Gravitation – CBSE NCERT Study Resources

Previous Chapter Next Chapter
Gravitation

Study Materials

9th

9th - Science

Gravitation

Jump to Question Sets

Chapter Overview: Gravitation

This chapter introduces the fundamental concept of gravitation, explaining the universal law of gravitation and its significance in understanding the motion of celestial bodies and objects on Earth. Key topics include gravitational force, free fall, mass, weight, and thrust and pressure.

Gravitation: The force of attraction between any two objects in the universe due to their masses.

Universal Law of Gravitation

Every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Mathematically, it is expressed as: F = G (m₁m₂)/r², where F is the gravitational force, G is the gravitational constant, m₁ and m₂ are masses, and r is the distance between them.

Free Fall

When an object falls towards the Earth under the influence of gravitational force alone, it is said to be in free fall. The acceleration due to gravity (g) on Earth is approximately 9.8 m/s².

Acceleration due to gravity (g): The uniform acceleration produced in a freely falling object due to the gravitational force of the Earth.

Mass and Weight

Mass: The amount of matter contained in an object. It remains constant everywhere.

Weight: The force with which an object is attracted towards the Earth. It varies with the value of g (W = m × g).

Thrust and Pressure

Thrust: The force acting perpendicularly on a surface.

Pressure: The thrust per unit area (P = Thrust/Area). Pressure depends on the area over which the force is distributed.

Pressure: The force exerted per unit area. Its SI unit is Pascal (Pa).

Buoyancy and Archimedes' Principle

Buoyancy is the upward force exerted by a fluid on an object immersed in it. Archimedes' Principle states that the buoyant force is equal to the weight of the displaced fluid.

Relative Density

Relative density is the ratio of the density of a substance to the density of water. It is a dimensionless quantity.

Relative Density: The ratio of the density of a substance to the density of water at 4°C.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:
Define gravitational force.
Answer:

Force attracting two masses.

Question 2:
What is the SI unit of gravitational force?
Answer:

Newton (N).

Question 3:
State Universal Law of Gravitation.
Answer:

Every mass attracts another with force ∝ product of masses and inversely ∝ square of distance.

Question 4:
What is acceleration due to gravity on Earth?
Answer:

9.8 m/s².

Question 5:
Why does a feather fall slower than a coin in air?
Answer:

Due to air resistance acting more on feather.

Question 6:
Name the force keeping planets in orbit around the Sun.
Answer:

Gravitational force.

Question 7:
What happens to weight if mass doubles?
Answer:

Weight also doubles.

Question 8:
Why is weightlessness experienced in space?
Answer:

No gravitational pull acts.

Question 9:
Calculate weight of 5 kg mass (g = 10 m/s²).
Answer:

50 N.

Question 10:
What is free fall?
Answer:

Motion under only gravity.

Question 11:
How does g change with altitude?
Answer:

g decreases with altitude.

Question 12:
Why is G called a universal constant?
Answer:

Same value everywhere.

Question 13:
Give one example of gravitational force from NCERT.
Answer:

Apple falling from tree.

Question 14:
What is the value of G?
Answer:

6.67 × 10⁻¹¹ Nm²/kg².

Question 15:
What is the SI unit of acceleration due to gravity?
Answer:

The SI unit of acceleration due to gravity is meter per second squared (m/s²).

Question 16:
State the universal law of gravitation.
Answer:

The universal law of gravitation states that every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Question 17:
Why does an object weigh less on the Moon than on Earth?
Answer:

An object weighs less on the Moon because the acceleration due to gravity on the Moon is about 1/6th of that on Earth, due to the Moon's smaller mass.

Question 18:
What is the value of acceleration due to gravity on Earth?
Answer:

The value of acceleration due to gravity on Earth is approximately 9.8 m/s² near the surface.

Question 19:
Name the force that keeps planets in orbit around the Sun.
Answer:

The gravitational force of the Sun keeps the planets in their orbits.

Question 20:
What happens to the gravitational force if the distance between two objects doubles?
Answer:

If the distance between two objects doubles, the gravitational force becomes one-fourth of its original value, as per the inverse square law.

Question 21:
Define weight of an object.
Answer:

The weight of an object is the force with which it is attracted towards the Earth (or any other celestial body) due to gravity. It is calculated as mass × acceleration due to gravity (W = m × g).

Question 22:
Why is the weight of an object zero in space?
Answer:

The weight of an object is zero in space because there is no gravitational force acting on it (or it is in a state of free-fall), making the acceleration due to gravity effectively zero.

Question 23:
What is the difference between mass and weight?
Answer:

Mass is the amount of matter in an object and is constant, while weight is the force exerted on the object due to gravity and varies with location.

Question 24:
How does the gravitational force change if the mass of one object is tripled?
Answer:

If the mass of one object is tripled, the gravitational force between the two objects also triples, assuming the distance remains the same.

Question 25:
What is the direction of the gravitational force acting on an object?
Answer:

The gravitational force acting on an object is always directed towards the center of the Earth (or the celestial body exerting the force).

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:
Define gravitational force and state its SI unit.
Answer:

The gravitational force is the attractive force between two objects with mass. It is responsible for keeping planets in orbit around the Sun and objects grounded on Earth.
The SI unit of gravitational force is the Newton (N).

Question 2:
What is the value of acceleration due to gravity (g) on Earth's surface?
Answer:

The value of acceleration due to gravity (g) on Earth's surface is approximately 9.8 m/s². This value may vary slightly depending on altitude and location.

Question 3:
State the universal law of gravitation in mathematical form.
Answer:

The universal law of gravitation states that the force (F) between two masses (m₁ and m₂) separated by distance (r) is:
F = G (m₁m₂)/r²
where G is the gravitational constant.

Question 4:
What happens to the gravitational force if the distance between two objects is doubled?
Answer:

According to the universal law of gravitation, the gravitational force becomes one-fourth of its original value when the distance between two objects is doubled, since force is inversely proportional to the square of the distance.

Question 5:
Define weight of an object and write its formula.
Answer:

The weight of an object is the force with which it is attracted towards the Earth due to gravity.
Formula: Weight (W) = mass (m) × acceleration due to gravity (g)

Question 6:
Why is the weight of an object less on the Moon than on Earth?
Answer:

The weight of an object is less on the Moon because the Moon's acceleration due to gravity (g) is about 1/6th of Earth's. Since weight depends on g, it decreases accordingly.

Question 7:
What is free fall? Give an example.
Answer:

Free fall is the motion of an object under the influence of gravity alone, with no other forces acting on it (like air resistance).
Example: A fruit falling from a tree (neglecting air resistance).

Question 8:
Why do astronauts float in space?
Answer:

Astronauts float in space because they are in a state of continuous free fall around the Earth (like a satellite), creating a sensation of weightlessness. The gravitational force still acts on them but they don't feel it due to constant acceleration.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:
Define gravitational force and explain why it is considered a universal force.
Answer:

The gravitational force is the attractive force between any two objects with mass. It is given by Newton's Law of Gravitation: F = G (m₁m₂)/r², where G is the gravitational constant, m₁ and m₂ are masses, and r is the distance between them.

It is called a universal force because it acts between all objects in the universe, regardless of their size or distance. This force governs planetary motion, tides, and even holds galaxies together.

Question 2:
Differentiate between mass and weight with suitable examples.
Answer:

Mass is the amount of matter in an object and remains constant everywhere (measured in kg). Weight is the force exerted by gravity on that mass (measured in N) and varies with location.

  • Example: A 10 kg object has the same mass on Earth and Moon, but its weight is ~98 N on Earth and ~16.3 N on the Moon due to different gravitational pulls.
Question 3:
Explain why the value of acceleration due to gravity (g) decreases with altitude.
Answer:

The value of g is derived from g = GM/R², where G is the gravitational constant, M is Earth's mass, and R is the distance from Earth's center.

As altitude increases, R increases, causing g to decrease because gravitational force weakens with distance. For example, g is ~9.8 m/s² at sea level but reduces to ~9.6 m/s² at 10 km altitude.

Question 4:
Describe how the motion of a freely falling body demonstrates uniform acceleration.
Answer:

A freely falling body accelerates uniformly under gravity (g = 9.8 m/s²), meaning its velocity increases by 9.8 m/s every second.


Proof:
v = u + gt (velocity increases linearly with time)
s = ut + ½gt² (displacement depends on t²)

This constant acceleration is independent of the object's mass, as proven by Galileo's experiments.

Question 5:
State the factors affecting the gravitational force between two objects and justify with the formula.
Answer:

From F = G (m₁m₂)/r², the factors are:

  • Masses (m₁, m₂): Force increases with mass (e.g., Earth-Sun force > Earth-Moon force).
  • Distance (r): Force decreases sharply with distance (e.g., doubling r reduces force to ¼).

Gravitational constant G is universal and fixed.

Question 6:
Why do astronauts appear weightless in space despite Earth's gravity being present?
Answer:

Astronauts appear weightless because they are in a state of free-fall along with their spacecraft (like the ISS).

Earth's gravity still acts on them (~90% of surface gravity at ISS altitude), but since both the astronaut and spacecraft accelerate at the same rate, no normal force (weight sensation) is felt. This creates the illusion of zero gravity.

Question 7:
Differentiate between mass and weight. How does weight change on the Moon compared to Earth?
Answer:

Mass is the amount of matter in an object (measured in kg), while weight is the force exerted by gravity on that mass (measured in N).

Weight changes with gravity, but mass remains constant. On the Moon, gravity is 1/6th of Earth's, so weight reduces to 1/6th, but mass stays the same.

Question 8:
State the universal law of gravitation and write its mathematical form. What does the symbol G represent?
Answer:

The universal law of gravitation states that every object attracts every other object with a force proportional to their masses and inversely proportional to the square of the distance between them.

Mathematically: F = G(m₁m₂)/r².
Here, G is the gravitational constant (6.67 × 10⁻¹¹ Nm²/kg²), a universal value determining the strength of gravity.

Question 9:
Why do objects float in water? Relate this to the concept of buoyant force and gravitational force.
Answer:

Objects float when the buoyant force (upward force exerted by water) equals the gravitational force (downward force due to weight).

If the object's density is less than water, buoyant force overcomes gravity, making it float. For example, a wooden block floats because it displaces water equal to its weight.

Question 10:
Calculate the gravitational force between two 5 kg masses placed 1 m apart. (Use G = 6.67 × 10⁻¹¹ Nm²/kg²)
Answer:

Using Newton's Law: F = G(m₁m₂)/r².
Given: m₁ = m₂ = 5 kg, r = 1 m.

Substitute values:
F = (6.67 × 10⁻¹¹ × 5 × 5) / (1)²
F = 1.6675 × 10⁻⁹ N.

The force is extremely small due to the tiny value of G.

Question 11:
Explain why the value of acceleration due to gravity (g) is not the same everywhere on Earth.
Answer:

The value of g varies because:

  • Earth is not a perfect sphere—it bulges at the equator, increasing distance from the center and reducing g.
  • Altitude: Higher places have weaker gravity due to increased distance from Earth's center.
  • Geological factors: Dense regions (like mountains) slightly increase g.

For example, g is 9.78 m/s² at the equator but 9.83 m/s² at the poles.

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:
Define gravitational force and explain how it varies with mass and distance. Provide an example from NCERT and a real-world application.
Answer:
Concept Overview

Gravitational force is the attraction between two objects with mass. We studied that it depends on their masses and distance.

Process Explanation

According to Newton's law, force increases with mass but decreases with distance squared (F = Gm₁m₂/r²). Our textbook shows Earth's gravity pulling objects downward.

Real-world Application

Example: Moon orbiting Earth (NCERT). Real-world: Tides caused by the Moon's gravity.

Question 2:
Describe free fall and derive the expression for acceleration due to gravity (g). How does it differ on the Moon?
Answer:
Concept Overview

Free fall is motion under only gravity. We learned all objects accelerate equally at g (9.8 m/s²).

Process Explanation

From Newton's second law and gravitation: g = GM/R² (M=Earth's mass, R=radius). NCERT demonstrates this with a feather and coin in vacuum.

Real-world Application

On the Moon, g is 1/6th of Earth's (Apollo astronauts' jumps).

Question 3:
Explain weight and mass with their differences. Why does weight change on different planets?
Answer:
Concept Overview

Mass is matter quantity (constant), while weight is gravitational force (W=mg).

Process Explanation

Our textbook shows a spring balance measuring weight. Weight changes with g (W ∝ g). On Jupiter, higher g increases weight.

Real-world Application

Astronauts weigh less on Moon (NCERT example). Spacecraft calculate fuel based on planetary g.

Question 4:
State universal law of gravitation mathematically. How does it explain Earth's gravitational pull?
Answer:
Concept Overview

The law states F=Gm₁m₂/r², where G is the gravitational constant (6.67×10⁻¹¹ Nm²/kg²).

Process Explanation

Earth's large mass creates strong pull (NCERT: apple falling). Gravity decreases with altitude as r increases.

Real-world Application

Satellites use this law for orbital motion. Geostationary satellites maintain fixed positions.

Question 5:
What is thrust and pressure? How does gravitational force affect pressure underwater?
Answer:
Concept Overview

Thrust is force perpendicular to surface, pressure is thrust per unit area (P=F/A).

Process Explanation

Our textbook shows a nail tip creating high pressure. Underwater, gravity increases pressure with depth (P = hρg).

Real-world Application

Submarines withstand high pressure (NCERT). Divers feel ear pain due to pressure changes.

Question 6:
Define gravitational force and explain how it varies with mass and distance using Newton's law of gravitation.
Answer:
Concept Overview

Gravitational force is the attractive force between two masses. Newton's law states it is directly proportional to the product of masses and inversely proportional to the square of the distance between them.

Process Explanation
  • Formula: F = G(m₁m₂)/r²
  • G is the gravitational constant (6.67×10⁻¹¹ Nm²/kg²)
Real-world Application

Our textbook shows Earth's gravity pulling objects downward. Similarly, the moon's gravity causes tides.

Question 7:
Describe how acceleration due to gravity (g) differs on Earth and the moon. Support your answer with calculations.
Answer:
Concept Overview

Acceleration due to gravity (g) is the force per unit mass exerted by a planet. It depends on the planet's mass and radius.

Process Explanation
  • Earth's g ≈ 9.8 m/s²
  • Moon's g ≈ 1.63 m/s² (1/6th of Earth's)
Real-world Application

Our textbook explains astronauts weigh less on the moon. A 60 kg person weighs only 98 N there vs 588 N on Earth.

Question 8:
Explain free fall with an example. Why do objects with different masses fall at the same rate in vacuum?
Answer:
Concept Overview

Free fall is motion under only gravity's influence. All objects accelerate equally in vacuum because g is independent of mass.

Process Explanation
  • Galileo demonstrated this at the Leaning Tower of Pisa
  • Air resistance causes variations in real-world falls
Real-world Application

Our textbook shows a feather and coin falling equally in a vacuum tube. Skydivers reach terminal velocity due to air resistance.

Question 9:
Compare mass and weight with three key differences. How would your weight change on Mars?
Answer:
Concept Overview

Mass is constant matter quantity (kg), while weight is gravitational force (N) that varies with location.

Process Explanation
  • Mass measured by balance; weight by spring scale
  • Weight = mass × g
  • Mars' g ≈ 3.7 m/s² (about 38% of Earth's)
Real-world Application

Our textbook mentions a 50 kg person would weigh 185 N on Mars vs 490 N on Earth.

Question 10:
Describe an experiment to verify that the acceleration due to gravity is 9.8 m/s² using simple pendulum.
Answer:
Concept Overview

A simple pendulum's time period relates to g through the formula T=2π√(L/g).

Process Explanation
  • Measure length (L) and time for 20 oscillations
  • Calculate T (time/20) and substitute in formula
Diagram
[Diagram: Pendulum with length L and bob]Real-world Application

Our textbook shows this experiment. Seismometers use similar principles to detect ground motion.

Question 11:
Define gravitational force and explain how it influences the motion of celestial bodies. Provide an example from NCERT and a real-world application.
Answer:
Concept Overview

Gravitational force is the attractive force between two masses. It keeps planets in orbit around the Sun.

Process Explanation

Our textbook shows that the Earth's gravity pulls objects toward its center, causing tides due to the Moon's pull.

Real-world Application

Satellites use gravitational force to stay in orbit, like communication satellites.

Question 12:
State Newton's Universal Law of Gravitation and derive its mathematical expression. How does it explain the weight of an object?
Answer:
Concept Overview

Newton's law states every mass attracts another with force proportional to their masses and inversely to distance squared.

Process Explanation

We studied F = G(m₁m₂)/r². Weight is F when one mass is Earth.

Real-world Application

This law helps calculate satellite orbits, like in ISRO missions.

Question 13:
Describe how acceleration due to gravity (g) varies with height and depth. Support your answer with an NCERT example.
Answer:
Concept Overview

'g' decreases with height as distance from Earth's center increases, and decreases with depth as mass below reduces.

Process Explanation

Our textbook shows 'g' is 9.8 m/s² at surface but less on mountains.

Real-world Application

This variation affects spacecraft launch calculations.

Question 14:
Explain why objects float or sink in fluids using Archimedes' Principle. Give one NCERT example and a daily-life observation.
Answer:
Concept Overview

Objects float if buoyant force equals weight, else sink. Archimedes' Principle quantifies this.

Process Explanation

We studied iron nails sink (density > water) while ships float (displaced water weight = ship weight).

Real-world Application

Hot-air balloons rise as heated air becomes less dense.

Question 15:
Differentiate between mass and weight. How would your weight change on the Moon compared to Earth? Use an NCERT activity to support your answer.
Answer:
Concept Overview

Mass is constant matter quantity, weight is gravitational force on that mass.

Process Explanation

Our textbook's spring balance activity shows weight = mass × g. Moon's 'g' is 1/6th Earth's.

Real-world Application

Astronauts weigh less on Moon but mass remains same.

Question 16:
Explain the concept of free fall and derive the expression for the acceleration due to gravity (g) using Newton's law of gravitation. Also, mention how the value of g varies with altitude and depth.
Answer:

Free fall is the motion of an object under the influence of gravitational force alone, without any other external forces like air resistance acting on it. During free fall, all objects, regardless of their masses, experience the same acceleration towards the Earth, known as the acceleration due to gravity (g).


To derive the expression for g, we use Newton's law of gravitation, which states that the gravitational force (F) between two masses (M and m) separated by a distance (r) is given by:


F = G * (M * m) / r²

Where G is the universal gravitational constant. For an object of mass m near the Earth's surface, the gravitational force is also equal to its weight:


F = m * g

Equating the two expressions:


m * g = G * (M * m) / r²

Canceling m from both sides, we get:


g = G * M / r²

Here, M is the mass of the Earth, and r is the distance from the center of the Earth to the object (approximately the Earth's radius near the surface).


The value of g varies with altitude and depth as follows:

  • With altitude: As we move above the Earth's surface, r increases, causing g to decrease because g is inversely proportional to .
  • With depth: Inside the Earth, only the mass enclosed within the radius from the center contributes to g. As depth increases, the effective mass decreases, reducing g linearly until it becomes zero at the Earth's center.
Question 17:
Explain the concept of free fall and derive the expression for the acceleration due to gravity (g) acting on an object during free fall near the Earth's surface. Also, mention how this value changes with altitude and depth.
Answer:

The term free fall refers to the motion of an object under the sole influence of gravity, without any other forces (like air resistance) acting on it. During free fall, all objects, regardless of their mass, experience the same acceleration towards the Earth, known as the acceleration due to gravity (g).


To derive the expression for g, we use Newton's Law of Universal Gravitation and Second Law of Motion:


1. Gravitational force (F) between Earth (mass M) and an object (mass m) is given by: F = G * (M * m) / R², where G is the gravitational constant and R is Earth's radius.


2. According to Newton's Second Law: F = m * a. Here, a = g (acceleration due to gravity).


3. Equating both expressions: m * g = G * (M * m) / R².


4. Simplifying, we get: g = G * M / R².


This shows that g depends on Earth's mass (M) and radius (R).


Variation of g with altitude and depth:

  • Altitude: As we go higher above Earth's surface, g decreases because the distance (R + h) from Earth's center increases. The formula becomes: g' = G * M / (R + h)².
  • Depth: Inside Earth, g decreases linearly with depth (d) because only the mass enclosed by the smaller radius (R - d) contributes. The formula is: g' = g * (1 - d/R).

Thus, g is maximum at Earth's surface and reduces both above and below it.

Question 18:
Explain the concept of free fall and derive the expression for the acceleration due to gravity (g) using Newton's law of gravitation. Also, discuss how the value of g varies with altitude and depth.
Answer:

The term free fall refers to the motion of an object under the sole influence of gravity, without any other forces (like air resistance) acting on it. During free fall, all objects, regardless of their mass, experience the same acceleration towards the Earth, known as the acceleration due to gravity (g).


To derive the expression for g, we use Newton's law of gravitation, which states that the force of attraction (F) between two masses (M and m) separated by a distance (r) is given by:
F = G * (M * m) / r²
Here, G is the universal gravitational constant.


According to Newton's second law of motion, F = m * a. For an object of mass m near the Earth's surface, this force is the weight of the object, and the acceleration is g. Thus:
m * g = G * (M * m) / r²
Canceling m from both sides, we get:
g = G * M / r²
Here, M is the mass of the Earth, and r is the distance from the object to the Earth's center (approximately the Earth's radius near the surface).


The value of g varies with altitude and depth as follows:

  • With altitude: As we move higher above the Earth's surface, r increases, causing g to decrease because g ∝ 1/r².
  • With depth: Below the Earth's surface, only the mass enclosed by the radius at that depth contributes to g. Since the effective mass decreases, g also decreases linearly with depth.


This understanding is crucial for applications like satellite motion, geophysical studies, and even everyday phenomena like weight variation at different locations.

Question 19:
Explain the concept of free fall and derive the expression for the acceleration due to gravity (g) using Newton's law of gravitation. Also, mention any two factors affecting the value of g.
Answer:

Free fall is the motion of an object under the influence of gravitational force alone, without any other forces (like air resistance) acting on it. During free fall, all objects, regardless of their mass, experience the same acceleration towards the Earth.


To derive the expression for g, we start with Newton's law of gravitation:


F = G * (M * m) / r²


Where:
F = Gravitational force
G = Universal gravitational constant
M = Mass of the Earth
m = Mass of the object
r = Distance between the object and Earth's center


From Newton's second law of motion, F = m * a. Here, a = g (acceleration due to gravity).


Equating both expressions:
m * g = G * (M * m) / r²
Canceling m from both sides:
g = G * M / r²


This is the expression for acceleration due to gravity.


Factors affecting g:

  • Altitude: As height increases, g decreases because r (distance from Earth's center) increases.
  • Shape of the Earth: Earth is not perfectly spherical, so g varies slightly at different latitudes.
Question 20:
Explain the concept of gravitational force and derive the expression for acceleration due to gravity (g) on the surface of the Earth using Newton's law of gravitation. Also, discuss how the value of g changes with altitude and depth.
Answer:

The gravitational force is the attractive force between two objects with mass. According to Newton's law of gravitation, the force (F) between two masses (m1 and m2) separated by a distance (r) is given by:

F = G (m1m2)/r2

where G is the universal gravitational constant.

To derive the acceleration due to gravity (g) on Earth's surface, consider Earth's mass (M) and radius (R). The force on an object of mass m is:

F = G (Mm)/R2

From Newton's second law (F = ma), the acceleration (g) is:

g = F/m = G (M)/R2

This is the expression for g on Earth's surface.

Variation of g:

  • With altitude: As height (h) increases, g decreases because the distance from Earth's center increases. The modified formula is:
    g' = g (R2)/(R + h)2
  • With depth: Inside Earth, g decreases linearly with depth (d) due to reduced effective mass. The formula becomes:
    g' = g (1 - d/R)

Understanding these variations helps in applications like satellite motion and geophysical studies.

Question 21:
Explain the concept of free fall and derive the expression for the acceleration due to gravity (g) using Newton's law of gravitation. Also, mention how the value of g changes with altitude and depth.
Answer:

The concept of free fall refers to the motion of an object under the sole influence of gravity, without any other forces (like air resistance) acting on it. During free fall, all objects, regardless of their mass, experience the same acceleration towards the Earth, known as the acceleration due to gravity (g).


To derive the expression for g, we start with Newton's law of gravitation:
F = G * (M * m) / r²
where:
- F is the gravitational force
- G is the universal gravitational constant
- M is the mass of the Earth
- m is the mass of the object
- r is the distance between the centers of the Earth and the object.


According to Newton's second law of motion:
F = m * a
For free fall, the acceleration (a) is g, so:
m * g = G * (M * m) / r²
Canceling 'm' from both sides:
g = G * M / r²


This shows that g depends on the mass of the Earth (M) and the distance from its center (r).


Variation of g:

  • With altitude: As altitude increases, r increases, causing g to decrease (since g ∝ 1/r²).
  • With depth: Inside the Earth, g decreases linearly with depth because only the mass enclosed by the smaller radius contributes to the gravitational force.

Question 22:
Explain the concept of free fall and derive the expression for the acceleration due to gravity (g) using Newton's law of gravitation. How does the value of g vary with altitude and depth?
Answer:

Free fall is the motion of an object under the influence of gravitational force alone, without any other forces (like air resistance) acting on it. During free fall, all objects, regardless of their masses, experience the same acceleration called the acceleration due to gravity (g).


To derive g using Newton's law of gravitation:
1. Newton's law states: F = G * (M * m) / r², where F is the gravitational force, G is the gravitational constant, M is Earth's mass, m is the object's mass, and r is the distance between their centers.
2. From Newton's second law: F = m * a. Here, a = g (acceleration due to gravity).
3. Equating both expressions: m * g = G * (M * m) / r².
4. Simplifying: g = G * M / r². This gives the acceleration due to gravity.


Variation of g:

  • With altitude: As height (h) increases, r (distance from Earth's center) increases, reducing g as per g' = g * (R² / (R + h)²), where R is Earth's radius.
  • With depth: Inside Earth, g decreases linearly with depth (d) as g' = g * (1 - d/R), since only the inner mass contributes to gravity.

Note: g is maximum at Earth's surface and zero at its center.

Question 23:
Explain the concept of free fall and derive the expression for acceleration due to gravity (g) using Newton's law of gravitation. How does the value of g vary with altitude and depth?
Answer:

Free fall refers to the motion of an object under the sole influence of gravity, without any other forces (like air resistance) acting on it. During free fall, all objects experience the same acceleration, known as acceleration due to gravity (g), regardless of their mass.


To derive g using Newton's law of gravitation:


1. Newton's law states: F = G (M × m) / r², where F is the gravitational force, G is the universal gravitational constant, M is Earth's mass, m is the object's mass, and r is the distance between their centers.
2. From Newton's second law: F = m × a (where a is acceleration).
3. Equating both: m × g = G (M × m) / r².
4. Simplifying: g = G M / r² (where r is Earth's radius at the surface).

Variation of g:

  • With altitude: As height increases, r increases, reducing g (since g ∝ 1/r²).
  • With depth: Inside Earth, only the mass enclosed by the radius contributes to g, so g decreases linearly with depth.
Question 24:
Define gravitational potential energy and derive its expression for an object at height h above Earth's surface. How does this energy change when the object is taken to infinity?
Answer:

Gravitational potential energy (U) is the energy possessed by an object due to its position in a gravitational field. It represents the work done to move the object from a reference point (usually Earth's surface) to its current position against gravity.


Derivation for height h:


1. Work done to lift an object of mass m to height h: W = Force × Displacement = m × g × h.
2. This work is stored as potential energy: U = m g h (valid for h << Earth's radius).
3. For larger heights, use the general formula: U = -G M m / (R + h), where R is Earth's radius.

Change at infinity:

  • As h → ∞, U → 0 (since gravitational force diminishes to zero).
  • The energy required to move the object to infinity is equal to the binding energy of the object-Earth system.
Question 25:
Explain the concept of free fall and derive the expression for acceleration due to gravity (g) using Newton's law of gravitation. How does this value vary with altitude and depth?
Answer:

Free fall refers to the motion of an object under the sole influence of gravity, without any other forces (like air resistance) acting on it. During free fall, all objects experience the same acceleration, known as acceleration due to gravity (g), regardless of their masses.


To derive g using Newton's law of gravitation:


1. Newton's law states: F = G (Mem)/r2, where:
- F = Gravitational force
- G = Universal gravitational constant
- Me = Mass of Earth
- m = Mass of object
- r = Distance between centers (Earth's radius for surface)


2. From Newton's second law: F = ma, where a = g for free fall.
Thus, mg = G (Mem)/r2.


3. Canceling m from both sides: g = G Me/r2.


Variation with altitude: As altitude increases, r increases, causing g to decrease (inverse square law).


Variation with depth: Inside Earth, only the mass enclosed by the radius contributes to g, so g decreases linearly with depth, reaching zero at Earth's center.

Question 26:
Define universal law of gravitation and explain its significance. Calculate the gravitational force between two objects of masses 50 kg and 100 kg separated by 2 meters. (Given: G = 6.67 × 10-11 Nm2/kg2)
Answer:

The universal law of gravitation states that every object in the universe attracts every other object with a force (F) that is:
1. Directly proportional to the product of their masses (m1 and m2).
2. Inversely proportional to the square of the distance (r) between their centers.


Mathematically: F = G (m1m2)/r2, where G is the gravitational constant.


Significance:
- Explains planetary motion and tides.
- Binds the solar system and galaxies.
- Determines weight of objects on Earth.


Calculation:
Given: m1 = 50 kg, m2 = 100 kg, r = 2 m, G = 6.67 × 10-11 Nm2/kg2.

1. Substitute values into the formula:
F = (6.67 × 10-11 × 50 × 100)/22.

2. Simplify numerator:
6.67 × 10-11 × 5000 = 3.335 × 10-7.

3. Divide by denominator:
3.335 × 10-7/4 = 8.3375 × 10-8 N.

Thus, the gravitational force is 8.34 × 10-8 N (very small due to weak gravitational force between everyday objects).

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:
A student drops a stone from a cliff and observes it takes 4 seconds to hit the ground. Calculate the height of the cliff using the equation of motion under gravity (g = 9.8 m/s²).
Answer:
Case Summary

A stone falls freely from a cliff, taking 4 seconds to reach the ground.

Scientific Principle

We studied that the distance (h) covered under free fall is given by h = ½gt², where g is acceleration due to gravity.

Solution Approach
  • Given: t = 4s, g = 9.8 m/s²
  • Using h = ½ × 9.8 × (4)² = 78.4 m

Thus, the cliff is 78.4 m high.

Question 2:
Two objects of masses 5 kg and 10 kg are dropped from the same height. Compare their time of fall and explain using gravitational principles.
Answer:
Case Summary

Two objects of different masses are dropped from the same height.

Scientific Principle

Our textbook shows that in a vacuum, all objects fall at the same rate regardless of mass, as acceleration due to gravity (g) is constant.

Solution Approach
  • Both objects experience the same g (9.8 m/s²).
  • Since initial velocity and height are equal, time of fall is identical.

This matches Galileo’s experiment.

Question 3:
A satellite orbits Earth at a constant speed. Explain why it doesn’t fall despite Earth’s gravity, referencing centripetal force.
Answer:
Case Summary

A satellite orbits Earth without falling.

Scientific Principle

We studied that gravity provides the centripetal force required for circular motion, balancing the satellite’s inertia.

Solution Approach
  • Gravity pulls the satellite toward Earth.
  • Its tangential speed creates a curved path, preventing a fall.

This is similar to the Moon’s orbit, as per NCERT.

Question 4:
A spring balance shows 50 N for an object on Earth. Predict its weight on the Moon (gₘ = 1/6 gₑ) and justify.
Answer:
Case Summary

An object weighs 50 N on Earth; we need its Moon weight.

Scientific Principle

Our textbook states weight = mass × g. On the Moon, g is 1/6th of Earth’s.

Solution Approach
  • Mass = 50 N / 9.8 ≈ 5.1 kg
  • Moon weight = 5.1 × (9.8/6) ≈ 8.33 N

This matches NCERT’s lunar weight example.

Question 5:
A student drops a stone from a cliff and observes it takes 4 seconds to hit the ground. Using gravitational acceleration (g = 9.8 m/s²), calculate the height of the cliff. Explain the universal law of gravitation in this context.
Answer:
Case Summary

A stone falls freely under gravity, taking 4s to reach the ground.

Scientific Principle
  • We studied that free fall follows s = ut + ½gt² (u=0).
  • Our textbook shows g is constant near Earth’s surface.
Solution Approach

Height = 0 + ½ × 9.8 × (4)² = 78.4m. The universal law explains Earth’s pull on the stone.

Question 6:
Two objects of masses 50kg and 100kg are placed 2m apart. Compute the gravitational force between them (G = 6.67 × 10⁻¹¹ Nm²/kg²). Relate this to the moon’s orbit around Earth.
Answer:
Case Summary

Gravitational force between two masses is calculated.

Scientific Principle
  • We studied F = G(m₁m₂)/r².
  • Our textbook links this to celestial motions like the moon’s orbit.
Solution Approach

F = (6.67 × 10⁻¹¹ × 50 × 100)/2² = 8.34 × 10⁻⁸ N. Similarly, Earth’s gravity keeps the moon in orbit.

Question 7:
A spring balance shows 60N for a 6kg object on Earth. Predict its reading on the moon (gₘ = gₑ/6). Explain how weight differs from mass.
Answer:
Case Summary

A 6kg object’s weight changes on the moon due to lower g.

Scientific Principle
  • We studied weight = mg.
  • Our textbook shows mass is constant, but weight varies with gravity.
Solution Approach

Moon’s weight = 6 × (9.8/6) = 9.8N. Mass remains 6kg, proving they’re distinct.

Question 8:
A satellite orbits Earth at a height where g is 4 m/s². If its weight is 2000N, find its mass. Compare this to the weightlessness astronauts feel.
Answer:
Case Summary

A satellite’s mass is derived from its weight in orbit.

Scientific Principle
  • We studied m = W/g.
  • Our textbook explains weightlessness occurs in free-fall orbits.
Solution Approach

Mass = 2000/4 = 500kg. Astronauts feel weightless as they and the satellite fall at the same rate.

Question 9:
A student drops a stone from a cliff and observes it takes 4 seconds to hit the ground. Calculate the height of the cliff using g = 9.8 m/s². Our textbook shows a similar problem with a freely falling object.
Answer:
Case Summary

A stone falls freely under gravity. Time taken = 4s.

Scientific Principle

We studied the equation of motion: h = ½gt², where h is height, g is acceleration due to gravity, and t is time.

Solution Approach
  • Substitute g = 9.8 m/s² and t = 4s into the equation.
  • Height h = ½ × 9.8 × (4)² = 78.4 m.
Question 10:
An astronaut on the Moon weighs less than on Earth. Explain why using the concept of gravitational force. Refer to NCERT's example of weight difference.
Answer:
Case Summary

Astronaut's weight varies on Moon and Earth.

Scientific Principle

We studied that weight depends on gravitational force (F = mg). Moon's g is 1/6th of Earth's.

Solution Approach
  • Gravitational force is weaker on Moon due to smaller mass.
  • As g decreases, weight (F) reduces. NCERT shows this with a 60 kg person weighing 10 kg on Moon.
Question 11:
A spring balance measures 50 N for a 5 kg object on Earth. Predict the reading if the same experiment is done on Mars (g = 3.7 m/s²). Use NCERT's method of calculating weight.
Answer:
Case Summary

Spring balance shows 50 N on Earth for 5 kg. Find reading on Mars.

Scientific Principle

We studied weight = mass × gravitational acceleration. NCERT uses W = mg.

Solution Approach
  • On Mars, g = 3.7 m/s².
  • Weight = 5 kg × 3.7 m/s² = 18.5 N. The balance will show 18.5 N.
Question 12:
Two objects of masses 10 kg and 20 kg are dropped from the same height. Compare their time to reach the ground ignoring air resistance. Refer to NCERT's example of free fall.
Answer:
Case Summary

Objects of 10 kg and 20 kg fall from the same height.

Scientific Principle

We studied that in free fall, time depends only on height and g, not mass. NCERT shows this with a feather and coin experiment.

Solution Approach
  • Both objects have same g and initial velocity (0).
  • They will reach the ground simultaneously, as confirmed by NCERT.
Question 13:

Rahul conducted an experiment to measure the acceleration due to gravity (g) using a simple pendulum. He observed that his calculated value of g was slightly less than the standard value (9.8 m/s²). Based on this case:

  • Identify two possible reasons for the observed discrepancy.
  • Explain how each factor affects the measurement of g.
Answer:

Possible reasons for discrepancy:

  • Air resistance: The pendulum bob experiences air resistance, which opposes its motion, causing the time period to increase slightly. Since g is inversely proportional to the square of the time period (T), a higher T leads to a lower calculated value of g.
  • String mass: If the string is not massless (as assumed in theory), it adds to the effective length of the pendulum, increasing T and reducing the calculated g.

Additional note: Other factors like improper measurement of pendulum length or human error in timing can also contribute.

Question 14:

Two astronauts, A and B, are floating in space near a planet. Astronaut A weighs 60 kg, and Astronaut B weighs 80 kg. Suddenly, the gravitational force of the planet pulls them toward its surface. Based on this scenario:

  • Compare the gravitational force acting on both astronauts.
  • Explain why they accelerate at the same rate despite their different masses.
Answer:

Comparison of gravitational force:

  • The gravitational force (F) is greater on Astronaut B because F = mg, where m is mass and g is acceleration due to gravity. Since Astronaut B has a higher mass, the force is larger.

Same acceleration reason:

  • From Newton's second law, a = F/m. Here, F is directly proportional to m, so the ratio F/m remains constant (equal to g). Thus, both astronauts accelerate at the same rate.

Key concept: This demonstrates the equivalence principle, where inertial mass and gravitational mass are identical.

Question 15:
A group of students conducted an experiment to measure the acceleration due to gravity (g) using a simple pendulum. They recorded the time for 20 oscillations of the pendulum for different lengths (50 cm, 75 cm, 100 cm).

Based on their observations:

  • What relationship between the length of the pendulum and its time period did they likely observe?
  • How can this experiment help verify the value of g?
Answer:

The students likely observed that as the length of the pendulum increases, the time period also increases. This is because the time period (T) of a simple pendulum is given by the formula:

T = 2π√(L/g)

where L is the length of the pendulum and g is the acceleration due to gravity.

To verify the value of g, they can use the following steps:

1. Measure the time for 20 oscillations (t) and calculate the time period (T = t/20).
2. Repeat this for different lengths (L).
3. Plot a graph of against L, which should give a straight line.
4. The slope of this line is equal to 4π²/g, from which g can be calculated as g = 4π²/slope.

This experiment demonstrates the direct relationship between the length and time period of a pendulum and provides a practical method to determine g accurately.

Question 16:
An astronaut on the Moon drops a hammer and a feather from the same height simultaneously. Both objects reach the lunar surface at the same time.

Explain this observation using the concept of gravitation. How would this scenario differ if the experiment were conducted on Earth?

Answer:

On the Moon, both the hammer and the feather fall at the same rate and reach the surface simultaneously because the acceleration due to gravity (g) is the same for all objects regardless of their mass. This is due to the absence of air resistance on the Moon, which is a vacuum. The gravitational force (F = mg) depends only on the mass of the object and the Moon's gravity, so both objects accelerate equally.

On Earth, however, the presence of air resistance affects the motion of objects differently based on their shape and mass. The feather, being lighter and having a larger surface area, experiences more air resistance compared to the hammer. As a result, the hammer would reach the ground first on Earth, while the feather would fall slower due to opposing air resistance.

This experiment highlights the principle that in the absence of air resistance, all objects fall at the same rate under gravity, as demonstrated by Galileo's famous experiment.

Question 17:
A student observes that a stone dropped from the top of a building takes 4 seconds to reach the ground. Using this information, calculate the height of the building. Assume acceleration due to gravity (g) = 9.8 m/s². Also, explain why the mass of the stone does not affect the time taken to reach the ground.
Answer:

To calculate the height of the building, we use the equation of motion:
s = ut + (1/2)gt²
Here, initial velocity (u) = 0 (since the stone is dropped), time (t) = 4 s, and g = 9.8 m/s².

Substituting the values:
s = 0 × 4 + (1/2) × 9.8 × (4)²
s = 0 + 0.5 × 9.8 × 16
s = 78.4 m

The height of the building is 78.4 meters.

The mass of the stone does not affect the time taken to reach the ground because the acceleration due to gravity (g) is independent of mass. All objects, regardless of their mass, experience the same acceleration (9.8 m/s²) in a vacuum (neglecting air resistance). This principle was demonstrated by Galileo and is a fundamental concept in gravitation.

Question 18:
Two objects of masses 5 kg and 10 kg are dropped simultaneously from the same height on Earth. Compare their acceleration and time taken to reach the ground. Justify your answer with relevant principles of gravitation.
Answer:

Both objects, regardless of their masses (5 kg and 10 kg), will have the same acceleration and take the same time to reach the ground when dropped from the same height on Earth.

  • Acceleration: The acceleration due to gravity (g) is constant for all objects near Earth's surface (≈ 9.8 m/s²). It does not depend on the mass of the object, as per Newton's Universal Law of Gravitation and Galileo's experiments.
  • Time taken: Since both objects start from rest (u = 0) and experience the same acceleration, their time of fall is identical. The equation s = ut + (1/2)gt² confirms that time (t) depends only on height (s) and acceleration (g), not mass.

This demonstrates the principle of equivalence in gravitation, where inertial and gravitational mass are equivalent, leading to identical motion under gravity for all objects in the absence of air resistance.

Question 19:
A group of students conducted an experiment to measure the acceleration due to gravity (g) using a simple pendulum. They varied the length of the pendulum and recorded the time period for 20 oscillations. Their data is shown below:

Length (m)Time for 20 oscillations (s)
0.528.2
1.040.0
1.549.0

Using the formula T = 2π√(L/g), help the students calculate the value of g for each length and analyze the consistency of their results.
Answer:

To calculate g, we first find the time period (T) for one oscillation by dividing the given time by 20. Then, we rearrange the formula T = 2π√(L/g) to solve for g:


For Length = 0.5 m:
Time period (T) = 28.2 s / 20 = 1.41 s
g = 4π²L / T² = (4 × 9.87 × 0.5) / (1.41)² ≈ 9.86 m/s²

For Length = 1.0 m:
Time period (T) = 40.0 s / 20 = 2.0 s
g = (4 × 9.87 × 1.0) / (2.0)² ≈ 9.87 m/s²

For Length = 1.5 m:
Time period (T) = 49.0 s / 20 = 2.45 s
g = (4 × 9.87 × 1.5) / (2.45)² ≈ 9.85 m/s²

The calculated values of g are consistent (~9.86 m/s²) and close to the standard value (9.8 m/s²). Minor variations may arise due to measurement errors or air resistance. The experiment confirms that g is independent of the pendulum's length.

Question 20:
On a school trip to the mountains, Riya noticed that her packet of chips puffed up as they ascended. Her teacher explained that this happens due to changes in atmospheric pressure. Relate this observation to the concept of gravitational force and explain why atmospheric pressure decreases with altitude.
Answer:

The puffing up of the chips packet is due to the decrease in atmospheric pressure at higher altitudes. Here's how it connects to gravitational force:


  • Gravity's Role: Earth's gravitational force pulls air molecules toward its surface, creating atmospheric pressure. At higher altitudes, the force weakens slightly because distance from Earth's center increases.
  • Air Density: As altitude increases, air molecules become less densely packed due to reduced gravitational pull, leading to lower pressure.
  • Chips Packet: The sealed packet contains air at ground-level pressure. At higher altitudes, the external pressure drops, causing the internal air to expand and puff up the packet.

This phenomenon demonstrates how gravitational force influences atmospheric pressure distribution. It also explains why mountaineers carry oxygen cylinders—the thinner air at high altitudes contains fewer oxygen molecules per unit volume.

Question 21:
A group of students conducted an experiment to measure the acceleration due to gravity (g) using a simple pendulum. They varied the length of the pendulum and recorded the time for 20 oscillations.

Based on their observations:

  • Length (L) = 100 cm, Time for 20 oscillations (T) = 40 s
  • Length (L) = 64 cm, Time for 20 oscillations (T) = 32 s

Using the formula T = 2π√(L/g), calculate the value of g for both cases and compare the results. Explain any discrepancies observed.

Answer:

To calculate g, we use the formula for the time period of a simple pendulum: T = 2π√(L/g), where T is the time period for one oscillation.


Step 1: Calculate Time Period (T)
For the first case (L = 100 cm, T for 20 oscillations = 40 s):
Time period (T) = Total time / Number of oscillations = 40 s / 20 = 2 s.

For the second case (L = 64 cm, T for 20 oscillations = 32 s):
Time period (T) = 32 s / 20 = 1.6 s.

Step 2: Rearrange the formula to solve for g
g = 4π²L / T².

Step 3: Calculate g for both cases
For L = 100 cm (1 m) and T = 2 s:
g = 4 × (3.14)² × 1 / (2)² = 4 × 9.86 × 1 / 4 = 9.86 m/s².

For L = 64 cm (0.64 m) and T = 1.6 s:
g = 4 × (3.14)² × 0.64 / (1.6)² = 4 × 9.86 × 0.64 / 2.56 ≈ 9.86 m/s².

Comparison and Discrepancies:
Both cases yield g ≈ 9.86 m/s², which is close to the standard value (9.8 m/s²). Minor discrepancies may arise due to:
  • Human error in measuring length or time.
  • Air resistance affecting pendulum motion.
  • Imperfections in the pendulum setup (e.g., string stiffness).
Question 22:
An astronaut on the Moon drops a hammer and a feather simultaneously from the same height. Both objects reach the lunar surface at the same time.

Explain this observation using the concept of gravitational acceleration. How would the result differ if the same experiment were conducted on Earth? Justify your answer.

Answer:

On the Moon, both the hammer and feather fall at the same rate because:

  • The Moon has no atmosphere, so there is no air resistance to oppose the motion of the objects.
  • Gravitational acceleration (g) is the same for all objects regardless of their mass, as per Galileo's law of falling bodies.
  • The Moon's g is about 1/6th of Earth's (1.63 m/s²), but it acts equally on both objects.

On Earth, the result would differ because:

  • Earth has an atmosphere, causing air resistance to act more significantly on the feather than the hammer.
  • The feather's larger surface area slows it down, while the hammer falls faster due to negligible air resistance.
  • In a vacuum (no air), both would fall at the same rate (9.8 m/s²), as demonstrated by the Apollo 15 mission on the Moon.

This experiment highlights the independence of gravitational acceleration from mass and the role of external forces like air resistance.

Chat on WhatsApp