NCERT Class 11 Physics Chapter 8 Notes Gravitation - Download PDF

NCERT Notes for Class 12 Chapter 7

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

$\begin{aligned} & F \propto \frac{m_1 m_2}{r^2} \\ & \therefore F=G \frac{m_1 m_2}{r^2}\end{aligned}$

(G is the universal constant of gravitation.)

Vector Form

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

$\vec{F}_{21}=-G \frac{m_1 m_2}{r^2} \hat{r_{12}}$

Where $\hat{r}_{12}$ is a unit vector drawn from particle m₁ to particle m₂

Similarly,

$\begin{aligned} & \vec{F}_{12}=-G \frac{m_1 m_2}{r^2} \hat{r_{21}} \end{aligned}$

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

$\frac{d y n e(\mathrm{~cm})^2}{g m^2}$

• 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)

$g=\frac{G M}{R^2}$

• 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

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

Variation in 'g' Due to Depth

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

$g=\frac{G M}{R^2}=\frac{4}{3} \pi \rho g R$

• 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 $g^{\prime}=g\left[1-\frac{d}{R}\right]$

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:

$\Rightarrow V=-\frac{G M}{r}$

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

$P . E .=-\frac{G M m}{r}$

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.

$\begin{aligned} & U=-\frac{G m_1 m_2}{r} \\ & \Rightarrow-\frac{G M m}{R+h}=U\end{aligned}$

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:

$\begin{aligned} & K . E .=\frac{1}{2} m v_c{ }^2 \\ & v_c=\sqrt{\frac{G M}{R+h}} \\ & \Rightarrow K \cdot E \cdot=\frac{1}{2} m\left(\frac{G M}{R+h}\right)=\frac{G M m}{2(R+h)}\end{aligned}$

3. Total Energy (T.E.):

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

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.
• The 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 v_o. 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 $\frac{G M m}{(R+h)^2}$

where M is the mass of the earth and Gis the constant of gravitation.

For circular motion,

Centrifugal force = Centripetal force

$\begin{aligned} & \frac{m v_o{ }^2}{(R+h)}=\frac{G M m}{(R+h)^2} \\ & \therefore v_o=\sqrt{\frac{G M}{(R+h)}}\end{aligned}$

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

$\begin{aligned} & \frac{1}{2} m v_e^2=\frac{G M m}{r} \\ & \Rightarrow v_e=\sqrt{\frac{2 G M}{R}}=\sqrt{2 g R}\end{aligned}$

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.)

Gravitation Previous year Question and Answer

Q1: If the mass of sun were ten times smaller and gravitational constant G were ten times larger in magnitudes

a) walking on ground would become more difficult

b) the acceleration due to gravity on earth will not change

c) raindrops will fall much faster

d) aeroplanes will have to travel much faster

Answer:

If $G^{\prime}=10 G$
And $M_s^{\prime}=\frac{M_s}{10}$
Then $g^{\prime}=\frac{G^{\prime} M_E}{r^2}=\frac{10 G M}{r^2}=10 g$
Weight of a person $=m g^{\prime}=10 \mathrm{mg}$
Hence statement (a) is correct, and statement (b) is incorrect
Fon man due to sun $=\frac{G M_s^{\prime} m}{r^2}=\frac{G M_s m}{10 r^2}$
The terminal velocity $v_T \alpha g^{\prime}$ or $\quad v_T \alpha 10 g$
Hence raindrops fall 10 times faster. Option statement (c) is correct
In order to maintain the speed after the increase in gravitational force, the aeroplane will have to travel faster.

Hence, answer are the option (a), (c) and (d)

Q2: If the law of gravitation, instead of being inverse-square law, becomes an inverse-cube-law

a) planets will not have elliptic orbits

b) circular orbits of planets is not possible

c) projectile motion of a stone thrown by hand on the surface of the earth will be approximately parabolic

d) there will be no gravitational force inside a spherical shell of uniform density

Answer:

If the law of gravitation suddenly changes and becomes an inverse cube law, then the law of areas still holds. This is because the force will still be central force, due to which angular momentum will remain constant.

Also, for a planet of mass $m$ revolving around the sun of mass $M$, we can write according to the question,

$
F=\frac{G M m}{a^3}=\frac{m v^2}{a}
$

$\left[\because\right.$ Centripetal force, $\left.m \omega^2 a=\frac{m v^2}{a}\right]$
where, $a$ is radius of orbiting planet.
Orbital speed, $v=\frac{\sqrt{G M}}{a} \Rightarrow v \propto \frac{1}{a}$
Time period of revolution of a planet,

$
T=\frac{2 \pi r}{v}
$

On putting the value of orbital speed, we get

$
\begin{aligned}
& T=\frac{2 \pi a}{\frac{\sqrt{G M}}{a}}=\frac{2 \pi a^2}{\sqrt{G M}} \\
& \Rightarrow T^2 \propto a^4
\end{aligned}
$

Hence, the orbit of the planet around sun will not be elliptical because for elliptical orbit $T^2 \propto a^3$.
Hence, the orbit of the planet around sun will not be circular because for circular orbit $T^2 \propto a^3$.
As force $F=\left(\frac{G M}{a^3}\right) m=g^{\prime} m$, where $g^{\prime}=\frac{G M}{a^3}$.
Since, $g^{\prime}=$ acceleration due to gravity is constant, hence path followed by a projectile will be approximately parabolic ( as $T \propto a^2$ ).

Hence, the answer are the options (a), (b) and (c).

Q3: Which of the following options is correct?

a) acceleration due to gravity decreases with increasing altitude

b) acceleration due to gravity increases with increasing depth

c) acceleration due to gravity increases with increasing latitude

d) acceleration due to gravity is independent of the mass of the earth

Answer:

g at an altitude $g\left(1-\frac{2 h}{R}\right)$

Acceleration due to gravity decreases with increasing altitude. Hence statement (a) is correct

$g^\lambda=g-\omega^2 R(\cos \lambda)^2$

As cos is a decreasing function, the value of gravity increases with an increase in latitude, i.e. from the equator to the pole. Hence, statement (c) is correct

$g=\frac{G M}{R^2}$ Where M is the mass of the earth. Hence, statement (d) is incorrect

Hence the answer are the options (a) and (c).

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