⁻¹¹ N m² kg⁻²)
💡 Example: Earth attracts an apple, and the apple attracts Earth with equal force, but Earth’s acceleration is negligible due to its huge mass.

🟡 Free Fall
When an object falls towards Earth due to gravity alone, it is called free fall.

Acceleration produced = g (acceleration due to gravity).
Value of g near Earth ≈ 9.8 m s⁻².
Formula:
g = G M / R²
M = mass of Earth
R = radius of Earth

🟣 Motion of Objects under Gravity
Equations of motion under gravity (taking g instead of a):
v = u + g t
s = u t + ½ g t²
v² − u² = 2 g s
💡 Example: A stone dropped from height accelerates at g = 9.8 m s⁻².

🔴 Mass and Weight
Mass:
Quantity of matter in body.
Scalar, constant, SI unit = kg.
Weight:
Force with which Earth attracts a body.
Formula: W = m g
Vector, depends on location (g value).
SI unit = N.
💡 Example: Mass of astronaut remains same on Moon, but weight is reduced as g_moon ≈ 1/6 g_earth.

🟤 Thrust and Pressure
Thrust: Force acting normally on a surface.
Pressure: Thrust per unit area.
Formula: P = F / A
SI unit: Pascal (Pa).
💡 Example: Sharp knife cuts easily (less area → more pressure).

🟠 Buoyancy and Archimedes’ Principle
Buoyancy: Upward force exerted by fluid on a body immersed in it.
Archimedes’ Principle:
A body immersed in fluid experiences an upward buoyant force equal to the weight of fluid displaced.
Applications:
Ships float.
Hydrometer measures density.
Submarines work on principle.

🟣 Relative Density
Relative density = Density of substance ÷ Density of water
Unitless.
Helps compare heaviness of substances.

🟢 Applications of Gravitation
🌍 Explains why objects fall.
🌑 Explains motion of Moon around Earth.
☀️ Explains planetary orbits.
💧 Explains tides in seas and oceans.

🟡 Summary
Newton’s law of gravitation: F = G m₁ m₂ / r².
Free fall occurs under gravity; g ≈ 9.8 m/s² near Earth.
Mass = constant, Weight = varies with g.
Pressure = Force / Area.
Archimedes’ principle explains buoyancy.
Relative density is ratio, no unit.

📝 Quick Recap
🔵 Universal law → F = G m₁ m₂ / r².
🟢 Free fall → acceleration = g.
🟡 Mass ≠ Weight.
🔴 Pressure inversely ∝ Area.
🟣 Archimedes’ principle → buoyant force = weight of displaced fluid.
🌍 Gravitation governs planetary & terrestrial phenomena.

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QUESTIONS FROM TEXTBOOK

🔵 Question 1
How does the gravitational force change if distance is reduced to half?
Answer
Given: distance becomes r/2
To find: new force compared to F
Idea: F is inversely proportional to square of distance
Steps:
Old force ∝ 1 / r^2
New distance = r/2
New force ∝ 1 / (r/2)^2 = 1 / (r^2/4) = 4 / r^2
Final: New force = 4 times the old force.

🟢 Question 2
If gravity acts on all masses, why doesn’t a heavy object fall faster than a light one?
Answer
Given: two objects of different masses in free fall
Idea: free-fall acceleration does not depend on mass
Steps:
For bodies near Earth, acceleration g = G M / R^2 (mass of body not present).
So in vacuum both fall with same g; any difference in air is due to air resistance.
Final: In free fall (no air), both fall together.

🟡 Question 3
Find gravitational force between Earth and a 1 kg object at Earth’s surface.
Data: M = 6 x10^24 kg, R = 6.4 x10^6 m, G = 6.67 x10^-11 N m^2/kg^2
Answer
Given: m = 1 kg, M, R, G as above
To find: F
Formula: F = G M m / R^2
Steps:
R^2 = (6.4 x10^6)^2 = 6.4^2 x10^12 = 40.96 x10^12 = 4.096 x10^13
Numerator = G M m = 6.67 x10^-11 x 6 x10^24 x 1
Combine numbers: 6.67 x 6 = 40.02
Combine powers: 10^-11 x 10^24 = 10^13
Numerator = 40.02 x10^13 = 4.002 x10^14
F = (4.002 x10^14) / (4.096 x10^13)
Divide = 9.77 (approximately)
Final: F ≈ 9.8 N.

🟣 Question 4
Does Earth pull the Moon with a greater, smaller, or equal force than the Moon pulls Earth? Why?
Answer
Given: Earth and Moon interact gravitationally
Idea: action–reaction pair
Final: Equal in magnitude and opposite in direction; they act on different bodies.

🔴 Question 5
If the Moon attracts Earth, why doesn’t Earth move toward the Moon?
Answer
Given: Earth mass is huge
Steps:
Both bodies accelerate toward the common centre of mass.
Earth’s acceleration = Force / Mass of Earth (very small).
Final: Earth does move, but the shift is tiny and not noticeable.

🟤 Question 6
How does force change if:
(i) one mass doubled, (ii) distance doubled/tripled, (iii) both masses doubled?
Answer
(i) Force becomes 2F.
(ii) Distance 2r → force F/4; distance 3r → force F/9.
(iii) Both masses doubled → product becomes 4 times → force 4F.

🟠 Question 7
State importance of universal law of gravitation.
Answer
Keeps planets and satellites in orbit; explains tides; explains fall of bodies and retention of atmosphere; guides satellite launching/orbits.

🟣 Question 8
What is the acceleration of free fall?
Answer
g ≈ 9.8 m/s^2 downward near Earth’s surface.

🟢 Question 9
What do we call the gravitational force between Earth and an object?
Answer
Weight (W = m g).

🟡 Question 10
Gold weighed at poles vs at equator — will weight be same?
Answer
Given: g at poles > g at equator
Steps:
Mass stays the same.
Weight = m g, so smaller g at equator → smaller weight.
Final: Weight at equator is slightly less (mass unchanged).

🔵 Question 11
Why does a flat sheet of paper fall slower than the same paper crumpled?
Answer
Given: same mass, different shape
Idea: air drag depends on exposed area
Final: Flat sheet has larger area → greater air resistance → smaller net acceleration → falls slower.

🟢 Question 12
On the Moon, gravity is one-sixth of Earth’s. Find weight of a 10 kg object on Earth and on Moon.
Answer
Given: m = 10 kg, g_E = 9.8 m/s^2, g_M = (1/6) g_E
Steps:
W_E = m g_E = 10 x 9.8 = 98 N
W_M = W_E / 6 = 98 / 6 = 16.3 N (approx)
Final: Earth 98 N; Moon 16.3 N.

🟡 Question 13
A ball is thrown vertically upward with speed 49 m/s. Find
(i) maximum height, (ii) total time of flight (take g = 9.8 m/s^2).
Answer
Given: u = 49 m/s, at top v = 0
To find: h_max, total time
Formulae: v^2 = u^2 – 2 g h; t_up = u / g
Steps:
0 = 49^2 – 2 x 9.8 x h
h = 49^2 / (2 x 9.8) = 2401 / 19.6 = 122.5 m
t_up = 49 / 9.8 = 5 s
t_total = 2 x t_up = 10 s
Final: Height 122.5 m; Total time 10 s.

🟣 Question 14
A stone is released from a height 19.6 m. Find its speed just before it hits the ground (g = 9.8 m/s^2).
Answer
Given: u = 0, s = 19.6 m
Formula: v^2 = u^2 + 2 g s
Steps:
v^2 = 0 + 2 x 9.8 x 19.6 = 384.16
v = square root of 384.16 = 19.6 m/s
Final: 19.6 m/s downward.

🔴 Question 15
A stone is thrown up with u = 40 m/s (g = 10 m/s^2). Find
(a) maximum height, (b) net displacement after return, (c) total distance.
Answer
Given: u = 40 m/s, g = 10 m/s^2
Steps for height:
0 = u^2 – 2 g h
h = 40^2 / (2 x 10) = 1600 / 20 = 80 m
Other results:
After return to the thrower: displacement = 0 m
Total distance = up + down = 80 + 80 = 160 m
Final: Height 80 m; Displacement 0 m; Distance 160 m.

🟤 Question 16
Find gravitational force between Earth and Sun.
Data: M_E = 6 x10^24 kg, M_S = 2 x10^30 kg, r = 1.5 x10^11 m, G = 6.67 x10^-11
Answer
Given: as above
Formula: F = G M_E M_S / r^2
Steps:
r^2 = (1.5 x10^11)^2 = 1.5^2 x10^22 = 2.25 x10^22
Multiply masses with G:
2a) Numbers: 6.67 x 6 x 2 = 80.04
2b) Powers: 10^-11 x 10^24 x 10^30 = 10^(43)
2c) Numerator = 80.04 x10^43 = 8.004 x10^44
Divide: F = (8.004 x10^44) / (2.25 x10^22)
Compute: 8.004 / 2.25 ≈ 3.56; powers: 10^(44-22) = 10^22
Final: F ≈ 3.56 x10^22 N.

🟠 Question 17
From a 100 m tower, one stone is dropped and another is thrown upward at 25 m/s at the same time. When and where do they meet? (g = 9.8 m/s^2)
Answer
Given: height = 100 m, u_up = 25 m/s
Let t = time to meet
Steps:
Downward stone distance = (1/2) g t^2
Upward stone distance from ground = u t – (1/2) g t^2
Sum of distances = 100
u t = 100 → t = 100 / 25 = 4 s
Height of meeting = u t – (1/2) g t^2 = 25 x 4 – 0.5 x 9.8 x 16
Height = 100 – 78.4 = 21.6 m above ground
Final: They meet after 4 s at 21.6 m above ground (78.4 m below the top).

🟣 Question 18
A ball returns to the thrower after 6 s. Find
(a) initial speed, (b) maximum height, (c) position after 4 s. (g = 9.8 m/s^2)
Answer
Given: total time = 6 s → time up = 3 s
(a) u = g x time up = 9.8 x 3 = 29.4 m/s upward
(b) h = u^2 / (2 g) = 29.4^2 / 19.6 = 44.1 m
(c) At t = 4 s: y = u t – (1/2) g t^2
y = 29.4 x 4 – 0.5 x 9.8 x 16
y = 117.6 – 78.4 = 39.2 m above the throw point
Final: u = 29.4 m/s; h = 44.1 m; at 4 s it is 39.2 m above the throw point.

🔵 Question 19
Direction of buoyant force on an immersed object?
Answer
Vertically upward through the centre of buoyancy.

🟢 Question 20
Why does a plastic block released under water rise to the surface?
Answer
Density of plastic < density of water → upthrust > weight → net upward force → it rises.

🟡 Question 21
A 50 g sample occupies 20 cm^3. Will it float or sink in water (1 g/cm^3)?
Answer
Given: mass = 50 g, volume = 20 cm^3
Steps:
Density = mass / volume = 50 / 20 = 2.5 g/cm^3
Compare: 2.5 > 1
Final: It sinks.

🟣 Question 22
A sealed packet has mass 500 g and volume 350 cm^3. In water (1 g/cm^3), will it float or sink? Also find mass of water displaced.
Answer
Steps:
Density of packet = 500 / 350 = 1.43 g/cm^3
Compare with water: 1.43 > 1 → sinks (fully submerged).
Volume of displaced water = 350 cm^3
Mass of displaced water = density x volume = 1 x 350 = 350 g
Final: Packet sinks; displaced water mass = 350 g.

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OTHER IMPORTANT QUESTIONS FOR EXAMS

Section A (Q1–Q20: MCQs, 1 mark each)
🔵 Q1. The force responsible for the revolution of planets around the Sun is:
Electrostatic force
Nuclear force
Gravitational force
Magnetic force
Answer: 3

🟢 Q2. Value of acceleration due to gravity (g) near the Earth’s surface is about:
9.8 m/s^2
6.67 m/s^2
3.6 m/s^2
1.6 m/s^2
Answer: 1

🟡 Q3. When a body is dropped from a certain height, neglecting air resistance, its velocity increases because of:
Inertia
Gravitational pull
Magnetic force
Frictional force
Answer: 2

🔴 Q4. Which of the following is a vector quantity?
Speed
Weight
Mass
Distance
Answer: 2

🟣 Q5. The weight of a body on the Moon is:
1/2 of Earth weight
1/6 of Earth weight
Same as Earth weight
Double of Earth weight
Answer: 2

🟤 Q6. The force that keeps the moon in its orbit around the Earth is:
Nuclear force
Gravitational force
Electrostatic force
Friction
Answer: 2

🟠 Q7. A ball is thrown vertically upward. At maximum height its velocity is:
Zero
g
Maximum
Cannot be determined
Answer: 1

🔵 Q8. SI unit of pressure is:
Dyne/cm^2
Newton
Pascal
kg m/s^2
Answer: 3

🟢 Q9. The value of g at the centre of Earth is:
9.8 m/s^2
Zero
Infinite
4.9 m/s^2
Answer: 2

🟡 Q10. The formula for gravitational force is:
F = G m1 m2 / r^2
F = m a
F = k q1 q2 / r^2
F = p h g
Answer: 1

Assertion–Reason (Q11–Q20)
🟣 Q11.
Assertion (A): Every object attracts every other object in the universe.
Reason (R): Gravitational force is a universal force.
Options:
Both A and R true; R correct explanation
Both A and R true; R not explanation
A true; R false
A false; R true
Answer: 1

🟤 Q12.
A: Weight of a body changes from place to place.
R: Weight depends on value of g.
Answer: 1

🟠 Q13.
A: Mass of a body changes on the Moon.
R: Mass depends on gravity.
Answer: 3

🔵 Q14.
A: Pressure = Force / Area.
R: For same force, smaller area → greater pressure.
Answer: 1

🟢 Q15.
A: Buoyant force acts upward on an object in fluid.
R: Pressure in fluid increases with depth.
Answer: 1

🟡 Q16.
A: Density = Mass / Volume.
R: SI unit of density is kg/m^3.
Answer: 1

🔴 Q17.
A: A ship made of iron floats on water.
R: It displaces water equal to its own weight.
Answer: 1

🟣 Q18.
A: Acceleration due to gravity decreases with height.
R: g ∝ 1 / (R + h)^2.
Answer: 1

🟤 Q19.
A: Atmospheric pressure is due to weight of air column.
R: Pressure decreases with altitude.
Answer: 2

🟠 Q20.
A: g is constant everywhere on Earth.
R: Earth is a perfect sphere.
Answer: 4

Section B (Q21–Q26: VSA, 2 marks)
🔵 Q21. Define free fall.
Answer: When objects fall under the influence of Earth’s gravity alone, without air resistance, it is free fall.

🟢 Q22. Write two differences between mass and weight.
Answer:
Mass is amount of matter; weight is force of gravity.
Mass constant everywhere; weight changes with g.

🟡 Q23. Write the SI units of:
(a) G (b) g.
Answer:
(a) G → N m^2 / kg^2
(b) g → m/s^2

🔴 Q24. Why does a body weigh less at the equator?
Answer: At equator, effective g is less due to Earth’s rotation and bulging.

🟣 Q25. State Archimedes’ principle.
Answer: Body immersed in fluid experiences upward thrust equal to weight of fluid displaced.

🟤 Q26. A ball of mass 1 kg has weight 9.8 N on Earth. Find its weight on Moon.
Answer: Weight = 9.8 / 6 = 1.63 N.

Section C (Q27–Q33: SA, 3 marks)
🟠 Q27. State the universal law of gravitation and give its importance.
Answer: F = G m1 m2 / r^2. It explains fall of bodies, planetary motion, tides, etc.

🔵 Q28. A 60 kg person weighs less on Moon. Calculate weight.
Steps:
Mass = 60 kg
g on Moon = g/6 = 9.8/6 = 1.63 m/s^2
W = m g = 60 x 1.63 = 97.8 N
Final: 97.8 N.

🟢 Q29. Why do raindrops become spherical?
Answer: Due to surface tension of water trying to minimize surface area.

🟡 Q30. Derive the formula for acceleration due to gravity at Earth’s surface.
Answer:
Force on mass m: F = G M m / R^2
Weight: W = m g
Equating: m g = G M m / R^2
g = G M / R^2

🔴 Q31. Calculate force between two bodies each of mass 100 kg, kept 1 m apart. (G = 6.67 x10^-11)
Steps:
m1 = m2 = 100 kg, r = 1 m
F = G m1 m2 / r^2
= 6.67 x10^-11 x 100 x 100 / 1
= 6.67 x10^-11 x 10^4
= 6.67 x10^-7 N
Final: 6.67 x10^-7 N.

🟣 Q32. Why does a stone fall faster in air than a feather?
Answer: Both have same g, but air resistance opposes feather more (larger area), so it falls slower.

🟤 Q33. A piece of iron of volume 0.05 m^3 has density 7800 kg/m^3. Find its mass and weight.
Steps:
Density = mass / volume
Mass = density x volume = 7800 x 0.05 = 390 kg
Weight = m g = 390 x 9.8 = 3822 N
Final: Mass 390 kg; Weight 3822 N.


Section D (Q34–Q36: LA, 5 marks each)
🔵 Q34. Derive the equations of motion for a body under free fall. Show each step clearly.
Answer:
Under free fall, acceleration = g.
First equation:
v = u + g t
(from definition of acceleration)
Second equation:
s = u t + ½ g t²
(using displacement formula with uniform acceleration)
Third equation:
v² − u² = 2 g s
(eliminating t between 1st and 2nd equation)
✔️ These three equations describe motion under gravity.

🟢 Q35. A ball is thrown vertically upwards with velocity 20 m/s. Calculate:
(i) Maximum height reached
(ii) Total time of flight (take g = 10 m/s²)
Answer:
u = 20 m/s, g = 10 m/s²
At max height, v = 0
v² = u² − 2 g h
0 = (20)² − 2 × 10 × h
400 = 20 h
h = 20 m
Time to reach max height:
v = u − g t
0 = 20 − 10 t
t = 2 s
Total time of flight = 2 × 2 = 4 s
✔️ Maximum height = 20 m, Total time = 4 s

🟡 Q36. A stone of mass 2 kg is dropped from a height of 50 m. Calculate:
(i) Time taken to reach ground
(ii) Velocity with which it strikes
(iii) Kinetic energy on striking ground (take g = 10 m/s²).
Answer:
u = 0, s = 50 m, g = 10 m/s²
s = ½ g t²
50 = ½ × 10 × t²
50 = 5 t²
t² = 10 → t = √10 ≈ 3.16 s
v = u + g t
v = 0 + 10 × 3.16
v ≈ 31.6 m/s
KE = ½ m v²
= ½ × 2 × (31.6)²
= 1 × 998 ≈ 998 J
✔️ Time = 3.16 s, Velocity ≈ 31.6 m/s, KE ≈ 998 J

Section E (Q37–Q39: Case/Source-based, 4 marks each)
🟣 Q37. Case: An astronaut takes a stone of mass 5 kg on the Earth and the Moon.
Questions:
(a) What will be its weight on Earth?
(b) What will be its weight on Moon?
(c) Why does his mass remain same on both?
(d) Give reason why astronauts can jump higher on the Moon.
Answer:
(a) W = m g = 5 × 9.8 = 49 N
(b) W_moon = 49 / 6 ≈ 8.17 N
(c) Mass is amount of matter, does not change with location.
(d) g on Moon is 1/6, hence weaker pull → higher jumps.

🟤 Q38. Case: A cork piece and an iron cube of same volume are immersed in water.
Questions:
(a) Which will float and why?
(b) Which will sink and why?
(c) Name the principle involved.
(d) State one practical use of this principle.
Answer:
(a) Cork floats because its density < water density. (b) Iron cube sinks because its density > water density.
(c) Archimedes’ Principle.
(d) Designing ships, hydrometers.

🟠 Q39. Case: A student drops two balls of different masses from same height in vacuum.
Questions:
(a) Which one will fall faster?
(b) Why do they reach ground together?
(c) What would happen if the same is done in air?
(d) State the value of g near Earth’s surface.
Answer:
(a) Both fall at same rate.
(b) In absence of air resistance, acceleration g is same for all bodies.
(c) In air, lighter body faces more air resistance, so falls slower.
(d) g = 9.8 m/s².

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ONE PAGE REVISION SHEET

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ACRONYMS

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MISCONCEPTIONS “ALERTS”

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KNOWLEDGE WITH FUN

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MIND MAPS

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