🪐 Chapter 12 · NCERT Science IX

GRAVITATION

The invisible force that holds planets in orbit, keeps us on Earth, and governs the fall of every apple — Newton's Universal Law, decoded.

F = G · m₁m₂ / r²
G6.67×10⁻¹¹
9.8 m/s²g on Earth
g/6g on Moon
★★★★★Exam Weight
📍 Chapter Snapshot — Universal Constants
G
Gravitational Constant
6.674 × 10⁻¹¹ N·m²/kg²
g
Acceleration due to Gravity
9.8 m/s² (on Earth surface)
Mₑ
Mass of Earth
5.972 × 10²⁴ kg
Rₑ
Radius of Earth
6.4 × 10⁶ m
🎯 Why This Chapter Matters for Exams
📌 Universal Law of Gravitation statement + formula is a must-memorise 3-mark theory.
📌 Difference between Mass and Weight is tested every year in both MCQ and SA sections.
📌 Deriving g from Universal Gravitation formula — standard 5-mark derivation question.
📌 Free fall numericals and Kepler's laws appear in NTSE and Olympiad selections.
⚗️ Important Formula Capsules
Universal GravitationF = G·m₁m₂ / r²N
g on Earthg = G·M / R²m/s²
WeightW = mgN
Buoyancy (Archimedes)F_b = ρ·V·gN
PressureP = F / APa (N/m²)
💡 Key Concept Highlights
Universal Law of GravitationFree FallMass vs WeightBuoyancyArchimedes' Principleg on MoonThrustPressureRelative DensityKepler's Laws
📚 What You Will Learn
  • 1Newton's Universal Law of Gravitation — statement, formula, and significance
  • 2Deriving the value of g and how it varies with altitude and depth
  • 3Free fall, equations of motion under gravity
  • 4Mass and weight — definitions, units, and differences
  • 5Thrust, pressure, buoyancy, and Archimedes' Principle with applications
🔗 Navigate to Chapter Resources
📄
Full Notes
🧩
MCQs Practice
✅
True / False
📝
NCERT Exercises
🏆 Exam Strategy & Preparation Tips
01
Derive g
The derivation of g = GM/R² from Universal Gravitation is tested as a 5-mark question — master it.
02
Mass vs Weight
Mass is constant everywhere; weight changes with g. A table comparison guarantees full marks.
03
Moon vs Earth g
g on moon = g/6. If W on Earth = 600 N, W on moon = 100 N — a classic 2-mark numerical.
04
Archimedes Proof
Verify Archimedes' Principle: weight of displaced fluid = buoyant force. Draw with beaker diagram.
Chapter 9 · CBSE · Class IX
🍏

Gravitation

Gravitation Universal Law of Gravitation Gravitational Force Centripetal Force Acceleration due to Gravity Free Fall Value of g Mass Weight Weight of an Object on Moon Thrust Pressure Pressure in Fluids Buoyancy Buoyant Force Objects Float or Sink Archimedes' Principle Relative Density Density Inverse Square Law SI Unit of Pressure Pascal Numericals
📋
Table of Contents
15 sections
📘 Definition
📖 Introduction
📌 Meaning of Gravity
🔷 Characteristics
Important Characteristics of Gravitation
🔷 Important Characteristics of Gravitation
  • Gravitation is always an attractive force. It never repels objects.
  • It acts between every pair of objects having mass.
  • It acts over very large distances and its range is considered infinite.
  • No medium is required for gravitational force to act. It can act even through vacuum.
  • The force becomes stronger when masses increase.
  • The force becomes weaker as the distance between objects increases.
  • Gravitational force obeys Newton's Third Law. If Earth attracts an apple, the apple also attracts Earth with an equal force in the opposite direction.
✏️ Everyday Examples of Gravitation
  1. An apple falls towards Earth.
  2. Rainwater falls from clouds.
  3. A football thrown upward returns to the ground.
  4. The Moon revolves around Earth.
  5. Earth revolves around the Sun.
  6. Artificial satellites remain in orbit.
  7. Sea tides occur due to the gravitational pull of the Moon and Sun.
🤔 Did You Know?
Why Don't We Feel Gravitational Attraction Between Two Persons?
Every person attracts every other person gravitationally. However, since the masses of human beings are very small, the force produced is extremely tiny and cannot be noticed in daily life.

Earth has an enormous mass \[ \left(5.97\times10^{24}\text{ kg}\right) \] therefore its gravitational pull on us is much larger than the pull between two people.
💡 Concept
Concept Builder
🏛️ Historical Background
Ancient astronomers observed that planets move in definite paths around the Sun, but the reason remained unknown for centuries.

Sir Isaac Newton explained this mystery by proposing the Universal Law of Gravitation. According to a famous story, seeing an apple fall from a tree inspired him to think that the same force pulling the apple downward might also be responsible for keeping the Moon in its orbit around Earth.

Newton concluded that the same gravitational force acts everywhere in the universe.
Newton's Revolutionary Idea
Before Newton, falling objects and planetary motion were considered unrelated phenomena.
Newton united them under one single law:
  • The force that causes an apple to fall is the same force that keeps the Moon in orbit.
  • The force that keeps planets revolving around the Sun is also gravitation.
This was one of the greatest scientific discoveries because it explained both earthly and celestial motions using one universal law.
🎨 SVG Diagram
Simple Concept Map
GRAVITATION Falling Objects Earth's Gravity Planetary Motion Apple, Rain, Stones TERRESTRIAL PHENOMENA Weight & Free Fall PHYSICAL PROPERTIES Moon & Satellites CELESTIAL DYNAMICS
✏️ Example
Solved Concept Example
Name three natural phenomena that occur due to gravitation.
  1. 1
    Recall common effects of gravitational force.
  2. 2
    Select any three correct examples.
  1. Falling of objects towards Earth.
  2. Revolution of planets around the Sun.
  3. Motion of the Moon around Earth.
🗒️ Think Beyond NCERT
Astronauts in orbit appear weightless not because there is no gravity. Gravity is still acting on them. They experience continuous free fall along with their spacecraft, producing the sensation of weightlessness.
⚡ Exam Tip
🗒️ Common Mistale
  • Writing that gravity and gravitation are exactly the same.
  • Thinking that only Earth exerts gravitational force.
  • Believing that only large bodies possess gravity.
  • Assuming gravitational force requires air or another medium.
  • Ignoring that gravitational attraction is mutual.
🌟 Board Important Definitions
📋 CBSE Competency-Based (HOTS) Case Study

During a science exhibition, Riya drops a steel ball and a wooden ball simultaneously from the same height. Both fall towards Earth. Her friend asks why both move downward even though they are made of different materials.

Questions

  1. Which force causes both balls to move downward?
  2. Does Earth attract only heavy objects?
  3. Do the balls also attract Earth?

Answers

  1. Earth's gravitational force.
  2. No. Earth attracts every object having mass.
  3. Yes. According to gravitation, attraction is mutual, although Earth's motion is too small to notice.
🗒️ Quick Reference
  • Every object having mass attracts every other object.
  • Gravitation acts throughout the universe.
  • Gravity is a special case of gravitation.
  • Gravitational force is always attractive.
  • It explains falling bodies, planetary motion, tides and satellite motion.
  • Newton proposed the Universal Law of Gravitation.
🍏

Universal Law of Gravitation

📋
Table of Contents
15 sections
📘 Definition
📘 Statement of the Universal Law of Gravitation
💡 Concept
🎨 SVG Diagram
The gravitational force acts along the line joining the centres of two bodies.
M m d F = G Mm d2
📐 Derivation
Mathematical Derivation
Let
  • Mass of first object = \(M\)
  • Mass of second object = \(m\)
  • Distance between their centres = \(d\)
  • Gravitational force between them = \(F\)

Experiments show that the gravitational force increases with the masses of both bodies.

Therefore, \[ F\propto M\times m \tag{1} \] It is also found that the force decreases rapidly as the distance between the bodies increases. \[ F\propto \frac{1}{d^2} \tag{2} \] Combining equations (1) and (2), \[F\propto \frac{Mm}{d^2}\] Replacing the proportionality sign by a constant, \[ \boxed{ F=G\frac{Mm}{d^2} } \] where
  • \(F\) = Gravitational force
  • \(M\) = Mass of first body
  • \(m\) = Mass of second body
  • \(d\) = Distance between their centres
  • \(G\) = Universal Gravitational Constant
Meaning of the Formula
Change Effect on Gravitational Force
Mass of either body doubles Force becomes twice.
Both masses double Force becomes four times.
Distance doubles Force becomes one-fourth.
Distance triples Force becomes one-ninth.
Distance becomes half Force becomes four times.
Inverse Square Law
The force depends upon the square of the distance. Therefore, even a small increase in distance causes a large decrease in gravitational force.

Mathematically, \[F\propto\frac{1}{d^2}\] This relationship is known as the Inverse Square Law.
🗒️ Universal Gravitational Constant (\(G\))
The proportionality constant \(G\) is called the Universal Gravitational Constant.

Its value remains the same everywhere in the universe irrespective of the place, planet or time.
Symbol \(G\)
SI Unit \(\mathrm{N\,m^2\,kg^{-2}}\)
CGS Unit \(\mathrm{dyne\,cm^2\,g^{-2}}\)
Accepted Value \[ \boxed{ G=6.67\times10^{-11}\; \mathrm{N\,m^2\,kg^{-2}} } \]
Note: Older textbooks may mention \(6.673\times10^{-11}\;\mathrm{N\,m^2\,kg^{-2}}\). For CBSE Class IX, write \[ 6.67\times10^{-11}\;\mathrm{N\,m^2\,kg^{-2}} \] unless a different value is specified in the question.
🌟 Physical Significance of \(G\)
The numerical value of \(G\) tells us the gravitational force between two bodies of mass \(1\;\mathrm{kg}\) each kept \(1\;\mathrm{m}\) apart.

Thus\[F=6.67\times10^{-11}\;\mathrm{N}\] showing that gravitational force between ordinary objects is extremely small.
🔷 Characteristics of Universal Gravitation
🔷 Characteristics
  • Acts between every pair of objects having mass.
  • Always attractive.
  • Acts along the line joining the centres of two bodies.
  • Acts through vacuum.
  • Has infinite range.
  • Independent of the nature or composition of the objects.
  • Obeys Newton's Third Law of Motion.
  • Becomes weaker very rapidly as distance increases.
⚠️ Limitations of the Formula
  • The formula is exact only for point masses or spherical bodies.
  • Distance must be measured between the centres of the two bodies.
  • The objects should not be extremely close compared to their sizes.
🌟 Importance of the Universal Law of Gravitation
🔢 Formula Summary
✏️ Example
Solved Example
If the distance between two bodies is doubled, how will the gravitational force change?
  1. 1
    Use inverse square law.
  2. 2
    Replace \(d\) by \(2d\).
\[ F=\frac{G Mm}{(2d)^2} \] \[ F=\frac{1}{4} \times \frac{G Mm}{d^2} \] Therefore, the gravitational force becomes\[\boxed{\frac{F}{4}}\]
If one mass becomes three times while the distance remains unchanged, how does the gravitational force change?
\[ F'=G\frac{(3M)m}{d^2} \] \[ F'=3F \] Therefore, the gravitational force becomes\[\boxed{3F}\]
⚡ Exam Tip
❌ Common Mistakes
  • Writing \(F\propto \frac{1}{d}\) instead of \(F\propto\frac{1}{d^2}\).
  • Confusing gravitational constant \(G\) with acceleration due to gravity \(g\).
  • Using surface-to-surface distance instead of centre-to-centre distance.
  • Writing the unit of \(G\) incorrectly.
📋 CBSE Competency-Based (HOTS) Case Study

Two satellites A and B have equal masses. Satellite B is moved to twice the distance from Earth compared to satellite A.

Questions

  1. Which satellite experiences greater gravitational force?
  2. How many times does the force change?
  3. Which law explains this?

Answers

  1. Satellite A.
  2. The force on B becomes one-fourth.
  3. Universal Law of Gravitation (Inverse Square Law).
⚡ Quick Revision
  • Every mass attracts every other mass.
  • \[ F=G\frac{Mm}{d^2} \]
  • Force is directly proportional to product of masses.
  • Force is inversely proportional to square of distance.
  • \[ G=6.67\times10^{-11}\; \mathrm{N\,m^2\,kg^{-2}} \]
  • Gravitational force is always attractive.
🍏

Worked Example – Gravitational Force Between Earth and Moon

📋
Table of Contents
9 sections
❓ Question
The mass of the Earth is \[ 6\times10^{24}\ \mathrm{kg} \] and the mass of the Moon is \[ 7.4\times10^{22}\ \mathrm{kg}. \] The average distance between the Earth and the Moon is \[ 3.84\times10^{5}\ \mathrm{km}. \] Calculate the gravitational force exerted by the Earth on the Moon.

Take \[ G=6.7\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}} \]
💡 Concept
🗺️ Roadmap
Roadmap for Solution
  1. Write the given quantities.
  2. Convert distance from kilometres to metres.
  3. Write the Universal Law of Gravitation.
  4. Substitute all values carefully.
  5. Simplify powers of 10 separately.
  6. Write the answer with proper SI unit.
🧩 Solution
Given:
Mass of Earth \[ M=6\times10^{24}\ \mathrm{kg} \]
Mass of Moon \[ m=7.4\times10^{22}\ \mathrm{kg} \]
Universal Gravitational Constant \[ G=6.7\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}} \]
Distance between Earth and Moon \[ d=3.84\times10^{5}\ \mathrm{km} \]
Part (a)
  1. Convert Distance into SI Unit
    \[ \begin{aligned} d &=3.84\times10^{5}\ \mathrm{km}\\ &=3.84\times10^{5}\times10^{3}\ \mathrm{m}\\ &=3.84\times10^{8}\ \mathrm{m} \end{aligned} \]
    Exam Tip: Never forget to convert kilometre into metre before using the gravitational formula.
  2. Write the Formula
    \[F=G\frac{Mm}{d^2}\]
  3. Substitute the Values
    \[ \begin{aligned} F &=6.7\times10^{-11} \times \frac{6\times10^{24}\times7.4\times10^{22}} {(3.84\times10^{8})^2} \end{aligned} \]
  4. Simplify
    \[ \small \begin{aligned} F &= \frac{6.7\times6\times7.4\times10^{-11+24+22}} {3.84^2\times10^{16}} \\ &= \frac{297.48\times10^{35}} {14.7456\times10^{16}} \\ &= \frac{297.48}{14.7456}\times10^{19} \\ &= 20.17\times10^{19} \\ &= 2.02\times10^{20}\ \mathrm{N} \end{aligned} \]
  5. Using the rounded value
    \[ d=3.8\times10^{8}\ \mathrm{m} \] gives \[ F\approx2.06\times10^{20}\ \mathrm{N}, \] which is the value generally obtained in NCERT-based solutions.
✅ Answer
Final Answert
Therefore, the gravitational force exerted by the Earth on the Moon is \[ \boxed{ F\approx2.06\times10^{20}\ \mathrm{N} } \]
🗒️ Verification of Unit
\[ \begin{aligned} F &= \mathrm{\left(N\,m^2\,kg^{-2}\right)} \times \frac{\mathrm{kg\times kg}} {\mathrm{m^2}} \\ &=\mathrm{N} \end{aligned} \] Hence, the obtained unit is the SI unit of force (Newton).
💡 Concept
Concept Check
📍 Key Learning Points
  • Always convert all quantities into SI units before substitution.
  • Distance must be measured between the centres of the two bodies.
  • Handle powers of ten separately to reduce calculation errors.
  • Keep at least three significant figures during intermediate calculations.
  • The force calculated is the mutual gravitational force between the Earth and the Moon.
❌ Common Mistakes
  • Using kilometre instead of metre.
  • Squaring only the numerical value and forgetting to square the power of ten.
  • Writing \[ d^2=(3.84)^2\times10^8 \] instead of \[ (3.84)^2\times10^{16}. \]
  • Using \[ F=\frac{GMm}{d} \] instead of \[ F=\frac{GMm}{d^2}. \]
🍏

Importance of the Universal Law of Gravitation

📋
Table of Contents
10 sections
📖 Introduction
🌟 Major Applications and Importance
🛠️ Real-Life Applications
Application Role of Gravitation
Satellite Television Satellites remain in stable orbit around Earth.
GPS Navigation Navigation satellites continuously orbit Earth.
Weather Forecasting Meteorological satellites monitor weather patterns.
Space Missions Rocket trajectories are calculated using gravitational laws.
Ocean Navigation Tides are predicted using gravitational effects of the Moon and the Sun.
💡 Concept Builder
🗒️ Board Examination Points
Most Frequently Asked Importance of the Universal Law of Gravitation
  1. Explains falling of bodies towards Earth.
  2. Explains the Moon's revolution around Earth.
  3. Explains planetary motion around the Sun.
  4. Explains ocean tides.
  5. Explains motion of artificial satellites.
🔤 Memory Trick
📋 CBSE Competency-Based (HOTS) Question

During a science exhibition, students observed that communication satellites continuously orbit the Earth without falling directly onto its surface. Another student said that satellites stay in space because there is no gravity.

Questions

  1. Is the student's statement correct?
  2. Which force keeps satellites in orbit?
  3. Name one modern technology that depends on artificial satellites.

Answers

  1. No. Gravity acts even at satellite heights.
  2. Earth's gravitational force provides the required centripetal force.
  3. GPS navigation, satellite communication, weather forecasting or television broadcasting.
⚡ Exam Tip
🗒️ Common Mistake
  • Writing only "objects fall on Earth" without explaining the reason.
  • Forgetting to mention ocean tides.
  • Confusing revolution with rotation.
  • Assuming satellites move without Earth's gravitational force.
  • Ignoring the importance of gravitation in modern space technology.
⚡ Quick Revision
  • Explains falling of bodies.
  • Explains Earth's gravity.
  • Explains the Moon's revolution.
  • Explains planetary motion.
  • Explains ocean tides.
  • Explains artificial satellites.
  • Supports GPS, communication and weather forecasting.
  • Forms the foundation of astronomy and space science.
🍏

Free Fall

📋
Table of Contents
12 sections
📘 Definition
💡 Concept of Free Fall
✏️ Examples of Free Fall
  • A stone dropped from the top of a building.
  • A ripe fruit falling from a tree.
  • A ball released from a height.
  • An object dropped inside a vacuum chamber.
  • Skydivers before opening their parachutes (approximately free fall).
🔷 Characteristics of Free Fall
🔷 Characteristics
Property Description
Force Acting Only gravitational force
Acceleration Constant and equal to \(g\)
Direction Towards the centre of the Earth
Nature of Motion Uniformly accelerated motion
Velocity Increases uniformly with time
📌 Velocity During Free Fall
📐 Derivation of Acceleration Due to Gravity (\(g\))
Let an object of mass \(m\) fall freely towards the Earth.

According to Newton's Second Law of Motion, \[F=ma\] During free fall,\[a=g\] Therefore, \[ F=mg \tag{1} \] According to the Universal Law of Gravitation, \[ F=G\frac{Mm}{d^2} \tag{2} \] where
  • \(M\) = Mass of the Earth
  • \(m\) = Mass of the object
  • \(d\) = Distance between the centres of the Earth and the object
Equating equations (1) and (2), \[mg=G\frac{Mm}{d^2}\] Dividing both sides by \(m\), \[g=\frac{GM}{d^2}\]
Important Observation: The mass of the falling object gets cancelled. Hence, acceleration due to gravity is independent of the mass of the object.
📌 Acceleration Due to Gravity Near the Earth's Surface
✏️ Example
Solved Example
A stone is dropped from the top of a tower. What force acts on it during free fall (ignoring air resistance)?
  1. 1
    Recall the definition of free fall.
  2. 2
    Identify the force acting on the object.
During free fall, only the Earth's gravitational force acts on the stone. Therefore, the stone accelerates downward with \[g=9.8\ \mathrm{m\,s^{-2}}\]
📋 CBSE Competency-Based (HOTS) Question

A feather and a coin are dropped simultaneously inside a vacuum chamber.

Questions

  1. Which object reaches the ground first?
  2. Why?
  3. Which scientific principle explains this?

Answers

  1. Both reach the ground at the same time.
  2. In vacuum, only gravity acts and both experience the same acceleration \(g\).
  3. Free fall and acceleration due to gravity.
⚡ Exam Tip
❌ Common Mistakes
  • Writing \(g\) as \(9.8\ \mathrm{m\,s^{-1}}\) instead of \(\mathrm{m\,s^{-2}}\).
  • Confusing gravitational constant \(G\) with acceleration due to gravity \(g\).
  • Thinking heavier objects fall faster in vacuum.
  • Forgetting that \(d=R\) only for objects near the Earth's surface.
  • Not cancelling the mass \(m\) during the derivation of \(g\).
🗒️ Quick Reviison
  • Free fall occurs under the influence of gravity alone.
  • \[ g\approx9.8\ \mathrm{m\,s^{-2}} \]
  • \[ g=\frac{GM}{d^2} \]
  • Near Earth's surface, \[ g=\frac{GM}{R^2} \]
  • All bodies fall with the same acceleration in vacuum.
  • Free fall is an example of uniformly accelerated motion.
🍏

Value of Acceleration Due to Gravity (\(g\))

📋
Table of Contents
9 sections
📖 Introduction
🔢 Important Formula
📐 Derivation
Given Data
Quantity Symbol Value
Universal Gravitational Constant \(G\) \[ 6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}} \]
Mass of the Earth \(M\) \[ 6\times10^{24}\ \mathrm{kg} \]
Radius of the Earth \(R\) \[ 6.4\times10^{6}\ \mathrm{m} \]
Formula
\[g=\frac{GM}{R^2}\]
Substituting the Values
\[ \small \begin{aligned} g &=6.67\times10^{-11} \times \frac{6\times10^{24}} {(6.4\times10^6)^2} \\ &= 6.67\times10^{-11} \times \frac{6\times10^{24}} {6.4\times6.4\times10^{12}} \\ &= \frac{6.67\times6\times10^{-11+24}} {40.96\times10^{12}} \\ &= \frac{40.02\times10^{13}} {40.96\times10^{12}} \\ &= \frac{40.02}{40.96}\times10 \\ &=0.977\times10 \\ &=9.77\ \mathrm{m\,s^{-2}} \\ &\approx9.8\ \mathrm{m\,s^{-2}} \end{aligned} \] Thus, every freely falling object near the Earth's surface accelerates downward at approximately \(9.8\ \mathrm{m\,s^{-2}}\).
Verification of SI Unit
From the formula, \[g=\frac{GM}{R^2}\] Unit of \(G\) is \[\mathrm{N\,m^2\,kg^{-2}}\] Since \[1\ \mathrm{N}=1\ \mathrm{kg\,m\,s^{-2}}\] Therefore, \[ \begin{aligned} g &= \frac{\mathrm{N\,m^2\,kg^{-2}}\times\mathrm{kg}} {\mathrm{m^2}} \\ &= \frac{\mathrm{kg\,m\,s^{-2}}\times\mathrm{m^2}} {\mathrm{kg}\times\mathrm{m^2}} \\ &= \mathrm{m\,s^{-2}} \end{aligned} \] Hence, the SI unit of acceleration due to gravity is \[ \boxed{\mathrm{m\,s^{-2}}} \]
Why is the Value Approximately 9.8?
The Earth is not a perfect sphere and its radius varies slightly from place to place. Moreover, factors such as altitude and rotation of the Earth also affect the value of \(g\). Therefore, the calculated value is approximately \[9.8\ \mathrm{m\,s^{-2}}\] near the Earth's surface.
💡 Concept Builder
✏️ Solved Concept Question
Two objects having masses 2 kg and 20 kg are dropped simultaneously from the same height in vacuum. Which one will reach the ground first?
  1. 1
    Recall the expression for \(g\).
  2. 2
    Check whether \(g\) depends upon the mass of the falling object.
Since \[g=\frac{GM}{R^2}\] the mass of the falling object does not appear in the formula. Therefore, both objects experience the same acceleration due to gravity and reach the ground simultaneously (ignoring air resistance).
📋 CBSE Competency-Based (HOTS) Question

A student claims that heavier objects always have a greater acceleration due to gravity because they possess more mass.

Questions

  1. Is the statement correct?
  2. Which formula justifies your answer?
  3. What actually determines the value of \(g\)?

Answers

  1. No.
  2. \[ g=\frac{GM}{R^2} \]
  3. The mass and radius of the Earth determine the value of \(g\), not the mass of the falling object.
⚡ Exam Tip
❌ Common Mistakes
  • Writing the unit of \(g\) as \(\mathrm{N}\).
  • Confusing \(G\) with \(g\).
  • Not squaring the Earth's radius while applying the formula.
  • Using the mass of the falling object in the final expression for \(g\).
  • Writing \(9.8\ \mathrm{m\,s^{-1}}\) instead of \(9.8\ \mathrm{m\,s^{-2}}\).
⚡ Quick Revision
  • \[ g=\frac{GM}{R^2} \]
  • \[ g\approx9.8\ \mathrm{m\,s^{-2}} \]
  • \(g\) depends on Earth's mass and radius.
  • \(g\) is independent of the mass of the falling object.
  • The SI unit of \(g\) is \(\mathrm{m\,s^{-2}}\).
🍏

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

📋
Table of Contents
11 sections
📖 Introduction
🗒️ Why Do All Objects Fall With The Same Acceleration?
According to Newton's Second Law, \[F=ma\] and according to the Universal Law of Gravitation, \[F=\frac{GMm}{R^2}\] Combining both equations, \[\begin{aligned} mg&=\frac{GMm}{R^2}\\ g&=\frac{GM}{R^2}\end{aligned} \] Since the mass of the falling object (\(m\)) is cancelled during derivation, every freely falling object experiences the same acceleration due to gravity, irrespective of its mass, shape or size (provided air resistance is neglected).
Conclusion: A feather and a hammer fall with the same acceleration in vacuum because both experience the same value of \(g\).
📌 Equations of Motion Under Gravity
🗂️ Types / Category
Different Cases of Motion Under Gravity
Case 1: Object Dropped from Rest
When an object is simply dropped,\[u=0\] Therefore, \[\boxed{v=gt}\] \[\boxed{s=\frac12gt^2}\] \[\boxed{v^2=2gs}\]
Case 2: Object Thrown Vertically Downward
Gravity and the initial velocity act in the same direction. Hence,\[v=u+gt\] The speed increases continuously because gravity accelerates the object.
Case 3: Object Thrown Vertically Upward
Initially the object moves upward, whereas gravity acts downward. Therefore, the object slows down while moving upward.

Taking upward direction as positive, \[ \boxed{v=u-gt} \] \[ \boxed{s=ut-\frac12gt^2} \] \[ \boxed{v^2=u^2-2gs} \]
📌
Note
Note: The sign of \(g\) depends upon the chosen direction. If downward is taken as positive, use \(+g\). If upward is taken as positive, use \(-g\).
🔎 Velocity-Time Behaviour
✏️ Example
Solved Examples
A stone is dropped from rest. Find its velocity after \(5\ \mathrm{s}\). Take \[ g=9.8\ \mathrm{m\,s^{-2}} \]
  1. 1
    Object is dropped, therefore \(u=0\).
  2. 2
    Use \[v=u+gt\][
\[ \begin{aligned} v &=0+9.8\times5\ &=49\ \mathrm{m\,s^{-1}} \end{aligned} \] Therefore, \[\boxed{v=49\ \mathrm{m\,s^{-1}}}\]
A ball is dropped from rest. How far will it fall in \(3\ \mathrm{s}\)?
\[ s=\frac12gt^2 \] \[ \begin{aligned} s &=\frac12\times9.8\times3^2\ &=4.9\times9\ &=44.1\ \mathrm{m} \end{aligned} \] Hence \[\boxed{s=44.1\ \mathrm{m}}\]
🛠️ Real-Life Applications
  • Calculating the height of buildings using falling objects.
  • Designing parachutes.
  • Rocket launching and landing.
  • Sports such as cricket, football and basketball.
  • Engineering safety calculations.
📋 CBSE Competency-Based (HOTS) Question

Two balls of masses \(2\ \mathrm{kg}\) and \(10\ \mathrm{kg}\) are dropped simultaneously from the same height in a vacuum.

Questions

  1. Which ball reaches the ground first?
  2. Which equation supports your answer?
  3. Why does mass not affect the acceleration?

Answers

  1. Both reach the ground simultaneously.
  2. \[ s=\frac12gt^2 \]
  3. Since \[ g=\frac{GM}{R^2}, \] acceleration due to gravity is independent of the mass of the falling object.
⚡ Exam Tip
❌ Common Mistakes
  • Using the wrong sign of \(g\).
  • Forgetting that \(u=0\) for a dropped object.
  • Using ordinary equations without considering the direction of motion.
  • Thinking heavier bodies fall faster in vacuum.
  • Writing \(g\) as \(\mathrm{m\,s^{-1}}\) instead of \(\mathrm{m\,s^{-2}}\).
⚡ Quick Revision
  • Free fall is uniformly accelerated motion.
  • \[ v=u+gt \]
  • \[ s=ut+\frac12gt^2 \]
  • \[ v^2=u^2+2gs \]
  • For upward motion, use negative sign for \(g\).
  • Acceleration due to gravity is independent of mass.
🍏

Example 1

📋
Table of Contents
4 sections
❓ Question
A car accidentally falls off a ledge and reaches the ground in \[ 0.5\ \mathrm{s} \] Assume \[ g=10\ \mathrm{m\,s^{-2}} \] for easy calculations. Calculate:
  1. Its speed just before striking the ground.
  2. Its average speed during the fall.
  3. The height of the ledge above the ground.
💡 Concept
Concept Used
🗺️ Roadmap
  1. Write the given data.
  2. \[v=u+gt.\]
  3. Calculate average speed.
  4. Find the height using either \[ s=ut+\frac12gt^2 \] or \[ v^2=u^2+2gh. \]
🧩 Solution
Given: Initial Velocity \[u=0\] Acceleration Due to Gravity\[g=10\ \mathrm{m\,s^{-2}}\] Time Taken\[t=0.5\ \mathrm{s}\]
Solution 1
Speed on Striking the Ground
  1. Using the first equation of motion,
    \[v=u+gt\]
  2. Substituting the values,
    \[ \begin{aligned} v &=0+10\times0.5\\ &=5\ \mathrm{m\,s^{-1}} \end{aligned} \]
  3. Answer (i)
    \[ \boxed{ v=5\ \mathrm{m\,s^{-1}} } \]
Solution 2
Average Speed During the Fall
  1. Since the motion is uniformly accelerated,
    \[ v_{\text{avg}} = \frac{u+v}{2} \]
  2. Substituting the values,
    \[ \begin{aligned} v_{\text{avg}} &=\frac{0+5}{2}\\ &=2.5\ \mathrm{m\,s^{-1}} \end{aligned} \]
  3. Answer (ii)
    \[ \boxed{ v_{\text{avg}}=2.5\ \mathrm{m\,s^{-1}} } \]
Solution 3
Height of the Ledge
  1. We may use either the second or the third equation of motion. Using
    \[s=ut+\frac12gt^2\]
  2. Since
    \[u=0,\] \[ \begin{aligned} h &=\frac12\times10\times(0.5)^2\\ &=5\times0.25\\ &=1.25\ \mathrm{m} \end{aligned} \]
  3. Alternatively,
    \[ v^2=u^2+2gh\] \[ \begin{aligned} 5^2 &=0+2\times10\times h\\ 25 &=20h\\ h &=\frac{25}{20}\\ &=1.25\ \mathrm{m} \end{aligned} \]
  4. Answer (iii)
    \[\boxed{h=1.25\ \mathrm{m}}\]
🍏

Example 2

📋
Table of Contents
10 sections
❓ Question
An object is thrown vertically upward and rises to a maximum height of \[ 10\ \mathrm{m}. \] Calculate:
  1. The initial velocity with which the object was thrown.
  2. The time taken to reach the highest point.
💡 Concept
Concept used
🗺️ Roadmap
  1. Choose upward direction as positive.
  2. Write the given quantities.
  3. Find the initial velocity using \[ v^2=u^2+2gh. \]
  4. Use \[ v=u+gt \] to calculate the time of ascent.
🗒️ Given
Maximum Height \[ h=10\ \mathrm{m} \]
Acceleration Due to Gravity \[ g=-9.8\ \mathrm{m\,s^{-2}} \]
Velocity at Highest Point \[ v=0 \]
📌 Sign Convention
🧩 Solution
Solution 1
Initial Velocity
  1. Using the third equation of motion,
    \[v^2=u^2+2gh\]
  2. Substituting the values,
    \[ \begin{aligned} 0 &=u^2+2(-9.8)(10) \ u^2 &=196 \ u &=\sqrt{196} \ &=14\ \mathrm{m\,s^{-1}} \end{aligned} \]
  3. Answer (i)
    \[ \boxed{ u=14\ \mathrm{m\,s^{-1}} } \]
Solution 2
Time Taken to Reach the Highest Point
  1. Using the first equation of motion,
    \[v=u+gt\]
  2. Substituting the values,
    \[ \begin{aligned} 0 &=14+(-9.8)t \\ 9.8t &=14 \\ t &=\frac{14}{9.8} \\ &=1.43\ \mathrm{s} \end{aligned} \]
  3. Answer (ii)
    \[ \boxed{ t\approx1.43\ \mathrm{s} } \]
💡 Concept Builder
⚡ Exam Tip
❌ Common Mistakes
  • Taking \[ g=+9.8\ \mathrm{m\,s^{-2}} \] while choosing upward as positive.
  • Assuming acceleration becomes zero at the highest point.
  • Using \[ u=0 \] instead of \[ v=0 \] at the highest point.
  • Ignoring the negative sign of gravity.
  • Forgetting SI units.
⚡ Quick Revision
  • At the highest point, \[ v=0. \]
  • Acceleration due to gravity always acts downward.
  • Velocity decreases uniformly during upward motion.
  • Use \[ v^2=u^2+2gh \] to calculate initial velocity.
  • Use \[ v=u+gt \] to calculate time.
🍏

Mass

📋
Table of Contents
13 sections
📘 Definition
💡 Concept
Concept of Mass
🔷 Characteristics of Mass
🔷 Characteristics
  • Mass is a scalar quantity; it has magnitude only.
  • Mass never becomes negative.
  • Mass remains constant everywhere in the universe.
  • Mass does not depend upon gravity.
  • Mass does not change with altitude or location.
  • Mass is an intrinsic property of matter.
  • SI Unit of Mass is: Kilogram \(\mathrm{kg}\)
📌 Measurement of Mass
🔗 Relations
Mass and Inertia
Inertia is the property by which an object resists any change in its state of rest or uniform motion. Since mass measures inertia,
  • A truck has greater inertia than a bicycle.
  • A cricket ball has greater inertia than a tennis ball.
  • A loaded suitcase is harder to move than an empty suitcase.
⚖️ Difference Between Mass and Matter
Matter Mass
Anything that occupies space and has mass. Amount of matter present in an object.
It is a substance. It is a measurable quantity.
Cannot be measured directly. Measured using a balance.
⚖️ Mass vs Weight
Mass Weight
Amount of matter. Gravitational force acting on a body.
Constant everywhere. Changes from place to place.
Scalar quantity. Vector quantity.
Measured in kilogram. Measured in newton.
Measured by beam balance. Measured by spring balance.
✏️ Real-Life Examples
  • A 5 kg bag of rice has the same mass on Earth and on the Moon.
  • A gold bar retains its mass even when taken into space.
  • An astronaut's mass remains constant although the astronaut experiences weightlessness.
✏️ Solved Example
A person has a mass of \[ 60\ \mathrm{kg}. \] What will be the person's mass on the Moon?
  1. 1
    Recall whether mass depends on gravity.
Mass is independent of gravity and location. Therefore, \[ \boxed{ \text{Mass on the Moon}=60\ \mathrm{kg} } \]
📋 CBSE Competency-Based (HOTS) Question

An astronaut carrying scientific equipment travels from Earth to the Moon. On reaching the Moon, he observes that his weight has decreased considerably.

Questions

  1. Does his mass change?
  2. Why or why not?
  3. Which physical quantity remains constant everywhere?

Answers

  1. No.
  2. Mass depends on the amount of matter, which does not change with location.
  3. Mass.
⚡ Exam Tip
❌ Common Mistakes
  • Confusing mass with weight.
  • Writing the SI unit of mass as newton.
  • Thinking mass decreases on the Moon.
  • Using a spring balance to measure mass.
  • Ignoring that mass is a scalar quantity.
⚡ Quick Revision
  • Mass is the amount of matter present in a body.
  • SI unit of mass is kilogram (\(\mathrm{kg}\)).
  • Mass is a scalar quantity.
  • Mass is independent of gravity.
  • Mass remains constant everywhere.
  • Mass is a measure of inertia.
🍏

Weight

📋
Table of Contents
13 sections
📘 Definition
💡 Concept of Weight
🔢 Formula for Weight
📌 Measurement of Weight
🔷 Characteristics of Weight
🔷 Characteristics
  • Weight is a force.
  • Weight is a vector quantity
  • Its direction is always towards the centre of the Earth.
  • Weight depends upon the value of \(g\).
  • Weight changes from one planet to another.
  • Weight becomes smaller where gravity is weaker.
⚖️ Variation of Weight
Since the value of \(g\) changes with location, the weight of an object also changes.
Location Weight Mass
Earth Maximum Constant
Moon About one-sixth of Earth's Constant
High Mountains Slightly less Constant
Space Station Very small (apparent weightlessness) Constant
Remember: Mass remains constant everywhere, but weight changes because the value of \[ g \] changes from place to place.
Weight on the Moon
The acceleration due to gravity on the Moon is approximately \(\frac{1}{6}\) of that on the Earth.

Therefore, \[ \boxed{ W_{\text{Moon}} = \frac{1}{6} W_{\text{Earth}} } \] Thus, a person weighing \[ 600\ \mathrm{N} \] on Earth will weigh only \[\frac{600}{6}=100\ \mathrm{N}\] on the Moon.
⚖️ Difference Between Mass and Weight
Mass Weight
Amount of matter. Force of gravity acting on a body.
Scalar quantity. Vector quantity.
Constant everywhere. Changes with gravity.
SI unit is kilogram. SI unit is newton.
Measured using beam balance. Measured using spring balance.
✏️ Example
Solved Example
Calculate the weight of a body having a mass of \[ 50\ \mathrm{kg} \] on the Earth. Take \[ g=9.8\ \mathrm{m\,s^{-2}}. \]
  1. 1
    Write the formula \[ W=mg. \]
  2. 2
    Substitute the values.
\[ \begin{aligned} W &=mg\ &=50\times9.8\ &=490\ \mathrm{N} \end{aligned} \] Therefore, \[ \boxed{ W=490\ \mathrm{N} } \]
✏️ Real-Life Examples
  • An astronaut weighs much less on the Moon than on the Earth.
  • Your school bag has the same mass everywhere but different weight on different planets.
  • A spring balance gives different readings on the Earth and the Moon.
📋 CBSE Competency-Based (HOTS) Question

A student with a mass of \[ 48\ \mathrm{kg} \] travels to the Moon. She notices that the reading on the spring balance becomes much smaller.

Questions

  1. Has her mass changed?
  2. Why has the spring balance reading decreased?
  3. Which quantity remains constant everywhere?

Answers

  1. No.
  2. The Moon's gravitational acceleration is much smaller than that of the Earth.
  3. Mass.
⚡ Exam Tip
❌ Common Mistakes
  • Writing the SI unit of weight as kilogram.
  • Confusing mass with weight.
  • Using beam balance to measure weight.
  • Thinking weight remains constant everywhere.
  • Writing \[ W=\frac{m}{g} \] instead of \[ W=mg. \]
⚡ Quick Revision
  • Weight is the gravitational force acting on a body.
  • \[ W=mg \]
  • SI unit of weight is newton (\(\mathrm{N}\)).
  • Weight is a vector quantity.
  • Weight changes from place to place.
  • Weight is measured using a spring balance.
🍏

Weight of an Object on the Moon

📋
Table of Contents
9 sections
📖 Introduction
📐 Weight of an Object on the Moon
Let
  • Mass of the object = \(m\)
  • Mass of the Moon = \(M_m\)
  • Radius of the Moon = \(R_m\)
  • Weight of the object on the Moon = \(W_m\)
According to the Universal Law of Gravitation, \[ \boxed{ W_m = G\frac{M_m m}{R_m^2} } \tag{1} \]
Weight of the Same Object on the Earth
Let
  • Mass of the Earth = \(M_e\)
  • Radius of the Earth = \(R_e\)
  • Weight of the object on the Earth = \(W_e\)
Then, \[ \boxed{ W_e = G\frac{M_e m}{R_e^2} } \tag{2} \]
Data Used
Celestial Body Mass (kg) Radius (m)
Earth \[ 5.98\times10^{24} \] \[ 6.37\times10^6 \]
Moon \[ 7.36\times10^{22} \] \[ 1.74\times10^6 \]
Dividing Equation (1) by Equation (2), \[ \frac{W_m}{W_e} = \frac{ G\frac{M_m m}{R_m^2} } { G\frac{M_e m}{R_e^2} } \] Cancelling the common quantities \(G\) and \(m\), \[ \frac{W_m}{W_e} = \frac{M_mR_e^2} {M_eR_m^2} \] Substituting the values, \[ \small \begin{aligned} \frac{W_m}{W_e} &= \frac{ 7.36\times10^{22}\times(6.37\times10^6)^2 } { 5.98\times10^{24}\times(1.74\times10^6)^2 } \\ &= \frac{ 7.36\times6.37^2 } { 5.98\times1.74^2 } \times10^{-2} \\ &= \frac{ 7.36\times40.58 } { 5.98\times3.03 } \times10^{-2} \\ &= \frac{298.7}{18.12} \times10^{-2} \\ &= 16.48\times10^{-2} \\ &= 0.165 \\ &\approx0.167 \\ &\approx\frac16 \end{aligned} \]
Final Relation
\[ \boxed{ \frac{W_m}{W_e} \approx \frac{1}{6} } \]
⚖️ >Comparison Between Earth and Moon
Property Earth Moon
Acceleration due to gravity \[ 9.8\ \mathrm{m\,s^{-2}} \] \[ 1.63\ \mathrm{m\,s^{-2}} \]
Weight of Object \(W\) \(\frac16W\)
Mass of Object Same Same
✏️ Example
Solved Example
A person weighs \[ 720\ \mathrm{N} \] on the Earth. Find the person's weight on the Moon.
Use the relation \[ W_m=\frac16W_e. \]
\[ \begin{aligned} W_m &=\frac16\times720 \\ &=120\ \mathrm{N} \end{aligned} \] Therefore, \[ \boxed{ W_m=120\ \mathrm{N} } \]
🛠️ Real-Life Applications
  • Astronauts can jump much higher on the Moon because their weight is much smaller.
  • Heavy objects become easier to lift on the Moon.
  • Space agencies use this principle while designing lunar missions.
📋 CBSE Competency-Based (HOTS) Question

During a lunar mission, an astronaut carries a toolbox weighing \[ 180\ \mathrm{N} \] on the Earth.

Questions

  1. What will be its weight on the Moon?
  2. Will its mass change?
  3. Why can astronauts jump higher on the Moon?

Answers

  1. \[ 30\ \mathrm{N} \]
  2. No.
  3. The Moon's gravitational pull is only about one-sixth of the Earth's.
⚡ Exam Tip
❌ Common Mistakes
  • Writing mass on the Moon as one-sixth of the Earth's mass.
  • Using kilogram instead of newton for weight.
  • Forgetting to cancel the common terms \(G\) and \(m\) during derivation.
  • Writing \[ W_m=6W_e. \]
  • Confusing acceleration due to gravity with gravitational constant.
⚡ Quick Revision
  • \[ W=mg \]
  • \[ W_m=\frac16W_e \]
  • Mass remains constant everywhere.
  • Weight depends on gravity.
  • Gravity on the Moon is approximately one-sixth of that on the Earth.
🍏

Example 3

📋
Table of Contents
12 sections
❓ Question
The mass of an object is \[ 10\ \mathrm{kg}. \] Calculate its weight on the Earth.
💡 Concept
🗺️ Roadmap
  1. Write the given values.
  2. Use the formula \[ W=mg. \]
  3. Substitute the values.
  4. Write the answer with the correct SI unit.
🗒️ Given
Mass of the Object \[ m=10\ \mathrm{kg} \]
Acceleration Due to Gravity \[ g=9.8\ \mathrm{m\,s^{-2}} \]
🔢 Formula
🧩 Solution
\[ \begin{aligned} W &=mg \\ &=10\times9.8 \\ &=98\ \mathrm{N} \end{aligned} \]
Verification of Unit
\[ \begin{aligned} W &=\mathrm{kg}\times\mathrm{m\,s^{-2}} \\ &=\mathrm{N} \end{aligned} \] Therefore, the SI unit of weight is \[ \boxed{\mathrm{Newton\;(N)}}. \]
🔍 Interpretation
The result \[ 98\ \mathrm{N} \] means that the Earth pulls the object downward with a gravitational force of 98 newtons.
Remember: The object's mass is 10 kg, whereas its weight is 98 N. These two quantities are different and must not be confused.
If the Same Object is Taken to the Moon
Since the Moon's gravity is approximately one-sixth that of the Earth, \[ W_{\text{Moon}} = \frac16\times98 \] \[ \boxed{ W_{\text{Moon}} \approx16.3\ \mathrm{N} } \] However, the mass of the object remains\[\boxed{10\ \mathrm{kg}}\]
✏️ Solved Concept Question
A body has a mass of \[ 20\ \mathrm{kg}. \] Calculate its weight on the Earth. Take \[ g=9.8\ \mathrm{m\,s^{-2}}. \]
\[ \begin{aligned} W &=20\times9.8 \\ &=196\ \mathrm{N} \end{aligned} \] Therefore, \[ \boxed{ W=196\ \mathrm{N} } \]
📋 CBSE Competency-Based (HOTS) Question

Ravi says that a school bag of mass \[ 8\ \mathrm{kg} \] has a weight of \[ 8\ \mathrm{N}. \] His friend disagrees.

Questions

  1. Who is correct?
  2. Find the actual weight of the bag.
  3. Why is kilogram not the unit of weight?

Answers

  1. His friend is correct.
  2. \[ W=8\times9.8=78.4\ \mathrm{N} \]
  3. Weight is a force, so its SI unit is newton.
⚡ Exam Tip
❌ Common Mistakes
  • Writing the answer in kilogram instead of newton.
  • Confusing mass with weight.
  • Using \[ W=\frac{m}{g} \] instead of \[ W=mg. \]
  • Forgetting to write the unit in the final answer.
⚡ Quick Revision
  • \[ W=mg \]
  • Weight is a force.
  • SI unit of weight is newton.
  • Mass remains constant everywhere.
  • Weight changes with gravity.
🍏

Example 4

📋
Table of Contents
9 sections
❓ Question
An object weighs \[ 10\ \mathrm{N} \] on the surface of the Earth. Calculate its weight on the surface of the Moon.
💡 Concept
🗺️ Roadmap
  1. Write the given weight on the Earth.
  2. Use the relation \[ W_m=\frac16W_e. \]
  3. Substitute the given value.
  4. Write the answer with the correct SI unit.
🗒️ Given
Weight on the Earth \[ W_e=10\ \mathrm{N} \]
🔢 Formula
🧩 Solution
\[ \begin{aligned} W_m &=\frac16\times10 \\ &=\frac{10}{6} \\ &=1.67\ \mathrm{N} \end{aligned} \]
✅ Final Answer
\[ \boxed{ W_m=1.67\ \mathrm{N} } \]
🗒️ Verification
Since the Moon's gravity is approximately one-sixth that of the Earth, the calculated weight should be much smaller than the Earth's weight. \[ 1.67\ \mathrm{N} < 10\ \mathrm{N} \] Hence, the answer is reasonable.
🔍 Interpretation
The object still contains the same amount of matter, but the Moon exerts a much weaker gravitational pull. Therefore:
  • Mass remains unchanged.
  • Weight decreases to approximately one-sixth.
Remember: If an object weighs \[ 10\ \mathrm{N} \] on the Earth, it does not mean that its mass is \[ 10\ \mathrm{kg}. \] Weight and mass are different physical quantities.
🍏

Thrust

📋
Table of Contents
11 sections
📘 Definition
💡 Concept
🔷 Characteristics of Thrust
🔷 Characteristics
  • Thrust is a force.
  • It always acts perpendicular to the surface.
  • Its SI unit is newton (\(\mathrm{N}\)).
  • Thrust may increase or decrease depending on the applied force.
  • Greater thrust generally produces greater pressure if the area remains unchanged.
⚖️ Difference Between Force and Thrust
Force Thrust
May act in any direction. Always acts perpendicular to a surface.
General push or pull. Normal component of force.
Measured in newton. Measured in newton.
✏️ Real-Life Examples
  • Pressing a nail into a wooden plank.
  • Standing on the floor.
  • A person sitting on a chair exerts thrust on the seat.
  • The tyres of a car exert thrust on the road.
  • A building exerts thrust on its foundation.
  • The bottom of a water tank exerts thrust on the ground.
🛠️ Applications of Thrust
  • Construction of buildings and bridges.
  • Design of dams.
  • Hydraulic machines.
  • Rocket propulsion.
  • Aircraft take-off and landing.
  • Heavy machinery and cranes.
✏️ Solved Example
A box exerts a force of \[ 200\ \mathrm{N} \] normally on the floor. Find the thrust acting on the floor.
  1. 1
    Recall the definition of thrust.
  2. 2
    Identify the force acting perpendicular to the surface.
Since the entire force acts perpendicular to the floor, \[ \boxed{ \text{Thrust}=200\ \mathrm{N} } \]
A person pushes a wall with a force of \[ 150\ \mathrm{N} \] perpendicular to the wall. What is the thrust exerted on the wall?
Since the applied force is perpendicular to the wall, \[ \boxed{ \text{Thrust}=150\ \mathrm{N} } \]
📋 CBSE Competency-Based (HOTS) Question

A student places a heavy box on the floor. Another student says that the box exerts only force but not thrust on the floor.

Questions

  1. Is the statement correct?
  2. What is thrust?
  3. Which physical quantity is produced due to thrust acting on a surface?

Answers

  1. No.
  2. Thrust is the force acting perpendicular to a surface.
  3. Pressure.
⚡ Exam Tip
❌ Common Mistakes
  • Confusing thrust with pressure.
  • Writing pascal as the SI unit of thrust.
  • Thinking thrust acts parallel to the surface.
  • Ignoring that only the perpendicular component of force is thrust.
🗒️ Quick Reference
  • Thrust is the force acting perpendicular to a surface.
  • SI unit of thrust is newton (\(\mathrm{N}\)).
  • \[ P=\frac{\text{Thrust}}{\text{Area}} \]
  • Greater thrust generally produces greater pressure if area is constant.
  • Thrust is responsible for producing pressure.
🍏

Pressure

📋
Table of Contents
11 sections
📘 Definition
💡 Concept
🔢 Formula for Pressure
🔷 Characteristics of Pressure
🔷 Characteristics
  • Pressure is a scalar quantity.
  • Pressure depends upon both force and area.
  • Pressure increases when the contact area decreases.
  • Pressure decreases when the contact area increases.
  • The SI unit of pressure is pascal.
📌 Effect of Area on Pressure
✏️ Pressure in Everyday Life
Situation Reason
Sharp knife cuts easily. Small contact area produces large pressure.
Nails have pointed ends. Large pressure helps them penetrate wood.
School bags have broad straps. Pressure on shoulders decreases.
Camels have broad feet. Pressure on sand is reduced.
Tractors use broad tyres. Pressure on soft soil decreases.
Snow shoes are wide. Pressure on snow decreases.
🎨 SVG Diagram
Same Mass (F) Force Spread Out Large Contact Area LOW PRESSURE Same Mass (F) Force Concentrated Small Contact Area HIGH PRESSURE
✏️ Example
Solved Example
A force of \[ 200\ \mathrm{N} \] acts normally on a surface of area \[ 4\ \mathrm{m^2}. \] Calculate the pressure produced.
  1. 1
    Write the formula.
  2. 2
    Substitute the given values.
  3. 3
    Write the answer in pascal.
\[ \begin{aligned} P &=\frac{F}{A} \\ &=\frac{200}{4} \\ &=50\ \mathrm{Pa} \end{aligned} \] Therefore, \[ \boxed{ P=50\ \mathrm{Pa} } \]
Why do sharp needles pierce cloth more easily than blunt needles?
Sharp needles have a much smaller contact area. For the same applied force, they produce greater pressure, allowing them to pierce the cloth easily.
📋 CBSE Competency-Based (HOTS) Question

During a trek in a snowy region, one student wears ordinary shoes while another wears wide snow shoes.

Questions

  1. Who will sink less into the snow?
  2. Which concept explains this?
  3. How does contact area affect pressure?

Answers

  1. The student wearing snow shoes.
  2. Pressure.
  3. Increasing the contact area decreases pressure.
⚡ Exam Tip
❌ Common Mistakes
  • Writing newton as the SI unit of pressure.
  • Confusing pressure with thrust.
  • Thinking larger area produces greater pressure.
  • Using area in \(\mathrm{cm^2}\) without converting to \(\mathrm{m^2}\) in numerical problems.
⚡ Quick Revision
  • \[ P=\frac{F}{A} \]
  • SI unit of pressure is pascal.
  • \[ 1\ \mathrm{Pa}=1\ \mathrm{N\,m^{-2}} \]
  • Greater force produces greater pressure.
  • Smaller contact area produces greater pressure.
  • Pressure explains the working of sharp tools, snow shoes and broad tyres.
🍏

Relation Between Thrust and Pressure

📋
Table of Contents
9 sections
📖 Introduction
🔗 Relations
Pressure is defined as the thrust acting normally on a unit area. \[ \boxed{ P=\frac{\text{Thrust}}{\text{Area}} } \] Using symbols, \[ \boxed{ P=\frac{F}{A} } \] where
Symbol Physical Quantity SI Unit
\(P\) Pressure Pascal (\(\mathrm{Pa}\))
\(F\) Thrust (Force) Newton (\(\mathrm{N}\))
\(A\) Area of contact \(\mathrm{m^2}\)
Rearranged Formulae
Depending on the unknown quantity, the formula can be rearranged as: \[ \boxed{ F=P\times A } \] \[ \boxed{ A=\frac{F}{P} } \] These three equations are frequently used in numerical problems.
📌 Effect of Thrust and Area on Pressure
✏️ Example
Solved Example
A thrust of \[ 500\ \mathrm{N} \] acts normally on an area of \[ 5\ \mathrm{m^2}. \] Calculate the pressure produced.
  1. 1
    Use \[ P=\frac{F}{A}. \]
  2. 2
    Substitute the values.
\[ \begin{aligned} P &=\frac{500}{5} \ &=100\ \mathrm{Pa} \end{aligned} \] Therefore, \[ \boxed{ P=100\ \mathrm{Pa} } \]
🛠️ Real-Life Applications
  • Sharp knives have a very small contact area, producing greater pressure.
  • Broad tyres of tractors reduce pressure on soft soil.
  • Snow shoes prevent sinking into snow by increasing the contact area.
  • Building foundations are made wide to reduce pressure on the ground.
  • School bags have broad straps to reduce pressure on the shoulders.
📋 CBSE Competency-Based (HOTS) Question

Two bricks of identical mass are placed on the ground. One brick rests on its largest face while the other rests on its smallest face.

Questions

  1. Which brick exerts greater pressure on the ground?
  2. Why?
  3. Does the thrust exerted by both bricks differ?

Answers

  1. The brick resting on its smallest face.
  2. It has a smaller contact area, so pressure is greater.
  3. No. Both exert the same thrust because their weights are equal.
⚡ Exam Tip
❌ Common Mistakes
  • Confusing thrust with pressure.
  • Using kilogram instead of newton for thrust.
  • Writing the SI unit of pressure as newton instead of pascal.
  • Forgetting to convert area into square metres.
  • Thinking greater area always produces greater pressure.
⚡ Quick Revision
  • \[ P=\frac{F}{A} \]
  • \[ F=PA \]
  • \[ A=\frac{F}{P} \]
  • Pressure increases with thrust.
  • Pressure decreases with increase in area.
  • SI unit of pressure is pascal (\(\mathrm{Pa}\)).
🍏

Pressure in Fluids

📋
Table of Contents
13 sections
📘 Definition
💡 Concept of Pressure in Fluids
🔢 Formula for Pressure in a Fluid
📌 Factors Affecting Pressure in Fluids
🔷 Characteristics of Pressure in Fluids
🔷 Characteristics
  • Fluids exert pressure in all directions.
  • Pressure always acts perpendicular to the surface.
  • Pressure increases with depth.
  • Pressure at the same depth is the same in all directions.
  • Pressure depends upon the density of the fluid.
  • Pressure is independent of the shape of the container.
📌 Pressure at Different Depths
🗒️ Pressure at different level
Pressure at different level
FLUID PRESSURE & OUTFLOW TRAJECTORY Hydrostatic Pressure: P = ρg·h SURFACE (h = 0) Ground Level h₁ (Shallow) h₂ (Mid) h₃ (Deep) v₁ (Low) v₂ (Med) v₃ (High) Range R₁ Range R₂ Longest Range R₃ PHYSICAL DYNAMICS 1. PRESSURE & VELOCITY As depth (h) increases, hydrostatic pressure and exit velocity (v) increase. v₃ > v₂ > v₁ 2. HORIZONTAL RANGE (R) Under high velocity, the stream from the deepest point achieves the furthest horizontal displacement. R₃ > R₂ > R₁ Velocity v ∝ √h
✏️ Real-Life Examples
  • Divers experience greater pressure as they dive deeper into water.
  • Dams are constructed thicker at the bottom because water pressure is greatest there.
  • Submarines are designed to withstand very high pressure at great depths.
  • Water from lower holes of a bottle comes out with greater force.
  • Swimming pools exert greater pressure at their deeper ends.
✏️ Example
Solved Example
Two holes are made in the side of a water bottle, one near the top and the other near the bottom. Which hole will eject water farther?
Recall how pressure changes with depth.
The lower hole is at a greater depth and therefore experiences greater pressure. Hence, water comes out with greater speed and travels farther.
📋 CBSE Competency-Based (HOTS) Question

Engineers always construct dams with a much thicker base than the top.

Questions

  1. Why is the lower portion made thicker?
  2. Which physical quantity is responsible for this design?
  3. How does pressure vary with depth?

Answers

  1. The bottom experiences greater water pressure.
  2. Fluid pressure.
  3. Pressure increases with depth.
⚡ Exam Tip
❌ Common Mistakes
  • Thinking pressure depends upon the shape of the container.
  • Confusing depth with volume of liquid.
  • Writing pressure decreases with depth.
  • Using kilogram instead of pascal as the unit of pressure.
  • Confusing pressure in fluids with atmospheric pressure.
⚡ Quick Revision
  • Both liquids and gases are fluids.
  • \[ P=\rho gh \]
  • Pressure increases with depth.
  • Pressure acts equally in all directions at the same depth.
  • Dams are thicker at the bottom because water pressure is greatest there.
  • SI unit of pressure is pascal (\(\mathrm{Pa}\)).
🍏

Buoyancy (Upthrust)

📋
Table of Contents
13 sections
🗒️ 
📘 Definition
💡 Concept
🤔 Did You Know?
Why Does Buoyancy Occur?
  • Pressure in a fluid increases with depth.
  • The bottom of an immersed object experiences greater pressure than its top.
  • This pressure difference creates an upward force.
  • This upward force is known as buoyancy or upthrust.
🔎 Factors Affecting Buoyant Force
🔷 Characteristics of Buoyancy
🔷 Characteristics
  • Buoyant force acts vertically upward.
  • It acts on every object immersed in a fluid.
  • It exists in both liquids and gases.
  • It opposes the weight of the object.
  • Greater density of the fluid produces greater buoyant force.
  • Larger immersed volume produces greater buoyant force.
🗒️ Floating And Sinking
Condition Result
Buoyant Force = Weight Object floats.
Buoyant Force < Weight Object sinks.
Buoyant Force > Weight Object rises until equilibrium is reached.
🛠️ Everyday Applications of Buoyancy
Application Role of Buoyancy
Ships Float because buoyant force balances their weight.
Submarines Control buoyancy to sink or float.
Life Jackets Increase buoyancy and keep a person afloat.
Hot Air Balloons Rise because warm air is less dense than surrounding air.
Fishing Floats Remain floating due to buoyancy.
✏️ Real-Life Examples
  • A wooden block floats on water.
  • An iron nail sinks in water.
  • A large steel ship floats although it is made of iron.
  • A swimmer feels lighter while swimming.
  • Oil floats on water because its density is lower.
✏️ Example
Solved Examples
Why does a person feel lighter while standing in a swimming pool?
  1. 1
    Identify the force exerted by water.
  2. 2
    Explain its effect on apparent weight.
Water exerts an upward buoyant force on the person's body. This upward force opposes the person's weight, reducing the apparent weight. Therefore, the person feels lighter while standing in water.
Why does a large steel ship float while a small iron nail sinks in water?
A ship has a hollow structure and displaces a very large volume of water, producing a large buoyant force equal to its weight. An iron nail displaces only a very small volume of water, so the buoyant force is less than its weight, causing it to sink.
📋 CBSE Competency-Based (HOTS) Question

Two objects of equal mass are placed in water. One floats while the other sinks.

Questions

  1. Which object experiences sufficient buoyant force to remain afloat?
  2. Why does the other object sink?
  3. Name the upward force exerted by water.

Answers

  1. The floating object.
  2. Its weight is greater than the buoyant force acting on it.
  3. Buoyant force (Upthrust).
⚡ Exam Tip
❌ Common Mistakes
  • Confusing buoyancy with pressure.
  • Thinking buoyancy acts downward.
  • Believing only liquids produce buoyancy. Gases also exert buoyant force.
  • Assuming all iron objects sink. Ships made of steel can float due to their shape.
  • Confusing actual weight with apparent weight.
⚡ Quick Revision
  • Buoyancy is the upward force exerted by a fluid.
  • It is also called upthrust.
  • It acts opposite to the weight of an object.
  • It depends upon the density of the fluid and the volume displaced.
  • Floating and sinking depend on the comparison between weight and buoyant force.
  • Life jackets, ships, submarines and hot-air balloons work on the principle of buoyancy.
🍏

Archimedes' Principle

📋
Table of Contents
12 sections
🖼️ Figure
Archimedes - the Greek mathematician
Archimedes (287 BC – 212 BC), the Greek mathematician and scientist who discovered the Principle of Buoyancy.
📘 Definition
💡 Concept
📌 Mathematical Expression
🗒️ How Archimedes' Principle Works
How Archimedes' Principle Works
  1. An object is immersed in a fluid.
  2. The object displaces some quantity of the fluid.
  3. The displaced fluid has a certain weight.
  4. The fluid exerts an upward buoyant force equal to this weight.
  5. The object floats or sinks depending on the comparison between its weight and the buoyant force.
🔎 Conditions for Floating and Sinking
🛠️ Applications of Archimedes' Principle
Application Explanation
Ships Ships float because they displace a large volume of water.
Submarines Control buoyancy by changing the amount of water in ballast tanks.
Hydrometers Measure the density of liquids using buoyancy.
Lactometers Measure the purity of milk.
Hot-air Balloons Rise due to buoyant force exerted by air.
Life Jackets Increase buoyancy and help people float.
✏️ Everyday Examples
  • A swimmer feels lighter while swimming.
  • Large ships float on water.
  • Ice floats because it is less dense than water.
  • Oil floats above water.
  • A cork floats whereas a stone sinks.
✏️ Example
Solved Example
Why does a swimmer feel lighter while swimming in a pool?
  1. 1
    Recall Archimedes' Principle.
  2. 2
    Identify the upward force acting on the swimmer.
Water exerts an upward buoyant force on the swimmer. According to Archimedes' Principle, this buoyant force equals the weight of the displaced water. The upward force reduces the swimmer's apparent weight, making the swimmer feel lighter.
Why can a steel ship float while a small iron nail sinks in water?
A ship is hollow and displaces a large volume of water, producing sufficient buoyant force to balance its weight. An iron nail displaces only a small amount of water, so the buoyant force is smaller than its weight, causing it to sink.
📋 CBSE Competency-Based (HOTS) Question

During a science exhibition, two identical sealed bottles are placed in water. One bottle is filled with sand while the other is filled with air.

Questions

  1. Which bottle is more likely to float?
  2. Which scientific principle explains this behaviour?
  3. What determines the buoyant force acting on each bottle?

Answers

  1. The bottle filled with air.
  2. Archimedes' Principle.
  3. The weight (or volume) of the fluid displaced.
⚡ Exam Tip
❌ Common Mistakes
  • Writing buoyant force equal to the weight of the object instead of the displaced fluid.
  • Confusing buoyant force with pressure.
  • Assuming only liquids obey Archimedes' Principle. It is valid for gases as well.
  • Ignoring the difference between actual weight and apparent weight.
⚡ Quick Revision
  • Archimedes' Principle explains buoyancy.
  • Buoyant force equals the weight of the displaced fluid.
  • Buoyant force acts vertically upward.
  • It explains floating and sinking.
  • Ships, submarines, hydrometers and balloons work on this principle.
  • Apparent loss of weight occurs because of buoyant force.
🍏

Important Points for Revision

📋
Table of Contents
2 sections
🌟 Important Points for Revision
⚡ Last-Minute Board Revision Tips
  • Memorise all important formulae with SI units.
  • Do not confuse mass with weight.
  • Remember that pressure depends on both force and area.
  • Pressure in liquids increases with depth.
  • Buoyant force always acts upward.
  • Archimedes' Principle states that buoyant force equals the weight of the displaced fluid.
  • Practice derivations of \(g\), weight on the Moon and numerical problems involving pressure.
  • Most competency-based questions are based on real-life applications such as dams, ships, submarines, balloons, snow shoes and hydraulic systems.
· Updated
⚖

NCERT · Class IX · Science · Chapter 9

Gravitation

The unseen thread that binds apples, oceans, moons and planets — explored through force, free fall, and fluids.

Two Masses, One Invisible Pull

m₁ m₂ F (attractive, mutual) distance, d F = G · m₁m₂ / d²

Every particle in the universe attracts every other particle — the force is mutual, always attractive, and acts along the line joining the two centres.

The Idea, In Plain Words

Newton's breakthrough was realising that the same force that pulls an apple to the ground also keeps the Moon orbiting Earth and Earth orbiting the Sun. He called this universal gravitation.

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 (F ∝ m₁m₂)
  • Inversely proportional to the square of the distance between their centres (F ∝ 1/d²)

Combining both gives F = G·m₁m₂/d², where G is the universal gravitational constant.

Value of G: 6.674 × 10⁻¹¹ N·m²/kg² — first measured experimentally by Henry Cavendish, independent of the medium, location, or the nature of the bodies.

Because G is so tiny, gravitational force between everyday objects (you and your chair) is immeasurably small — it only becomes significant when at least one mass is planet-sized.

This single law also explains Kepler's laws of planetary motion: planets move in elliptical orbits because the Sun's gravity continuously bends their straight-line motion into a curve.

Falling Freely Under Gravity

v increases uniformly (g) g = 9.8 m/s² (Earth, average) Independent of the object's own mass v = u + gt s = ut + ½gt² v² = u² + 2gs (g is + for fall, − for rise)

In free fall, only gravity acts — air resistance ignored. All objects, light or heavy, gain speed at the same rate, g.

From Force To Acceleration

Apply the universal law with one of the masses being a planet of mass M and radius R, and the second mass being any small object m on its surface:

F = GMm/R² = mg  ⟹  g = GM/R²

This is the acceleration due to gravity — the acceleration produced in a freely falling body purely because of the planet's pull. Notice m cancels out: g does not depend on the falling object's own mass, only on the planet.

  • g varies with location: slightly higher at the poles, lower at the equator (Earth is not a perfect sphere — R is smaller at poles).
  • g decreases with height and depth from Earth's surface.
  • g on the Moon ≈ 1/6th of g on Earth, because the Moon's mass and radius are both much smaller.
Mass vs Weight — don't confuse them: Mass (kg) is the quantity of matter, constant everywhere. Weight (N) = mg, the force of gravity on that mass — it changes from planet to planet, even though mass does not.

A 60 kg person always has 60 kg of mass, but weighs about 588 N on Earth and only about 98 N on the Moon.

Floating, Sinking & Buoyant Force

fluid surface object Buoyant force (F_B) Weight (mg) displaces fluid equal to its volume

If F_B equals weight → floats; if F_B is less than weight → sinks. Archimedes' principle quantifies the upward push exactly.

Thrust, Pressure & The Push From Below

Thrust is simply the force acting perpendicular to a surface. Pressure is thrust spread over area:

Pressure = Thrust / Area  (unit: Pa = N/m²)

The same thrust feels different depending on area — a sharp knife (small area) cuts easily; a flat hand (large area) does not, for the same push.

Inside a fluid, pressure increases with depth: P = h·ρ·g, where h is depth and ρ is fluid density. This pressure acts in all directions, which is why a submerged object feels an upward push (greater pressure on its bottom face than its top face) — this net upward force is buoyancy.

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

Whether an object floats or sinks depends on comparing its own density with the fluid's density:

  • Object density < fluid density → floats (partially submerged)
  • Object density = fluid density → floats fully submerged, balanced
  • Object density > fluid density → sinks

Relative density = density of a substance ÷ density of water — a pure ratio, no units, which is why it is used to quickly compare materials.

Rule-Based Step Solver

Pick a problem type, fill in the known values, and the solver walks through every step — formula selection, substitution, unit handling, and the final answer. Runs entirely in your browser; no internet or API needed.

Tip: leave the field you want to find blank — the solver detects the unknown automatically when possible, or simply asks for the values it needs.

Your step-by-step solution will appear here. Choose a problem type and enter the known values to begin.

Concept 1

F = G · m₁m₂ / d²

Universal Law of Gravitation

G = 6.674 × 10⁻¹¹ N·m²/kg². F in newton, m in kg, d in metre.

Concept 1

F ∝ m₁m₂  ·  F ∝ 1/d²

Proportionality Relations

Doubling either mass doubles F. Doubling distance reduces F to ¼ (inverse square).

Concept 2

g = G · M / R²

Acceleration Due to Gravity

M = mass of planet, R = radius of planet. Independent of falling body's mass.

Concept 2

v = u + gt

Velocity in Free Fall

u = initial velocity. Take g positive for falling, negative for rising.

Concept 2

s = ut + ½gt²

Distance in Free Fall

For a body dropped from rest, u = 0, so s = ½gt².

Concept 2

v² = u² + 2gs

Velocity–Distance Relation

Useful when time is not known or not asked.

Concept 2

W = m · g

Weight

Weight is a force (unit: newton); mass (kg) stays constant everywhere.

Concept 3

P = Thrust / Area

Pressure

Unit: pascal (Pa) = N/m². Same thrust on smaller area gives larger pressure.

Concept 3

P = h · ρ · g

Pressure Inside a Fluid

h = depth, ρ = fluid density. Pressure acts equally in all directions at a point.

Concept 3

F_B = ρ_fluid · V_displaced · g

Buoyant Force (Archimedes' Principle)

Equal to the weight of fluid displaced — not the weight of the object itself.

Concept 3

RD = ρ_substance / ρ_water

Relative Density

A pure ratio — no unit. ρ_water = 1000 kg/m³ (or 1 g/cm³).

Concept 3

ρ = mass / volume

Density

SI unit kg/m³; commonly used as g/cm³ in problems (1 g/cm³ = 1000 kg/m³).

✓ Smart Ticks & Tips

  • Always convert distance to metres and mass to kilograms before using F = Gm₁m₂/d² — G is defined for SI units only.
  • When a body is "dropped" or "released," its initial velocity u = 0. When it is "thrown up," its final velocity at the highest point v = 0.
  • Remember the sign convention: take the direction of motion as positive. For a body thrown upward, g is negative (deceleration); for a falling body, g is positive.
  • g on the Moon ≈ g/6 on Earth — a fast shortcut for "weight on Moon" questions without recomputing from G, M, R.
  • To check float-or-sink instantly, compare densities, not weights: if object density < fluid density, it floats — regardless of how heavy the object is in absolute terms (e.g., a huge steel ship still floats because its overall average density is less than water's).
  • Pressure questions: identify whether you need thrust/area (solids) or hρg (fluids) — they look similar but use different given quantities.
  • 1 g/cm³ = 1000 kg/m³ — a very common unit-conversion trap in density and relative density problems.
  • In F ∝ 1/d², if distance becomes "half," force becomes 4 times, not 2 times — square the ratio, don't just invert it.

✕ Common Mistakes To Avoid

  • Confusing mass and weight: writing "weight = 60 kg" instead of "mass = 60 kg, weight = 588 N." Weight always needs a force unit (N), mass needs kg.
  • Forgetting to square the distance in F = Gm₁m₂/d² — a very frequent slip under exam pressure.
  • Using g = 9.8 m/s² for the Moon or another planet — g changes with the planet's own mass and radius; never assume Earth's g elsewhere.
  • Mixing up radius and diameter when computing g = GM/R² — R must be the radius, not the diameter, of the planet.
  • Sign errors in equations of motion: not flipping the sign of g for upward motion, leading to wrong (often negative-looking) answers that get blindly accepted.
  • Treating buoyant force as the object's own weight: F_B depends only on the fluid displaced, not on what the object is made of or how heavy it is.
  • Forgetting units mid-calculation — especially mixing g/cm³ and kg/m³ in the same expression without converting first.
  • Assuming heavier objects fall faster: in the absence of air resistance, all objects fall with the same acceleration g, regardless of mass (a classic misconception traced back to pre-Galilean physics).

Q1

Two satellites of mass 80 kg and 120 kg are placed 5 m apart in a lab for a thought experiment. Calculate the gravitational force of attraction between them. (Take G = 6.674 × 10⁻¹¹ N·m²/kg²)

Step 1 — Identify knowns: m₁ = 80 kg, m₂ = 120 kg, d = 5 m, G = 6.674 × 10⁻¹¹ N·m²/kg².

Step 2 — Write the formula: F = G·m₁m₂/d²

Step 3 — Substitute: F = (6.674 × 10⁻¹¹ × 80 × 120) / 5²

Step 4 — Simplify numerator: 6.674 × 10⁻¹¹ × 9600 = 6.407 × 10⁻⁷

Step 5 — Divide by d² = 25: F = 6.407 × 10⁻⁷ / 25 = 2.563 × 10⁻⁸ N

Answer: F ≈ 2.56 × 10⁻⁸ N — extremely small, confirming why everyday gravitational pulls between objects go unnoticed.

Q2

The gravitational force between two fixed point masses is F. If the distance between them is reduced to one-third of the original distance, find the new force in terms of F.

Step 1 — Recall proportionality: F ∝ 1/d² (masses unchanged, so they drop out of the ratio).

Step 2 — Set up ratio: F_new / F_old = (d_old / d_new)²

Step 3 — Substitute d_new = d_old/3: F_new / F = (d / (d/3))² = (3)² = 9

Step 4 — Solve: F_new = 9F

Answer: The new force becomes 9 times the original force F.

Q3

A planet has twice the mass of Earth but the same radius as Earth. Compare the gravitational force exerted by this planet on a 1 kg object at its surface with the force exerted by Earth on the same object.

Step 1 — Write force on Earth: F_Earth = G·M·1/R²

Step 2 — Write force on the new planet: mass = 2M, radius unchanged = R, so F_planet = G·(2M)·1/R² = 2 × (G·M/R²)

Step 3 — Compare: F_planet = 2 × F_Earth

Answer: The force (and hence g) on the new planet is twice that on Earth — an object would feel twice as heavy there.

Q4

A stone is dropped from the top of a tower and reaches the ground in 4 s. Find (a) the height of the tower and (b) the velocity with which it strikes the ground. (Take g = 10 m/s²)

Step 1 — List knowns: u = 0 (dropped), t = 4 s, g = 10 m/s².

Step 2 — Find height using s = ut + ½gt²: s = 0 + ½ × 10 × 4² = ½ × 10 × 16 = 80 m

Step 3 — Find final velocity using v = u + gt: v = 0 + 10 × 4 = 40 m/s

Answer: Height of tower = 80 m; velocity on striking ground = 40 m/s.

Q5

A ball is thrown vertically upward with an initial velocity of 30 m/s. Find the maximum height it reaches and the total time it stays in the air before returning to the thrower's hand. (Take g = 10 m/s²)

Step 1 — List knowns (upward motion, g acts against motion): u = 30 m/s, v = 0 at highest point, g = −10 m/s².

Step 2 — Find max height using v² = u² + 2gs: 0 = 30² + 2(−10)s ⟹ 0 = 900 − 20s ⟹ s = 45 m

Step 3 — Find time to reach top using v = u + gt: 0 = 30 − 10t ⟹ t = 3 s

Step 4 — Total time in air = 2 × time to top (by symmetry of upward/downward journey) = 2 × 3 = 6 s

Answer: Maximum height = 45 m; total time in air = 6 s.

Q6

An astronaut has a mass of 75 kg. Calculate her weight on Earth and on the Moon. (Take g_Earth = 9.8 m/s², g_Moon = 1/6 of g_Earth)

Step 1 — Weight on Earth using W = mg: W_Earth = 75 × 9.8 = 735 N

Step 2 — Find g on Moon: g_Moon = 9.8 / 6 ≈ 1.633 m/s²

Step 3 — Weight on Moon: W_Moon = 75 × 1.633 ≈ 122.5 N

Answer: Weight on Earth ≈ 735 N; weight on Moon ≈ 122.5 N. Mass stays 75 kg in both places — only weight changes.

Q7

A box of base area 0.5 m² exerts a thrust of 600 N on the floor. Calculate the pressure exerted on the floor. If the same box is turned so its base area becomes 0.2 m², find the new pressure.

Step 1 — Original pressure using P = Thrust/Area: P₁ = 600 / 0.5 = 1200 Pa

Step 2 — Thrust stays the same (same weight) when turned; only area changes to 0.2 m²: P₂ = 600 / 0.2 = 3000 Pa

Answer: Original pressure = 1200 Pa; new pressure = 3000 Pa. Smaller area → larger pressure for the same thrust.

Q8

Find the pressure exerted by water at a depth of 12 m below the surface of a lake. (Density of water ρ = 1000 kg/m³, g = 10 m/s². Ignore atmospheric pressure.)

Step 1 — Write the formula: P = h · ρ · g

Step 2 — Substitute: P = 12 × 1000 × 10

Step 3 — Simplify: P = 120000 Pa = 1.2 × 10⁵ Pa

Answer: Pressure at 12 m depth = 1.2 × 10⁵ Pa (this is in addition to atmospheric pressure pressing down from above).

Q9

A solid block has a mass of 270 g and a volume of 100 cm³. Find its density and relative density, and state whether it will float or sink in water. (Density of water = 1 g/cm³)

Step 1 — Find density using ρ = mass/volume: ρ = 270 g / 100 cm³ = 2.7 g/cm³

Step 2 — Find relative density using RD = ρ_substance / ρ_water: RD = 2.7 / 1 = 2.7 (no unit)

Step 3 — Compare with water's density: Since 2.7 g/cm³ > 1 g/cm³ (the block is denser than water)…

Answer: Density = 2.7 g/cm³, Relative density = 2.7. Since the block's density exceeds water's, it will sink.

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ground

Orbital radius (R_earth + altitude): 6771 km

Approx. orbital speed: 7.67 km/s

Approx. orbital period: 92.4 minutes

Uses v = √(GM/r). Lower orbits need higher speed to balance Earth's stronger pull there — that's why the ISS (~400 km) circles Earth in about 90 minutes, while geostationary satellites (~36,000 km) take 24 hours.

Earth

Status: Floating

Fraction submerged: 60%

Academia Aeternum · NCERT Class IX Science · Chapter 9: Gravitation
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ACADEMIA AETERNUM तमसो मा ज्योतिर्गमय · Est. 2025
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    Frequently Asked Questions

    Gravitation is the force of attraction between any two objects in the universe.

    Sir Isaac Newton formulated the Universal Law of Gravitation.

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

    \(F=G \frac{Mm}{d^2}\), where \(G\) is gravitational constant, \(M\) and \(m\) are masses, and \(d\) is distance.

    Free fall is the motion of an object under gravity only, without air resistance.

    Because gravitational acceleration \(g\) is uniform near Earth's surface.

    It keeps planets orbiting the Sun and moons orbiting planets, maintaining system stability.

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