NCERT Class 11 Physics Chapter 8 Notes Gravitation - Download PDF

An apple falls directly to the ground when it is dropped from a tree. The force of gravity, which controls everything from falling items to planetary motion, was discovered by Sir Isaac Newton as a result of this simple observation.

Welcome to the complete Class 11 Gravitation Notes on Careers360, which have been carefully created by subject matter experts. For students studying for board examinations and competitive tests like JEE and NEET, these Chapter 7 Physics notes are a vital resource.

Our detailed Class 11 Gravitation notes cover the complex nature of gravitational forces and describe how they affect everything from planets and stars to meteoroids and galaxies.

In addition, these organised CBSE Class 11 Physics Chapter 7 notes are accessible offline for smooth learning thanks to their PDF format.

Also, students can refer,

• NCERT Solutions for Class 12 Physics Chapter 8: Gravitation
• NCERT Exemplar Solution Class 12 Physics Chapter 8: Gravitation

Newton's Law of Gravitation

Every particle of matter attracts each other with a force proportional to the product of their masses and inversely proportional to the square of the distance between them.

Mathematical Form

If m₁ and m₂ are the particle’s masses and the distance between them is r, the force of attraction F among the particles is given by

$\begin{array}{rl}& F\propto \frac{m_1{m}_2}{r^2}\\ & \therefore F=G\frac{m_1{m}_2}{r^2}\end{array}$

(G is the universal constant of gravitation.)

Vector Form

The force F₂₁ exerted on the particle m₂ by the particle m₁ is given by:

${\overrightarrow{F}}_{21}=-G\frac{m_1{m}_2}{r^2}{r}_{12}$

Where r₁₂ cap is a unit vector drawn from particle m₁ to particle m₂

Similarly,

$\begin{array}{rl}& {\overrightarrow{F}}_{21}=-G\frac{m_1{m}_2}{r^2}{r}_{\widehat{12}}\\ & {\overrightarrow{F}}_{12}=G\frac{m_1{m}_2}{r^2}{r}_{\widehat{12}}\end{array}$

Universal Constant for Gravitation

The universal gravitation constant is given as

$G=\frac{F{r}^2}{m_1{m}_2}$

S.I. unit

$\frac{\text{ newton }(\text{ metre }{)}^2}{(\mathrm{kg}{)}^2}$

C.G.S. unit

$\frac{dyne(\mathrm{cm}{)}^2}{g{m}^2}$

Value of G

$G=6.67\times {10}^{-11}\mathrm{Nm}/{\mathrm{kg}}^2$

Dimensions of G

$[G]=\frac{\left[M^1{L}^1{T}^2\right]\left[M^0{L}^0{T}^0\right]}{\left[M^2{L}^0{T}^0\right]}=\left[M^{-1}{L}^3{T}^{-2}\right]$

Acceleration Due to Gravity

Acceleration due to gravity, denoted as g, is a fundamental concept in physics that describes the acceleration of objects due to the gravitational pull of the Earth. This acceleration is approximately
9.8 m/s2 near the Earth's surface. It means that in the absence of air resistance, any object will increase its velocity by about 9.8m/s for every second it falls.

Variation in 'g' Due to Height

Value of g at the surface of the earth (at distance r=R from earth centre)

$g=\frac{GM}{R^2}$

Value of g at height h from the surface of the earth (at a general distance r=R+h from earth centre)

$g^{\prime }\propto \frac{1}{r^2}$

Where $r=R+h$

• Value of g when h < < R

$\begin{array}{rl}& g^{\prime }=g{\left(\frac{R}{R+h}\right)}^2=g{(1+\frac{h}{R})}^{-2}\\ & g^{\prime }=g[1-\frac{2h}{R}]\end{array}$

Variation in 'g' Due to Depth

Value of g at the surface of the earth (at d=0)

$g=\frac{GM}{R^2}=\frac{4}{3}\pi \rho gR$

Value of g at depth d from the surface of the earth (at a general distance r=(R-d) from earth centre)=g'

And $g^{\prime }\propto (R-d)$

This means Value of g' decreases on going below the surface of the earth.

So $g^{\prime }=g[1-\frac{d}{R}]$

Gravitational Field

The space that surrounds any mass is known as a gravitational field. Any additional mass brought into this space is subjected to gravitational attraction. In a word, the area in which any mass field of gravity experiences a gravitational pull.

Gravitational Potential

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

1. At a distance r from a point mass M, the gravitational potential (V) is given by:

$\Rightarrow V=-\frac{GM}{r}$

2. The potential energy of a unit mass is converted from the work done on it. As a result, the gravitational potential at any point equals the potential energy of a unit mass positioned there.

3. The gravitational potential energy (P.E.) of a small point mass m put in a gravitational field at a position where the gravitational potential is V is given by

$P.E.=-\frac{GMm}{r}$

Satellite

A satellite is a smaller body that spins around a larger body under the influence of its gravitation. It could be natural or man-made.

1. Because the moon revolves around the earth, it is considered a satellite. Jupiter has a total of sixteen satellites orbiting it. These satellites are referred to as natural satellites.

2. A human-built artificial satellite is one that has been sent into a circular orbit. The first satellite, SPUTNIK–I, was launched by the Soviet Union, whereas the first satellite, ARYABHATTA, was launched by India.

Minimum Two-Stage Rocket is Used to Project a Satellite in a Circular Orbit Around a Planet

Assume that a satellite is launched from the earth's surface using a single-stage launching method. Once the rocket's fuel is ignited, the rocket begins to ascend. The rocket reaches its maximum velocity when all of the fuels have been used up.

1. The rocket escapes into space with the satellite if its maximum velocity is equal to or greater than the escape velocity, exceeding the earth's gravitational force.

2. If the rocket's greatest velocity is less than escape velocity, it will be unable to defy the earth's gravitational attraction, and both the rocket and the satellite will eventually crash to the ground.

As a result, a single-stage rocket will not be able to place a satellite into a circular orbit around the Earth. As a result, a launching device is required to place a satellite into a circular orbit around the Earth.

• Expressions for Different Energies of Satellite

1. Potential Energy (P.E.):

The satellite is at a distance (R + h) from the centre of the earth.

$\begin{array}{rl}& U=-\frac{G{m}_1{m}_2}{r}\\ & \Rightarrow -\frac{GMm}{R+h}=U\end{array}$

2. Kinetic Energy (K.E.):

The Revolution of satellites around circular orbit is having critical velocity (v_c). Hence its kinetic energy is given by:

$\begin{array}{rl}& K.E.=\frac{1}{2}m{v}_c{}^2\\ & v_c=\sqrt{\frac{GM}{R+h}}\\ & \Rightarrow K\cdot E\cdot =\frac{1}{2}m\left(\frac{GM}{R+h}\right)=\frac{GMm}{2(R+h)}\end{array}$

3. Total Energy (T.E.):

$T.E.=-\frac{GMm}{R+h}+\frac{GMm}{2(R+h)}=-\frac{GMm}{2(R+h)}$

4. Binding Energy (B.E.)

T.E. = -(B.E.)

Kepler’s law

Kepler gave three laws of planetary motion. They are:

• First law: All planets revolve in an elliptical orbit with the Sun as the focus.

• Second law: Irrespective of the orbit, the planet covers equal areas in equal intervals of time.

• Third law: The orbital period of the planet is proportional to the orbit’s size. The cube of the mean distance of a planet from the Sun is proportional to the square of its orbital period T.

$r^3/T^2=$ constant

Different Cases of Projection:

When a satellite is lifted to a specific height above the earth and then projected horizontally, the following four scenarios may occur, depending on the amount of horizontal velocity.

1. If the projection velocity is smaller than the critical velocity, the satellite will move in an elliptical orbit, but the point of projection will be apogee, and the spacecraft will approach the earth with its perigee point at 180o. If it enters the atmosphere as it approaches perigee, it loses energy and spirals downward. It will continue to fly in an elliptical orbit if it does not enter the atmosphere.

2. If the projected velocity achieves the critical velocity, the satellite will move in a circular orbit around the planet.

3. The satellite will be in an elliptical orbit with an apogee greater than the predicted height if the projected velocity is more than the critical velocity but less than the escape velocity.

4. The satellite goes in a parabolic path if the projection velocity is equal to the escape velocity.

5. If the projection velocity is greater than the escape velocity, the orbit will become hyperbolic, escaping the earth's gravitational attraction and continuing to travel indefinitely.

Orbital Velocity

The orbital velocity of a satellite is the horizontal velocity with which it must be propelled from a point above the earth's surface in order to circle in a circular orbit around the earth.

An Expression for the Critical Velocity of a Satellite Revolving Around the Earth

Assume that a mass m satellite is elevated to a height h above the earth's surface and then projected horizontally with an orbital velocity ve. The satellite starts moving around the globe in a circular orbit with a radius of R + h, where R is the earth's radius.

The gravitational force acting on the satellite is

$\frac{GMm}{(R+h{)}^2}$

where Mis the mass of the earth and Gis the constant of gravitation.

For circular motion,

Centrifugal force = Centripetal force

$\begin{array}{rl}& \frac{m{v}_c{}^2}{(R+h)}=\frac{GMm}{(R+h{)}^2}\\ & \therefore v_c=\sqrt{\frac{GM}{(R+h)}}\end{array}$

Escape Velocity of a Body

The escape velocity is the smallest velocity at which a body can be launched from the earth's surface and yet escape the planet's gravitational field.

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

When the body is at rest on Earth’s surface, Escape velocity is:

Kinetic Energy of projection = Binding Energy

$\begin{array}{rl}& \frac{1}{2}m{v}_e^2=\frac{GMm}{r}\\ & \Rightarrow v_e=\sqrt{\frac{2GM}{R}}=\sqrt{2gR}\end{array}$

Weightlessness

1. A feeling of weightlessness is similar to a moving satellite since it alludes to the gravitational force that draws a body towards the earth's centre. It isn't due to the fact that the weight is zero.

2. A gravitational pull exerts on an astronaut when he is on the surface of the earth. This gravitational force is equal to an astronaut's weight, and an astronaut exerts this force on the earth's surface. The earth's surface produces an equal and opposite reaction, and he feels his weight on the ground as a result of this reaction.

3. An astronaut on an orbiting spacecraft experiences the same acceleration towards the earth's centre as the satellite, and this acceleration is equal to the acceleration due to the earth's gravity.

As a result, the astronaut has no impact on the satellite's floor. The astronaut, of course, is unaffected by the floor's reaction force. Because there is no reaction, the astronaut has a sense of weightlessness. (In other words, he has no concept of how heavy he is.)

Importance of NCERT Class 11 Physics Chapter 8 Notes

Students can clearly understand the basic ideas of the Gravitation chapter by using the NCERT notes for Class 11 Physics Chapter 7, which offer a thorough study of the chapter. These Class 11 Gravitation notes are very helpful for a variety of competitive exams, including VITEEE, BITSAT, JEE Main, NEET, and others, in addition to board exams. They provide a strong conceptual foundation by carefully going over all of the important subjects listed in the CBSE Physics syllabus.

Furthermore, these CBSE Class 11 Physics Chapter 7 notes are easily accessible in PDF format, enabling students to study offline at any time and from any location, increasing the flexibility and effectiveness of exam preparation.

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