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

Gravitational Potential Energy

  • Potential energy is due to the virtue of position of the object.
  • Gravitational Potential Energy is due to the potential energy of a body arising out of the force of gravity.
  • Consider a particle which is at a point P above the surface of earth and when it falls on the surface of earth at position Q, the particle is changing its position because of force of gravity.
  • The change in potential energy from position P to Q is same as the work done by the gravity.
  • It depends on the height above the ground and mass of the body.
Class 11 Physics Gravitation
Class 11 Physics Gravitation

Stationary roller-coaster

Expression for Gravitational Potential Energy

Case1:- ‘g’ is constant.

  • Consider an object of mass ‘m’ at point A on the surface of earth.
  • Work done will be given as :
  • WBA=F X displacement where F = gravitational force exerted towards the earth)
  • =mg(h2-h1) (body is brought from position A to B)
  • =mgh2-mgh1
  • where
    • VA=potential energy at point A
    • VB= potential energy at point B
  • From above equation we can say that the work done in moving the particle is just the difference of potential energy between its final and initial positions.

Case2:-‘g’ is not constant.

  • Calculate Work done in lifting a particle from r = r1 to r = r2 (r2> r1) along a vertical path,
  • We will get , W=V (r2) – V (r1)

Conclusion: –

  • In general the gravitational potential energy at a distance ‘r’ is given by :

V(r) = -GMem/r + Vo

  • where
  • V(r) = potential energy at distance ‘r’
  • Vo = At this point gravitational potential energy is zero.
  • Gravitational potential energy is ∝ to the mass of the particle.

Gravitational Potential ( Class 11 Physics Gravitation )

  • Gravitational Potential is defined as the potential energy of a particle of unit mass at that point due to the gravitational force exerted byearth.
  • Gravitational potential energy of a unit mass is known as gravitational potential.
  • Mathematically:
  • Gpotential= -GM/R

Problem: ( Gravitation )

Choose the correct alternative:

Acceleration due to gravity increases/decreases with increasing altitude.

Acceleration due to gravity increases/decreases with increasing depth. (assume the earth to be a sphere of uniform density).


Acceleration due to gravity is independent of mass of the earth/mass of the body.


The formula –G M(1/r2– 1/r1) is more/less accurate than the formula mg(r2– r1) for the difference of potential energy between two points r2and r1 distance away from the centre of the earth.




(c)Mass of the body



  • Acceleration due to gravity at depth h is given by the relation:

gh = (1- 2h/Re)g


Re = Radius of the Earth, g = acceleration due to gravity on the surface of the earth.

It is clear from the given relation that acceleration due to gravity decreases with an increase in height.

  • Acceleration due to gravity at depth d is given by the relation:


It is clear from the given relation that acceleration due to gravity decreases with an increase in depth.

  • Acceleration due to gravity of body of mass m is given by the relation: g=GM/r2


G = Universal gravitational constant

M = Mass of the Earth

R = Radius of the Earth

Hence, it can be inferred that acceleration due to gravity is independent of the mass of the body.

  • Gravitational potential energy of two points r2 and r1 distance away from the centre of the Earth is respectively given by:

V (r1) = – G mM/r1

V (r2) = -G mM/r2


Difference in potential energy, V = V(r2) – V(r1) =-GmM (1/r2 – 1/r1)

Hence, this formula is more accurate than the formula mg (r2– r1).

Problem:- ( Class 11 Physics Gravitation )

Two earth satellites A and B each of mass m are to be launched into circular orbits earth’s surface at altitudes 6400km and 1.92X104 km resp. The radius of the Earth is 6400km.Find (a) The ratio of their potential energies and (b) the ratio of their kinetic energies. Which one has greater total energy?


  • ma = mass of satellite A

mb=mass of satellite B

ha=6400km, hb=1.92X104 km


Potential Energy = -GMem/ (Re+h)

For A (P.E)A = -GMem/ (6400+6400)

            =-GMem/12800 —(1) 

For B(P.E)B = -GMem/(6400+1.92X104)

                        -GMem/ (6400+19200)   —(2)

Divide 1 by 2 we will get

(P.E)A /(P.E)B = 2 : 1

  • (K.E)A = GMm/2×12800 (3)

(K.E)B= GMm/2(1.92 X10+6400)   (4)

Dividing (3) by (4)

(K.E)A/(K.E)= GMm/(12800)  x 2(1.92 X10+6400)/ GMm

(K.E)A/ (K.E)= 2:1

  • Total Energy of A = – GMm/2r r=12800km

Total Energy of B = – GMm/2r  r=(1.92×104 +6400)km

Total energy of B is greater than A.

Problem: – A rocket is fired vertically with a speed of 5 km s–1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0 × 1024 kg; mean radius of the earth = 6.4 × 106m;  G= 6.67 × 10–11 N m2 kg–2.

Answer: Velocity of the rocket, v = 5 km/s = 5 × 103 m/s

Mass of the Earth, Me = 6.0 × 1024 kg

Radius of the Earth, Re=6.4 × 106m

Height reached by rocket mass, m = h

At the surface of the Earth,

Total energy of the rocket = Kinetic energy + Potential energy

= 1/2mv2+(-GMem/Re)

At highest point h,


And Potential Energy = – (GMem/Re+h)

Total energy of the rocket

=0+ – (GMem/Re+h)


Total energy of the rocket

From the law of conservation of energy, we have

Total energy of the rocket at the Earth’s surface = Total energy at height h.

1/2mv2+(-GMem/Re) = – GMem/Re+h

1/2v2 = GMe(1/ Re – 1/ Re+h)

By calculating

1/2v2 = gReh/Re+h

Where g=GM/Re2 = 9.8m/s2


v2(Re+h) = (2gReH)

v2Re=h (2gRe-v2)

h=Re-v2/ (2gRe-v2)

=6.4 × 106x(5×103)2/2×9.8×6.4×106-(5×103)2


Height achieved by the rocket with respect to the centre of the Earth


=6.4 × 106x1.6×106


The distance of the rocket is 8 × 106 m from the centre of the Earth.

Problem: – Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the mid-point of the line joining the centres of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?


  • 0;
  • –2.7 × 10–8 J /kg;
  • Yes;
  • Unstable


Class 11 Physics Gravitation

Any object placed at point X will be in equilibrium state, but the equilibrium is unstable. This is because any change in the position of the object will change the effective force in that direction.

Problem:- ( Class 11 Physics Gravitation )

Two stars each of one solar mass (= 2× 1030 kg) are approaching each other for a head on collision. When they are a distance 109 km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 104 km. Assume the stars to remain undistorted until they collide. (Use the known value of G).


Mass of each star, M = 2 × 1030 kg

Radius of each star, R = 104 km = 107 m

Distance between the stars, r = 109 km = 1012m

For negligible speeds, v = 0 total energy of two stars separated at distance r

=-GMM/r + 1/2mv2

=-GMM/r + 0   (i)

Now, consider the case when the stars are about to collide:

Velocity of the stars = v

Distance between the centres of the stars = 2R

Total kinetic energy of both stars = 1/2Mv2 + 1/2Mv2 = Mv2

Total potential energy of both stars =-GMM/2R

Total energy of the two stars =Mv2 – GMM/2R (ii)

Using the law of conservation of energy, we can write:

Mv2 – GMM/2R =- GMM/r

v2=-GM /r + GM/2R = GM (-1/r + 1/2R)

=6.67×10-11 x 2 x1030(-1/1012 + 1/2×107)


=13.34×1019x 5 x 10-8


v=√6.67×1012= 2.58×106m/s

Problem:- A 400kg satellite is in circular orbit of radius 2RE about the Earth. How much energy is required to transfer it to circular orbit of radius 4RE?What are the changes in the kinetic and potential energies?


Ei= – GMem/2RE

E= – GMem/4RE

ΔE = Ef – Ei

 =- GMem/2RE(1/4-1/2)

ΔE =GMem/8 RE

In terms of ‘g’

ΔE = gm RE/8

By putting the values and calculating

ΔE = 3.13 x 109J

The energy which is required to transfer the satellite to circular orbit of radius 4RE is 3.13 x 109J.

Change in Kinetic energy Δk=kf – ki

Δk = 3.13 x 109J

Change in Potential energy ΔV= 2xΔE = -6.25x109J

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