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10

Gravitation

Class 9 Science  ·  NCERT Solutions  ·  15 Questions

15 Questions & Answers
Chapter 10: Gravitation — NCERT Solutions. This chapter covers the universal law of gravitation, free fall, acceleration due to gravity (g), mass vs. weight, thrust and pressure, buoyancy, Archimedes' principle, and relative density. Key formulas: F = Gm1m2/r2  |  g = GM/R2  |  W = mg  |  P = F/A  |  G = 6.674 × 10−11 N m2 kg−2
Q1

State the universal law of gravitation.

Answer

Statement: Every object in the universe attracts every other object 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 acts along the line joining the centres of the two objects.

Mathematically, if two objects of masses m1 and m2 are separated by a distance r, the gravitational force F between them is:

F = G × m1 × m2 / r2

Where G is the Universal Gravitational Constant:

G = 6.674 × 10−11 N m2 kg−2

This law was given by Sir Isaac Newton in 1687. It applies universally — between planets, stars, and everyday objects. The force is always attractive (pulls the two masses toward each other) and acts along the line joining their centres. The value of G was first determined experimentally by Henry Cavendish in 1798.

Q2

Write the formula to find the magnitude of the gravitational force between the Earth and an object on the surface of the Earth.

Answer

The gravitational force between the Earth and an object on its surface is given by Newton's Law of Gravitation applied with the Earth's parameters:

F = G × M × m / R2

Where:

• F = gravitational force between Earth and the object (in N)
• G = Universal Gravitational Constant = 6.674 × 10−11 N m2 kg−2
• M = mass of the Earth = 6 × 1024 kg
• m = mass of the object (in kg)
• R = radius of the Earth = 6.4 × 106 m (the distance from the object to the Earth's centre, approximately equal to Earth's radius for objects on the surface)

This gravitational force is what we experience as the weight of the object (W = mg), where g = GM/R2 ≈ 9.8 m/s2.

Q3

What do you mean by free fall?

Answer

Free fall is the motion of an object when it falls solely under the influence of the Earth's gravitational force, with no other force (such as air resistance, upthrust, or any contact force) acting on it.

Key characteristics of free fall:

(i) Uniform acceleration: A freely falling object accelerates downward with the acceleration due to gravity, g ≈ 9.8 m/s2 (near Earth's surface). The equations of motion apply with a = g.

v = u + gt  |  h = ut + ½gt2  |  v2 = u2 + 2gh

(ii) Mass-independent: All objects in free fall have the same acceleration (g) regardless of their mass. A feather and a hammer dropped in a vacuum (no air) will fall at the same rate — this was famously demonstrated on the Moon by Apollo astronauts.

(iii) Weightlessness: An object in free fall appears weightless because there is no normal reaction force from any surface. This is why astronauts in orbit (which is essentially continuous free fall) experience weightlessness.

Example: A stone released from a height falls freely to the ground (neglecting air resistance).

Q4

What do you mean by acceleration due to gravity?

Answer

Acceleration due to gravity (g) is the acceleration experienced by a freely falling object due to the gravitational force of the Earth. Near the Earth's surface, g = 9.8 m/s2, directed downward (toward the centre of the Earth).

It is derived from Newton's Law of Gravitation and Newton's Second Law:

F = GMm/R2 = mg
g = GM / R2

Where M = mass of Earth, R = radius of Earth, G = gravitational constant.

Key facts about g:

(i) g decreases as we go higher above the Earth's surface (R increases in the formula).
(ii) g decreases slightly as we go inside the Earth (towards the centre, g = 0 at centre).
(iii) g is slightly greater at the poles (Earth is slightly flatter, so R is smaller at poles) than at the equator.
(iv) g on the Moon ≈ 1.6 m/s2 (about 1/6th of Earth's g) due to the Moon's smaller mass and radius.
(v) g is a vector quantity directed toward the Earth's centre (downward).

Q5

What are the differences between the mass of an object and its weight?

Answer

Mass:

(i) The amount of matter contained in an object.
(ii) A scalar quantity (no direction).
(iii) Measured in kilograms (kg).
(iv) Constant everywhere in the universe — mass does not change with location, altitude, or gravitational field.
(v) Measured using a balance (equal-arm balance) by comparing with standard masses.
(vi) Mass can never be zero for a physical object.

Weight:

(i) The gravitational force exerted on an object by a planet or celestial body.
(ii) A vector quantity (directed toward the centre of the Earth / downward).
(iii) Measured in Newtons (N).
(iv) Varies with location — weight is more at the poles than at the equator, less on the Moon (1/6 of Earth), and zero in deep space far from any massive body.
(v) Measured using a spring balance (weighing scale).
(vi) Calculated as: W = mg

W = m × g

Example: A person of mass 60 kg has a weight of 60 × 9.8 = 588 N on Earth, and 60 × 1.6 = 96 N on the Moon — but their mass remains 60 kg on both.

Q6

Why is the weight of an object on the Moon (1/6)th its weight on the Earth?

Answer

Weight = mg, so weight depends on the acceleration due to gravity (g). The value of g depends on the mass (M) and radius (R) of the body: g = GM/R2.

The Moon has:

• Much smaller mass than Earth (Mmoon ≈ 7.35 × 1022 kg vs. Mearth ≈ 6 × 1024 kg)
• Smaller radius than Earth (Rmoon ≈ 1.74 × 106 m vs. Rearth ≈ 6.4 × 106 m)

Calculating g on Moon:

gmoon = G × Mmoon / Rmoon2 ≈ 1.6 m/s2

Since gmoon ≈ 1.6 m/s2 and gearth ≈ 9.8 m/s2:

gmoon / gearth = 1.6 / 9.8 ≈ 1/6

Since W = mg, and mass (m) is the same on Earth and Moon:

Wmoon = m × gmoon = m × (gearth/6) = Wearth/6

Therefore, the weight of an object on the Moon is approximately one-sixth of its weight on Earth. A person who weighs 600 N on Earth would weigh only 100 N on the Moon.

Q7

A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate (i) the maximum height to which it rises, (ii) the total time it takes to return to the surface of the Earth.

Answer

Given: Initial velocity u = 49 m/s (upward). g = 9.8 m/s2 (acting downward). At maximum height, final velocity v = 0.

(i) Maximum height — using v2 = u2 − 2gh:

02 = 492 − 2 × 9.8 × h
0 = 2401 − 19.6h
h = 2401 / 19.6 = 122.5 m

(ii) Total time — time to reach maximum height using v = u − gt:

0 = 49 − 9.8 × tup
tup = 49 / 9.8 = 5 s

By symmetry of projectile motion, the time to fall back = time to go up = 5 s.

Total time = tup + tdown = 5 + 5 = 10 s

The ball reaches a maximum height of 122.5 m and takes a total time of 10 s to return to the ground.

Q8

A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.

Answer

Given: Initial velocity u = 0 (released from rest). Height h = 19.6 m. g = 9.8 m/s2.

Using v2 = u2 + 2gh:

v2 = 02 + 2 × 9.8 × 19.6
v2 = 2 × 9.8 × 19.6 = 384.16
v = √384.16 = 19.6 m/s

The final velocity of the stone just before it touches the ground is 19.6 m/s (directed downward).

Q9

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?

Answer

Given: u = 40 m/s (upward). v = 0 at maximum height. g = 10 m/s2 (taking deceleration as −10 m/s2 during ascent).

Maximum height — using v2 = u2 − 2gh:

0 = 402 − 2 × 10 × h
0 = 1600 − 20h
h = 1600 / 20 = 80 m

Net displacement:
The stone goes up 80 m and then comes back down 80 m to its starting point (same initial and final position).
Net displacement = 0 m (initial position = final position).

Total distance covered:

Total distance = 80 m (up) + 80 m (down) = 160 m

Maximum height = 80 m. Net displacement = 0. Total distance = 160 m.

Q10

Calculate the force of gravitation between the Earth and the Sun, given that the mass of Earth = 6 × 1024 kg and of the Sun = 2 × 1030 kg. The average distance between the two is 1.5 × 1011 m.

Answer

Given: MEarth = 6 × 1024 kg. MSun = 2 × 1030 kg. r = 1.5 × 1011 m. G = 6.674 × 10−11 N m2 kg−2.

Using F = G × MEarth × MSun / r2:

F = (6.674 × 10−11 × 6 × 1024 × 2 × 1030) / (1.5 × 1011)2
Numerator = 6.674 × 6 × 2 × 10(−11+24+30) = 80.088 × 1043
Denominator = (1.5)2 × 1022 = 2.25 × 1022
F = (80.088 × 1043) / (2.25 × 1022) = 35.6 × 1021 ≈ 3.56 × 1022 N

The gravitational force between the Earth and the Sun is approximately 3.56 × 1022 N. This enormous attractive force keeps the Earth in its orbit around the Sun.

Q11

A man weighs 600 N on the Earth. What is his mass? (Take g = 10 m/s2.) If he were taken to the Moon, his weight would be 100 N. What is his mass on the Moon? What is the acceleration due to gravity on the Moon?

Answer

Mass on Earth:

W = mg  ⇒  m = W/g = 600/10 = 60 kg

Mass on Moon:
Mass is a fundamental property of matter and does not change with location. Therefore, his mass on the Moon is also 60 kg.

Acceleration due to gravity on the Moon:

Wmoon = m × gmoon
gmoon = Wmoon / m = 100 / 60 = 1.67 m/s2

Mass on Earth = Mass on Moon = 60 kg. Acceleration due to gravity on Moon = 1.67 m/s2 (which is approximately 1/6 of Earth's g = 10 m/s2).

Q12

What is thrust and pressure? Give their SI units.

Answer

Thrust:
The force acting on an object perpendicular (normal) to its surface is called thrust. It is the total force applied on a surface. Thrust is a vector quantity measured in Newtons (N).

Example: The weight of a book lying on a table is the thrust it exerts on the table surface.

Pressure:
Pressure is defined as the thrust (force) acting per unit area of the surface on which it acts.

Pressure (P) = Thrust (F) / Area (A)

SI unit of pressure is Pascal (Pa), where 1 Pa = 1 N/m2.

Key concept: The same thrust (force) produces greater pressure on a smaller area and lesser pressure on a larger area. This is why:

• A sharp knife (small area) cuts better than a blunt knife (larger area) — same force, higher pressure.
• A bed of nails is safe to lie on because the total weight is distributed over many nail points, reducing pressure per nail.
• Wide tyres on heavy vehicles reduce pressure on the road surface.

Q13

A block of wood is kept on a tabletop. The mass of the wooden block is 5 kg and its dimensions are 40 cm × 20 cm × 10 cm. Find the pressure exerted by the wooden block on the table if it is made to lie on the table with its faces of dimensions (a) 20 cm × 10 cm and (b) 40 cm × 20 cm.

Answer

Given: Mass m = 5 kg. g = 9.8 m/s2. Thrust (weight) = mg = 5 × 9.8 = 49 N.

(a) Face with dimensions 20 cm × 10 cm:

Area = 20 cm × 10 cm = 200 cm2 = 200 × 10−4 m2 = 0.02 m2
P = Thrust / Area = 49 / 0.02 = 2450 Pa

(b) Face with dimensions 40 cm × 20 cm:

Area = 40 cm × 20 cm = 800 cm2 = 800 × 10−4 m2 = 0.08 m2
P = Thrust / Area = 49 / 0.08 = 612.5 Pa

Conclusion: When the block lies on its smallest face (20 × 10 cm), it exerts a greater pressure (2450 Pa) compared to when it lies on its largest face (40 × 20 cm), which gives lesser pressure (612.5 Pa). The thrust (weight = 49 N) is the same in both cases.

Q14

State Archimedes' principle.

Answer

Archimedes' Principle: When a body is immersed wholly or partially in a fluid (liquid or gas), it experiences an upward buoyant force (upthrust) equal to the weight of the fluid displaced by the body.

Buoyant Force = Weight of fluid displaced = ρfluid × Vdisplaced × g

Where ρfluid is the density of the fluid, Vdisplaced is the volume of fluid displaced, and g is the acceleration due to gravity.

Consequences and applications:

(i) Apparent loss of weight: A body immersed in a fluid appears lighter (loses weight = buoyant force). This is why heavy objects feel lighter when lifted in water.
(ii) Floating: An object floats when buoyant force = weight of object (i.e., density of object ≤ density of fluid).
(iii) Sinking: An object sinks when its weight > buoyant force (density of object > density of fluid).
(iv) Applications: Design of ships and submarines (by changing their density using ballast tanks), hydrometers (to measure density of liquids), lactometers (to test purity of milk).

This principle was discovered by the ancient Greek scientist Archimedes (~250 BCE), according to legend while taking a bath.

Q15

What is relative density?

Answer

Relative density (also called specific gravity) of a substance is the ratio of the density of the substance to the density of water at 4°C (where water has its maximum density of 1000 kg/m3).

Relative Density = Density of substance / Density of water at 4°C

Key properties of relative density:

(i) Dimensionless (no unit): Since both densities are in the same unit (kg/m3), their ratio has no unit. This makes relative density very convenient for comparing materials.

(ii) Relation to floating/sinking:
• If relative density > 1: the substance is denser than water and will sink in water. (e.g., iron with RD = 7.8)
• If relative density < 1: the substance is less dense than water and will float in water. (e.g., wood with RD ≈ 0.5–0.8)
• If relative density = 1: the substance has the same density as water (e.g., pure water itself).

(iii) Practical application: Relative density can be measured using Archimedes' principle:

Relative Density = Weight in air / (Weight in air − Weight in water)

It is also used in hydrometers (instruments that directly measure the relative density of liquids) for checking the density of battery acid, milk (lactometer), or soil solutions.