CBSE Class 11 Physics Revision Notes Chapter 8

Class 11 Physics Revision Notes for Chapter 8 – Gravitation

The Chapter Gravitation is important for examinations since most questions are asked from this chapter. It is necessary for students to understand it from the basics. Extramarks Revision Notes provide clear and concise definitions, explanations and equations for different topics under Class 11 Physics Chapter 8 which will aid students in understanding the chapter better and scoring well in exams. 

1. Introduction

The cosmos is made up of galaxies, stars, planets, comets, asteroids and meteoroids. The force that holds them together is known as gravitational force. Gravitation is a natural phenomenon that attracts material objects.

Sir Isaac Newton, an English physicist, released Principia Mathematica in 1687 A.D., which describes the inverse-square law of gravitation.

2. Newton’s Law of Gravitation

2.1 Definition

Every particle of matter attracts every other particle of matter with a force proportional to their mass product and inversely proportional to the square of their spacing.

2.2 Mathematical Form

If the masses of the particles are m1 and m2, and the distance between them is r, the force of attraction F between the particles is given by the following formula:

F∝m1m2/r2

∴F=Gm1m2/r,

where G is the universal constant of gravitation. 

2.3 Vector Form

Newton’s law of gravitation is represented in vector form as given below.  The force (F→21) exerted by the particle m1 on the particle m2 can be calculated as follows:

F→21 =−Gm1m2/r2 r12 …… (1),

where (r12) is a unit vector connecting particles m1 and m2.

Similarly, the force (F→12) exerted on particle m1 by particle m2 can be calculated as follows:

F→12=+Gm1m2/r2 r12  ….. (2),

where (r12) is a unit vector connecting particles m1 and m2.

From (1), (2),

F→12 = −F→21

3. Universal Constant for Gravitation(G)

G=Fr2/m1m2

If, m1 = m2 = 1, and r = 1 then G = F.

The attraction force between two unit masses separated by a unit distance is mathematically equal to the universal gravitational constant.

3.1. Unit

SI Unit:

Newton(metre)2/(kilogram)2=Nm2/kg2

CGS Unit: 

dynecm2/gm2

3.2. Value of G

G=6.67×10−11Nm2/kg2

Dimensions: [G] = [F][r2]/[m1m2] = [M1L1T2][M0L3=2T0]/ [M2L3=0T0] = [M−1L3T−2]

4. Variation in ‘g’

4.1. The Acceleration Due to Gravity at a Height h above the Earth’s Surface.

Let M and R represent the mass and radius of the Earth respectively, and g represent the acceleration due to gravity at the surface. Assume that an object of mass m is placed on the surface of the earth.

mg=GMm/R2

⇒g=GM/R2 ….(1)

where mg is the weight of the body equal to the gravitational force acting on it.

Assume the body is raised to a height h above the earth’s surface; the body’s weight is now mg, and the gravitational force acting on it is:

mgh=GMm/(R+h)2

gh=GM/(R+h)2 ….(2) 

By dividing (1) and (2), we get,  

gh/g=R2/(R+h)2

⇒gh=[R2/(R+h)2]g.

4.2.  Acceleration due to Gravity at a Very Small Height

gh= g R+hR2

⇒gh=g(1−hR)−2 

⇒gh = g (1− 2hR+h2/R2…….) 

If h << R, we get:  

 gh=g(1− 2hR).

4.3. Effect of Depth on  Acceleration due to Gravity

g in terms of ρ:  

g=GM/R2 

If ρ is the density of the material of the earth, then:  

M=4/3πR3ρ                                                                              

g=G×4/3πR3ρ/R2                                                               

g=4/3πGRρ …..(1) 

Let gd denote the gravitational acceleration at point B at depth x under the earth’s surface. There is a body at point B that will only be subjected to force as a result of the earth’s part OB (R-d) radius.

Now, M′=4/3π(R-x)3ρ

Or       

gd=4/3πG(R−d)ρ ……(2)

Now,

gd/g = 4/3πG(R−d)ρ / 4/3πGRρ

 =R−d/R orgd=g(1−d/R) …..(3) 

Hence, the acceleration because of gravity decreases with depth.

4.4. Variation of ‘g’ with Latitude due to the Rotational Motion of the Earth

The mrω2cosλ force radiates outward as a result of the earth’s rotation. Because of this, the net attraction force generated by the particle’s earth and pointed toward the planet’s centre is provided by the following formula:

mg′=mg−mrω2cosλ.

g′=g−rω2cosλ 

Now, r=Rcosλ

Then, 

g′=g−(Rcosλ)ω2cosλ

 ∴g′=g−Rω2cos2λ

Effective acceleration at a point ‘P’ due to gravity is given by the following formula: 

g′=g−Rω2cos2λ

Thus the value of ‘g’ changes with ‘λ’ and ‘ω’. 

5. Satellite

5.1. A satellite is any smaller body that orbits around another larger body under the influence of its gravitation. It could be either natural or man-made.

• Given that it revolves around the earth, the moon is a satellite of that body. There are sixteen satellites orbiting Jupiter. These satellites are referred to as natural satellites.
• A satellite classified as artificial is one that was constructed by humans and sent into orbit. India launched the first satellite, ARYABHATTA, while the USSR launched the first satellite, SPUTNIK-I.

5.2. A two-stage rocket is used to launch a satellite into a circular orbit around a planet.

• If the maximum velocity of the rocket is equal to or greater than the escape velocity, the rocket escapes into space with the satellite, overcoming the gravitational effect of the Earth.
• If the maximum velocity of the rocket is less than the escape velocity, it will be unable to resist the earth’s gravitational pull, and both the rocket and the satellite will eventually fall to the earth’s surface due to gravity.
• As a result, a single-stage rocket cannot place a satellite in a global circular orbit.

5.3. Different Cases of Projection

• The satellite moves in an elliptical orbit with apogee as the point of projection if the projection velocity is less than the critical velocity; otherwise, the satellite moves in a circular orbit with 180o for the perigee point. If it enters the atmosphere as perigee approaches, it will deplete its energy and fall. If it does not enter the atmosphere, it will continue to orbit in an elliptical pattern.
• If the projection velocity equals the critical velocity, the satellite will orbit the planet in a circular orbit.
• If the projection velocity is greater than the critical velocity but less than the escape velocity, the satellite will be in an elliptical orbit with an apogee greater than the projected height.
• If the projection velocity equals the escape velocity, the satellite follows a parabolic route.
• If the velocity of the projection is greater than the escape velocity, the orbit will become hyperbolic, escaping the earth’s gravitational pull and continuing to travel indefinitely.

6. Orbital Velocity

The satellite’s orbital velocity is the horizontal speed at which it must be projected from a point above the surface of the earth in order to travel in a circular orbit around the planet. 

7. Height of the Satellite Above the Surface of the Earth

With an orbital velocity of vc and a mass of m, imagine a satellite rotating in a circular path above the earth’s surface at a height of h. The duration needed for one earthly rotation is known as a satellite’s period or periodic time (T). Let M and R represent the mass and radius of the earth, respectively.

The radius (r) of the circular orbit is r=R+h.

For the circular motion,

vc=√GM/r  …..(1)

If T is the period of revolution of the satellite,  

Period(T)= circumference of orbit / critical velocity = 2πr/vc

T=2π√r3/GM

Squaring the equation, we get,

T2 α r3

8. Gravitational Field

The space that surrounds any mass is known as a gravitational field. Any additional mass added into this space experiences a gravitational pull. In essence, the area in which any mass field of gravity experiences a gravitational pull.

9. Gravitational Intensity

The gravitational intensity at a given location is described as the force acting on a unit mass placed there.

1. The following formula determines the gravitational intensity (E) at a point r from a point mass M: E=GM/r2 (where G is the constant of gravitation)
2. The force exerted on a point mass, m, in the presence of an intense gravitational field, E, is given by the formula: F=mE.

10. Gravitational Potential

The gravitational potential at that point is defined as the work necessary to transfer a unit mass from infinity to any point in a gravitational field.

1. The gravitational potential (V) at a position r away from a point mass M can be calculated as follows:

⇒V=−GM/r

(Where G is the gravitational constant)

1. Work performed on a unit mass is turned into potential energy. As a result, the gravitational potential at any point equals the potential energy of a unit mass placed there.
2. When a small point mass m is placed in a gravitational field at a location where the gravitational potential is V, the gravitational potential energy (P.E.) is given by: 

P.E.= mass × gravitational potential

 =mV

P.E.=−GMm/r

11. Escape Velocity of a Body

The least velocity at which a body must be launched from the surface of the earth in order to escape the planet’s gravitational field is known as the escape velocity.

As a result, if a body or satellite is given the escape velocity, its kinetic energy of projection equals its binding energy.

Kinetic Energy of projection = Binding Energy

1/2mv2e=GMm/r

⇒ve=√2GM/R.

12. Communication Satellite

A geostationary satellite, also known as a communication satellite, revolves around the globe in a circular orbit similar to that of the planet itself and has the same period of revolution (i.e. 1 day = 24 hours = 86400 seconds).

The satellite appears stationary from the surface of the earth because its relative velocity to the globe is zero. Hence, it is called a geostationary satellite or a geosynchronous satellite.

The height of the communication satellite above the earth’s surface is approximately 36000 km, and its period of revolution is 24 hours or 2460 60 seconds.

Uses

• For transmitting television signals across long distances on the earth’s surface.
• Telecommunication.
• Forecasting the weather.
• For photographing astronomical objects.
• Solar and cosmic radiation research

13. Weightlessness

A body’s weight refers to the gravitational force that pushes it towards the centre of the Earth; a sense of weightlessness is similar to a moving satellite. It’s not because the weight is nil.

A gravitational pull acts on an astronaut when he is on the earth’s surface. This gravitational force is the weight of an astronaut, and an astronaut exerts it on the earth’s surface. He feels his weight on the earth as a result of the equal and opposite reaction of the earth’s surface.

For an astronaut on an orbiting satellite, both the satellite and the astronaut experience the same acceleration towards the centre of the planet, which is equivalent to the acceleration caused by the earth’s gravity.

As a result, the astronaut does not influence the satellite’s floor. Naturally, the astronaut is unaffected by the reaction force of the floor. Because there is no reaction, the astronaut has a sense of weightlessness. (That is, he has no concept of how heavy he is.)

14. Kepler’s Laws

14.1. Law of Orbit

Each planet moves in an elliptical orbit around the sun, with the sun at one of the foci of the ellipse. An ellipse’s eccentricity is defined as the ratio of the distances SO and AO, i.e.

e=SO/AO

e=SO/a

⇒SO=ea

The closest encounter with the sun at F1 is AS miles. This is referred to as perigee. Apogee is the planet’s greatest distance (BS) from the sun.

Perigee (AS) = AO – OS = a − − ea = a (1 − − e)

Apogee (BS) = OB + OS = a + ea = a (1 + e)

14.2 Law of Area

The line connecting the sun and a planet sweep out equal portions at equal time intervals. A planet takes the same amount of time to get from A to B as it does from C to D.

Area velocity= Area swept/time

 =½r(rdθ)/dt=½r2 dθ/dt=constant

Hence, ½r2ω=constant.

14.3. Law of Periods

The square of the time required for the planet to complete one rotation around the sun is proportional to the cube of the semimajor axis of the elliptical orbit.

Centripetal force = Gravitational force

mv2/R=GMm/R2⇒GM/R=v2

Now, the velocity of the planet is, 

v = circumference of the circular orbit / Time Period=2πR/T

Thus,

GM/R=4π2R2/T2

∴T2∝R3  or  T2/R3=constant.