Course Content
Class 11 Physics Chapter 1 Physical World
Section Name Topic Name 1 Physical World 1.1 What is physics? 1.2 Scope and excitement of physics 1.3 Physics, technology and society 1.4 Fundamental forces in nature 1.5 Nature of physical laws
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Class 11 Physics Chapter 4 Motion In A Plane
4 Motion in a plane 4.1 Introduction 4.2 Scalars and vectors 4.3 Multiplication of vectors by real numbers 4.4 Addition and subtraction of vectors – graphical method 4.5 Resolution of vectors 4.6 Vector addition – analytical method 4.7 Motion in a plane 4.8 Motion in a plane with constant acceleration 4.9 Relative velocity in two dimensions 4.10 Projectile motion 4.11 Uniform circular motion
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Class 11 Physics Chapter 5 Laws of motion
Section Name Topic Name 5 Laws of motion 5.1 Introduction 5.2 Aristotle’s fallacy 5.3 The law of inertia 5.4 Newton’s first law of motion 5.5 Newton’s second law of motion 5.6 Newton’s third law of motion 5.7 Conservation of momentum 5.8 Equilibrium of a particle 5.9 Common forces in mechanics 5.10 Circular motion 5.11 Solving problems in mechanics
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Class 11 Physics Chapter 6 Work Energy and Power
Section Name Topic Name 6 Work Energy and power 6.1 Introduction 6.2 Notions of work and kinetic energy : The work-energy theorem 6.3 Work 6.4 Kinetic energy 6.5 Work done by a variable force 6.6 The work-energy theorem for a variable force 6.7 The concept of potential energy 6.8 The conservation of mechanical energy 6.9 The potential energy of a spring 6.10 Various forms of energy : the law of conservation of energy 6.11 Power 6.12 Collisions
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Class 11 Physics Chapter 7 Rotation motion
Topics Introduction Centre of mass Motion of COM Linear Momentum of System of Particles Vector Product Angular velocity Torque & Angular Momentum Conservation of Angular Momentum Equilibrium of Rigid Body Centre of Gravity Moment of Inertia Theorem of perpendicular axis Theorem of parallel axis Moment of Inertia of Objects Kinematics of Rotational Motion about a Fixed Axis Dynamics of Rotational Motion about a Fixed Axis Angular Momentum In Case of Rotation about a Fixed Axis Rolling motion
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Class 11 Physics Chapter 9 mechanics properties of solid
Section Name Topic Name 9 Mechanical Properties Of Solids 9.1 Introduction 9.2 Elastic behaviour of solids 9.3 Stress and strain 9.4 Hooke’s law 9.5 Stress-strain curve 9.6 Elastic moduli 9.7 Applications of elastic behaviour of materials
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Class 11 Physics Chapter 11 Thermal Properties of matter
Section Name Topic Name 11 Thermal Properties of matter 11.1 Introduction 11.2 Temperature and heat 11.3 Measurement of temperature 11.4 Ideal-gas equation and absolute temperature 11.5 Thermal expansion 11.6 Specific heat capacity 11.7 Calorimetry 11.8 Change of state 11.9 Heat transfer 11.10 Newton’s law of cooling
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Class 11 Physics Chapter 14 Oscillations
Section Name Topic Name 14 Oscillations 14.1 Introduction 14.2 Periodic and oscilatory motions 14.3 Simple harmonic motion 14.4 Simple harmonic motion and uniform circular motion 14.5 Velocity and acceleration in simple harmonic motion 14.6 Force law for simple harmonic motion 14.7 Energy in simple harmonic motion 14.8 Some systems executing Simple Harmonic Motion 14.9 Damped simple harmonic motion 14.10 Forced oscillations and resonance
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Class 11th Physics Online Class For 100% Result
About Lesson

Escape Velocity ( Class 11 Physics Gravitation )

  • Escape velocity is the minimum velocity that a body must attain to escape the gravitational field of the earth.
  • Suppose if we throw a ball,it will fall back. This is happening due to the force of gravitation exerted on the ball by the surface of the earth due to which the ball is attracted towards the surface of the earth.
  • If we increase the velocity to such an extent that the object which is thrown up will never fall back.This velocity is known as escape velocity.
Class 11 Physics Gravitation

The same ball is thrown with a velocity that itescapes the force of gravitation of earth and does not come back. This velocity is known as escape velocity.

  • Mathematically:-
  • Suppose we throw a ball and the initial velocity of the ball is equal to the escape velocity such that ball never comes back.
  • Final Position will be infinity.
  • At Final Position: At Infinity
  • Total Energy (∞) = kinetic Energy (∞) + PotentialEnergy (∞)
  • KineticEnergy (∞) = ½ mvf2 where vf=final velocity
  • Potential Energy (∞) = -GMm/r + V0
  • where M=mass of the earth, m= mass of the ball,

V0=potential energy at surface of earth, r=∞ r=distance from the centre of the earth.

  • Therefore: – Potential Energy (∞) =0
  • Total Energy (∞) =½ mvf2 (1)
  • At initial position:-
  • E. = 1/2mvi2
  • E= -GMm/ (Re+h) + V0
  • Where h= height of the ball from the surface of the earth.
  • Total Energy (initial) = 1/2mvi2– GMm/ (Re+h) (2)
  • According to law of conservation of energy
  • Total Energy (∞) =Total Energy (initial)
  • ½ mvf2 = 1/2mvi2 – GMm/ (Re+h)
  • As L.H.S = positive
  • 1/2mvi2 – GMm/ (Re+h) ≥ 0
  • 1/2mvi2 = GMm/ (Re+h)
  • By calculating
  • vi2 = 2GM/ (Re+h)
  • Assume Ball is thrown from earth surface h<<Re
  • This implies Re+h is same as Re as we can neglect h.
  • Therefore,vi2 = 2GM/ (Re)
  • Or vi = √(2GM/Re)
  • This is the initial velocity with which if the ball is thrown it will never fall back on the earth surface.

In terms of ‘g’

  • g = GM/Re2
  • Escape velocity can be written as

Ve= √2gRe

Example of Escape Velocity: No atmosphere on moon

  • Earth Escape velocity= Ve= √2gRe
  • Moon Escape velocity = Ve= √2gmRm where Rm is the radius of the moon.
  • gm=1/6 ge ; Rm = 1/4 Re
  • (Ve)moon= √2 gmRm = √2xg/6x Re/4
  • After calculating we will get:

(Ve)moon = 1/5 (Ve)earth = 2.3km/s

  • As this velocity is very less, the molecules cannot accumulate on the moon so there is no atmosphere on the moon.

Problem: – Does the escape speed of a body from the earth depend on

(a) the mass of the body,

(b) the location from where it is projected,

(c) the direction of projection,

(d) the height of the location from where the body is launched?

Answer

 (a) No

(b) No

(c) No

(d) Yes

Escape velocity of a body from the Earth is given by the relation:

Ve= √2gRe

g = Acceleration due to gravity

R = Radius of the Earth

It is clear from equation (i) that escape velocity is independent of the mass of the body and the direction of its projection. However, it depends on gravitational potential at the point from where the body is launched. Since this potential marginally depends on the height of the point, escape velocity also marginally depends on these factors.

Problem: A comet orbits the Sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.

Answer:-

 (a) No

(b) No

(c) Yes

(d) No

(e) No

(f) Yes

Angular momentum and total energy at all points of the orbit of a comet moving in a highly elliptical orbit around the Sun are constant. Its linear speed, angular speed, kinetic and potential energy varies from point to point in the orbit.

Problem: – The escape speed of a projectile on the earth’s surface is 11.2 km s–1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.

Answer:-

Escape velocity of a projectile from the Earth, vesc = 11.2 km/s

Projection velocity of the projectile, vp = 3vesc

Mass of the projectile = m

Velocity of the projectile far away from the Earth = vf

Total energy of the projectile on the Earth=1/2mp2 – 1/2mvesc2

Gravitational potential energy of the projectile far away from the Earth is zero.

Total energy of the projectile far away from the Earth =1/2mvf2

From the law of conservation of energy, we have,

1/2mp2 – 1/2mvesc2 = 1/2mvf2

vf = √vp2-vesc2

=√(3 vesc)2 –(vesc)2

=√8 vesc

=31.68km/s

Problem:- ( Class 11 Physics Gravitation )

Calculate the escape velocity on the surface of the moon? Given that the radius of the moon is 1.7×106m and the mass of the moon is 1022kg.

Answer:-

Ve= √2GM/Re

Rm = 1.7×106m

M= 1022kg

G=6.67×10-11Nm2/kg2

V= √2×6.67×10-11x1022/1.7×106

=2.4×103m/s

The escape velocity on the surface of the moon is 2.4×103m/s.

Problem:- What is the escape velocity from Jupiter given that the mass is 300 times that of the Earth’s and its radius is 10times larger?

Answer:-

MJ=300Me

RJ=10Re

Ve= √2GMe/Re = Ve= √2GMJ/RJ

= √2Gx300Me/10x Re

After putting the values,

=√30×11.2m/s

=61.3km/s

The escape velocity on the surface of the Jupiter is 61.3km/s.

Class 11 Physics Gravitation

Artificial Satellites:

  • Human built objects orbiting the earth for practical uses. There are several purposes which these satellites serve.
  • Example:- Practical Uses of Artificial satellites
  • Communication
  • Television broadcasts
  • Weather observation
  • Military support
  • Navigation
  • Scientific research
Class 11 Physics Gravitation
Class 11 Physics Gravitation

 

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