Course Content
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
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
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
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
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
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
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
Class 11th Physics Online Class For 100% Result
About Lesson

Planetary Motion ( Class 11 Physics Gravitation )

  • Ptolemy was the first scientist who studied the planetary motion.
  • He gave geocentric model. It means all the planets, stars and sun revolve around the earth and earth is at the centre.
  • Heliocentric model was proposed by some Indian astronomers.
  • According to which all planets revolve around the sun.
  • Nicholas proposed the Nicholas Copernicus modelaccording to which all planets move in circles around the sun.
  • After Nicholas one more scientist named Tycho Brahe did lot of observations on planets.
  • Finally came Johannes Kepler who used Tycho Brahe observations and he gave Kepler’s 3 laws of Gravitation.
  • These 3 laws became the basis of Newton’s Universal law of Gravitation.
Class 11 Physics Gravitation
Class 11 Physics Gravitation


Kepler’s 1st law Vs. Copernicus Model

  • According to Copernicus planets move in circular motion whereas according to Kepler planets revolve in elliptical orbit around the sun.
  • Copernicus model is based on one special case because circle is a special case of ellipse whereas Kepler’s laws aremore of ageneral form.
  • Kepler’s law also tells us about the orbits which planets follow.

To Show ellipse is a special form of Circle

  • Select two points F1 and F2.
  • Take a pieceof string and fix its ends at F1 and F2.
  • Stretch the string taut with the help of a pencil and then draw a curve by moving the pencil keeping the string taut throughout. Fig. (a).
  • The resulting closed curve is an ellipse. For any point T on the ellipse, the sum of distances from F1 and F2 is a constant. F1,F2 are called the foci.
  • Join the points F1 and F2,and extend the line to intersect the ellipse at points P and A as shown in Fig. (a).
  • The centre point of the line PA is the centre of the ellipse O and the length PO = AO, which is also known as the semimajor axis of the ellipse.
  • For a circle, the two foci merge onto one and the semi-major axis becomes the radius of the circle.
Class 11 Physics Gravitation

Kepler’s 2nd law: Law of Areas

Statement:-The line that joins a planet to the sun sweeps out equal areas in equal intervals of time.

  • Area covered by the planet while revolving around the sun will be equal in equal intervals of time. This means the rate of change of area with time is constant.
  • Suppose position and momentum of planet is denoted by ‘r’ and ‘p’ and the time taken will be Δt.
  • ΔA=1/2xrxvΔt (where vΔt is distance travelled by a planet in Δt time.)
  • ΔA / Δt =1/2(rxv)
  • where
  • (Linear momentum) p=mv or we can write as
  • =p/m
  • =1/2m(rxp)
  • =1/2 L/2m where L= angular momentum(It is constant for any central force)

ΔA / Δt = constant (This means equal areas are covered in equal intervals of time).

Kepler’s 3rdLaw: Law of periods

Statement: –

  • According to this law the square of time period of a planet is ∝ to the cube of the semi-major axisof its orbit.
  • Suppose earth is revolving around the sun then the square of the time period (time taken to complete one revolution around sun) is ∝ to the cube of the semi major axis.
  • It is known as Law of Periods as it is dependent on the time period of planets.
  • Derivation of 3rd Law: assumption: The path of the planet is circular.
  • Let m=mass of planet
  • M= mass of sun
  • According to Newton’s Law of Gravitation:
  • F= GMm/r2
  • Fc=mv2/r
  • where
  • Fc =centripetal force which helps the planet to move around sun in elliptical order.
  • F = Fc
  • GMm/r2= mv2/r where r=radius of the circle
  • GM/r= v2 (1)
  • v= 2 πr/T
  • Squaring both the sides the above equation
  • v2=4 π2r2/T2
  • putting the value (1)
  • GM/r=4 πr2/T2
  • T2= (4 π2 r3/GM) where (4 π2/GM) = constant
  • T2=r3 (In ellipse semi-major axis is same as radius of the circle)

Problem: – Suppose there existed a planet that went around the sun twice as fast as the earth.What would be its orbital size as compared to that of the earth?

Answer: Lesser by a factor of 0.63

Time taken by the Earth to complete one revolution around the Sun,

T= 1 year

Orbital radius of the Earth in its orbit, Re = 1 AU

Time taken by the planet to complete one revolution around the Sun,

Tp=1/2Te=1/2 year

Orbital radius of the planet = Rp

From Kepler’s third law of planetary motion, we can write:

(Rp /Re)3 = (Tp/Te)2

Rp /R= (Tp/Te) 2/3

            =((1/2)/1) 2/3 =(0.5) 2/3 = 0.63

Hence, the orbital radius of the planet will be 0.63 times smaller than that of the Earth.

Problem: A Saturn year is 29.5 times the earth year. How far is the Saturn from the sun, if theearth is 1.50 ×108 km away from the sun?


 Answer: Distance of the Earth from the Sun, Re = 1.5 × 108 km = 1.5 × 1011 m

Time period of the Earth = Te

Time period of Saturn, Ts = 29. 5 Te

Distance of Saturn from the Sun = Rs

From Kepler’s third law of planetary motion, we have

T2= (4 π2 r3/GM) 1/2

For Saturn and Sun, we can write

(Rs3/ Re3) = Ts2/ Te2

Rs = Re(Ts/ Te)2/3

=1.5 × 1011(29. 5) 2/3

=1.5 × 1011x9.55


Hence, the distance between Saturn and the Sun is 1.432×1012m.

Problem:-Let the speed of the planet at the perihelion P in Fig. be vP and


Sun-planet distance SP is rP. Relate {rP, vP} to the corresponding quantities at


aphelion {rA, vA}. Will the planet take equal times to traverse BAC and CPB?

Answer The magnitude of the angular momentum at P is Lp = mp rp vp, since inspection tells us that rp and vp are mutually perpendicular. Similarly,

 LA = mp rA vA. Fromangular momentum conservation

mp rp vp = mp rA vA


Since rA> rp, vp> vA .

The area SBAC bounded by the ellipse andthe radius vectors SB and SC is larger than SBPCin Fig. From Kepler’s second law, equal areasare swept in equal times. Hence the planet willtake a longer time to traverse BAC than CPB.

Wisdom TechSavvy Academy