NCERT Notes For Class 11 Physics Chapter 8 Gravitation

Class 11 Physics Chapter 8 Gravitation

NCERT Notes For Class 11 Physics Chapter 8 Gravitation, (Physics) exam are Students are taught thru NCERT books in some of state board and CBSE Schools.  As the chapter involves an end, there is an exercise provided to assist students prepare for evaluation.  Students need to clear up those exercises very well because the questions withinside the very last asked from those. 

Sometimes, students get stuck withinside the exercises and are not able to clear up all of the questions.  To assist students solve all of the questions and maintain their studies with out a doubt, we have provided step by step NCERT Notes for the students for all classes.  These answers will similarly help students in scoring better marks with the assist of properly illustrated Notes as a way to similarly assist the students and answering the questions right.

NCERT Notes For Class 11 Physics Chapter 8 Gravitation

Class 11 Physics Chapter 8 Gravitation




  • Ptolemy introduced the ‘geocentric’ model to explain the motion of planets.
  • Copernicus putforward the ‘heliocentric theory’ in which the sun is at the centre of the planetary system.


Law of orbits

• All planets move in elliptical orbits with the Sun situated at one of the foci of the ellipse.



Law of areas

  • The line that joins any planet to the sun sweeps equal areas in equal intervals of time.
  • The planets moves slower when they are away from the sun
  • The planets moves faster when they are near to the sun
  • The law of areas is a consequence of conservation of angular momentum.

Law of periods

• The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of the ellipse traced out by the planet.

T2α a3


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

  • Mathematically,

  • where G is theuniversal gravitational constant
  • The value of the gravitational constant G is experimentally determined by English scientist Henry Cavendish in 1798.

G = 6.67×10-11 N m2/kg2


Acceleration due to gravity on the surface

  • The gravitational force acting on a body on the surface of earth is given by

Where G- gravitational constant, M- mass of earth, m- mass of the body, R- radius of the earth.

  • The weight experience by the body is F = mg, where g – acceleration due to gravity
  • Thus ,

  • Therefore

  • The mass of the earth can be calculated using the values of acceleration due to gravity, G and radius of earth.
  • This is the reason for the statement “Cavendish weighed the earth”.

Variation of acceleration due to gravity with height

  • The gravitational force on the mass m at a height h above the surface of the earth is
  • The weight of the body at the height h is mgh , where gh is the acceleration due to gravity at height.

  • Using binomial expression and neglecting the higher order terms we get

  • Thus, for small heights h above the value of g decreases


Variation of g with depth

  • .If ‘ρ’ is the mean density of earth, then mass of earth is
  • Mass = Volume x Density, ie

  • Similarly mass of the small sphere of radius R-d is

• Thus

  • Thus, as we go down below earth’s surface, the acceleration due gravity decreases.
  • At the centre of the earth acceleration due to gravity is zero.


  • The region around a mass where its gravitational force is experienced is called

its gravitational field.

  • The gravitational potential at point in the gravitational field is the work done in bringing a unit mass from infinity to that point in the field.
  • The gravitational potential at a distance r from a mass M is given by

  • If we are considering earth, M is the mass of the earth.


  • The gravitational potential energy at a point is the work done to bring a mass from infinity to that point.
  • The gravitational potential energy associated with two particles of masses m1 and m[1] separated by distance by a distance r is given by


• The minimum vertical velocity that has to be imparted to a body on the earth’s surface , so that it escapes from the earth’s gravitational pull and never returns to the earth is called escape velocity.

Expression for escape velocity

  • To escape from the earth’s gravitational field the total energy of the body must be greater than or equal to zero

  • Therefore, the velocity of escape is given by

  • If the body is thrown from near to the surface of earth, then h=0 , therefore

  • This is called the escape speed, sometimes loosely called the escape velocity.
  • On the surface of earth the escape speed is

11.2km/s approximately.

  • The escape velocity does not depend on mass of the body.

Why moon has no atmosphere?

  • The escape speed for the moon is 2.3 km/s.
  • The rms velocities of gas molecules on the surface of moon are greater than its escape speed.
  • Therefore gas molecules escape from the moon and hence moon has no atmosphere.


  • Earth satellites are objects which revolve around the earth.
  • The natural satellite of earth is moon ( Time period 27.3 days aprox.).
  • Examples for artificial (manmade) satellites are Sputnik, Aryabatta, INSAT etc.
  • Artificial satellites are used in fields like telecommunication, geophysics and meteorology.


• The velocity of a satellite in its orbit is called orbital velocity.

Expression for orbital velocity

  • The centripetal force required for the orbital motion of the satellite is provided by the gravitational force between satellite and the planet.
  • Thus

Where m- mass of the satellite, v –velocity

  • Therefore, the orbital velocity is given by

  • A satellite very close to earth’s surface (h=0) is referred to as an earth satellite.
  • Thus, the orbital velocity of an earth satellite is given by



  • Time period of satellite is the time for it to go once fully in its orbit.
  • Period , T = ( Distance travelled in one revolution) / (Orbital velocity)
  • Thus


  • Here GM = gR 2 ,.
  • For a satellite very close to the surface of earth h can be neglected in comparison to radius.
  • The period when h=0 is given by

  • Substituting the numerical values , we get T= 85 minutes approx.


  • The kinetic energy of the satellite in a circular orbit with speed v is

  • Substituting for orbital velocity , we get

• The total energy of an circularly orbiting satellite is negative.


  • Satellites in a circular orbits around the earth in the equatorial plane with T = 24 hours are called Geostationary Satellites.
  • They appears stationary with respect to earth is called geo-stationary satellite.
  • The direction of motion of the satellite is same as that of earth, i.e., from west to east.
  • The height of the orbit of geostationary satellite is 36,000km approx.
  • These satellites are mainly used for communication purpose like T V and radio broadcasting and weather forecasting.


  • These are low altitude (h is approximately equal to 500 to 800 km) satellites, but they go around the poles of the earth in a north-south direction.
  • Since its time period is around 100 minutes it crosses any altitude many times a day.
  • These satellites are used for studying the weather, environment and spying.
  • Polar satellites are extremely useful for remote sensing, meteorology as well as for environmental studies of the earth.


  • Weight of an object is the force with which the earth attracts it.
  • In a satellite around the earth, every part and parcel of the satellite has acceleration towards the center of the earth which is exactly the value of earth’s acceleration due to gravity at that position.
  • Thus in the satellite everything inside it is in a state of free fall.
  • Thus, in a manned satellite, people inside experience no gravity.
  • When an object is in free fall, it is weightless and this phenomenon is usually called the phenomenon of weightlessness.

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