Syjc chap2-gravitation

1

GravitationProf. Mukesh N.

TekwaniDepartment of PhysicsIsmail Yusuf College,

Mumbaimukeshtekwani@hotmail.

com

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Isaac Newton

Prof. Mukesh N Tekwani

Prof. Mukesh N Tekwani 3

Newton’s Law of GravitationEvery particle of matter attracts every

other particle with a force which is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.


Newton’s Law of Gravitation

G is called the Universal Gravitational Constant

G = 6.67 x 10-11 N m2/kg2

G is constant throughout the Universe and G does not depend on the medium between the masses


Difference between G and gG g

G is the Universal Gravitational Constant

g is acceleration due to gravity

G = 6.67 x 10-11 N m2/kg2

Approx value g = 9.8 m / s2.Value of g varies from one place to another on the Earth.

Constant throughout the Universe

Changes every place on a planet. E.g., on the Moon, the value of g is 1/6th of that on the Earth’s surface.


Relation between G and gLet M = mass of the Earthm = mass of an object on the

surface of the Earthg = acceleration due to gravity

on the Earth’s surfaceR = radius of the Earth


Relation between G and g

m

M


Relation between G and gm

M

Weight of the object is the gravitational force acting on it.

………………………….(1)


Relation between G and gAt height h from the surface of the Earth’s surface, acceleration due to gravity is gh

At height h, Weight of object = gravitational force

h

m

M

………………………….(2)


Relation between G and gDividing (2) by (1) we get,

h

m

M Thus, g is independent of the mass of the object.


Projection of a SatelliteWhy is it necessary to have at least a two stage rocket to launch a satellite?

A rocket with at least two stages is required to launch a satellite becauseThe first stage is used to carry the satellite up to

the desired height.

In the second stage, rocket is turned horizontally (through 90 degrees) and the satellite is fired with the proper horizontal velocity to perform circular motion around the earth.


Critical Velocity of a SatelliteThe horizontal velocity with which a satellite should be projected from a point above the earth's surface, so that it orbits in a circular path around the earth is called the orbital velocity or critical velocity (Vc) of the satellite.


Kepler’s Laws of MotionBorn: December 27, 1571

Died: November 15, 1630

German Mathematician, AstronomerAstrologer.


Kepler’s First Law – Law of OrbitEvery planet revolves in an elliptical orbit around the Sun, with the Sun situated at one focus of the ellipse.


Kepler’s Second Lawor Law of Equal Areas

The radius vector drawn from the Sun to any planet sweeps out equal areas in equal intervals of time. This law is called the law of areas.

The areal velocity of the radius vector is constant.


Kepler’s Law of Equal Areas


Kepler’s Laws


Kepler’s Third Law - Laws of PeriodThe square of period of revolution of the planet around the Sun is directly proportional to the cube of the semi-major axis of the elliptical orbit.

T2 α r3

According to this law, when the planet is closest to the Sun, its speed is maximum and when it is farthest from the Sun, its speed is minimum.


Critical Velocity of a Satellite

R

Mr

h

vcLet

M = mass of the EarthR = radius of Earthm = mass of satelliteh = height of the satellite above Earth’s r = R + h, where r is the distance of the satellite from the center of the EarthVc = critical velocity of the satellite


Critical Velocity of a SatelliteThe centripetal force necessary for the circular motion of the satellite around the Earth is provided by the gravitational force of attraction between the Earth and the satellite.Centripetal force = gravitational force


Critical Velocity of a Satellite

Factors on which Critical Velocity of a satellite depends:1. Mass of the planet2. Radius of the planet3. Height of the satellite

Critical velocity is not dependent on the mass of the satellite as m does not appear in the above equation

……………………………………………….(1)


Critical Velocity of a SatelliteBut we know that

Substituting this value in eqn (1), we get,

…………………………………..(2)


Critical Velocity of a SatelliteAssignment 1:

Modify eqn (2) to find the critical velocity of a satellite orbiting very close to the surface of the Earth (h << R)

Assignment 2:

How does the critical velocity (or orbital velocity) of a satellite vary with an increase in the height of the satellite above the Earth’s surface?


Time Period of a SatelliteThe time taken by a satellite to complete one revolution around the earth is called its periodic time or time period.


Time Period of a Satellite

R

Mr

h

vcLet



Time Period of a SatelliteDistance covered by the satellite in 1 revolution

= Circumference of the circle

Time taken to cover this distance is the time period

Critical speed Vc =

Vc =

But


Time Period of a Satellite

Squaring both sides, we get

As is a constant,

so we get,

T2 α r3

Thus, the square of the period of revolution is directly proportional to the cube of the radius of its orbit.


Time Period of a SatelliteFactors on which Time Period of a satellite depends:1. Mass of the planet2. Radius of the planet, and 3. Height of the satellite from the planet’s

surface

Period of the satellite does not depend on the mass of the satellite.Assignment:

(1)Obtain an expression for the time period of a satellite in terms of gh.

(2)For a satellite close to the earth, calculate the period of revolution in minutes.


Binding Energy of a SatelliteDefinition: The minimum amount of energy required to remove a satellite from the earth’s gravitational influence is called as binding energy of the satellite.

A satellite revolving around the Earth has Kinetic energy, andPotential energy


What is kinetic energy?The energy possessed by a body

due to its motion is called its kinetic energy.

If m = mass of an object, and v = its velocity

K.E = (1/2) mv2


What is potential energy?The energy possessed by a body due

to its position is called its potential energy.

If m = mass of an object, and h = its height above the surface

P.E. = mgh

Quiz: Does a body in motion have potential energy?


Binding Energy of a Satellite

R

Mr

h

vcLet



Binding Energy of a SatelliteThe critical velocity is given by

Kinetic energy of motion KE =

Substituting (1) in (2), we get,

KE =

……………………………………………………...(1)

…..……...(2)

..………………………………………………..(3)


Binding Energy of a SatelliteThe gravitational potential at a distance r from the centre of the Earth is given by:

GP =

Potential energy = gravitational potential x mass of the satellite

Therefore, PE = ..…….………………………..(4)


Binding Energy of a SatelliteThe total energy of the satellite is given by

TE = KE + PE

TE = +

TE =

The negative sign indicates that the satellite is bound to the Earth due to the gravitational force of the Earth.

..…….…………………………………………..(5)


Binding Energy of a SatelliteTo free the satellite from the Earth’s gravitational influence, an amount of energy equal to its total energy must be supplied. This is called the binding energy of the satellite.

Therefore, BE =

Where r = R + h

Assignment: Calculate the BE of a satellite at rest on the surface of the Earth.


Weightlessness in a Satellite1. The weight of a body is the

gravitational force exerted on it by the Earth.

2. When a person stands on a floor, he exerts a force on the floor. The floor in turn exerts a force (normal reaction) on the person.

3. This normal reaction is equal to the weight of the person.

4. The person has a feeling of weight due to this normal reaction.


Weightlessness in a Satellite5. Consider an astronaut of mass m, in a

satellite that is moving around the Earth in a circular orbit.

6. There is a centripetal force on the satellite and the astronaut. Thus, both are attracted towards the Earth with the same acceleration, due to the Earth’s gravitational force.

7. So the astronaut is not able to exert a force on the floor of the satellite & the satellite in turn cannot exert normal reaction on the astronaut. This causes the “feeling” of weightlessness.


Weightlessness in a Satellite8. We must remember that the mass of

the astronaut DOES NOT become zero.

9. This condition of weightlessness is also known (incorrectly) as zero gravity condition.

10.But, weightlessness does not mean the absence of gravity.


Escape Velocity of a SatelliteThe minimum velocity with which a body should be projected from the surface of the earth so that it escapes the gravitational field of the earth is called the escape velocity of the body.


Escape Velocity of a SatelliteConsider a satellite of mass m,

stationary on the surface of the Earth.The binding energy of the satellite, on

the surface of the Earth, is given by BE =

To escape from the Earth’s influence, energy must be provided to the satellite in the form of kinetic energy.

R


Escape Velocity of a SatelliteTherefore, KE of satellite = BEKE =

Therefore, = R

R


Numerical ProblemsObtain an equation for the

escape velocity of a body from the surface of a planet of radius R and mean density ρ.

What would be the duration of the year if the distance between the Earth and Sun gets doubled?


Numerical ProblemsCalculate the height of a communications

satellite from the surface of the Earth. Values of G, M and R are as given in the text book. (These values will also be provided in the question paper)

A body weighs 4.5 kgwt on the surface of Earth. How much will it weigh on the surface of a planet whose mass is (1/9)th that of the Earth’s mass and radius is half that of the Earth?


Variation of g with Altitudem

M

Weight of the object is the gravitational force acting on it.

………………………….(1)


Variation of g with AltitudeAt height h from the surface of the Earth’s surface, acceleration due to gravity is gh

At height h, Weight of object = gravitational force

h

m

M

………………………….(2)


Variation of g with AltitudeDividing (2) by (1) we get,

h

m

M Thus, g is independent of the mass of the object.

From this eqn. it is clear that the acceleration due to gravity decreases as altitude of the body from the earth’s surface increases.


Variation of g with AltitudeBy Binomial expansion:

Since the higher powers of h/R are neglected.


Variation of g due to depthThe acceleration due to gravity on the surface of the earth is given by:

Consider the earth as a sphere of density ρ

Mass = Volume x Density


Variation of g due to depthMass = Volume x Density

Mass =

…………………………………………. (1)


Variation of g due to depthConsider a point P at the depth d below the surface of the earth. At this point, let the acceleration due to gravity be gd.

Distance of point P from centre of Earth is (R – d)


Variation of g due to depthThe acceleration due to gravity at point P due to sphere of radius (R –d) is

Here, M’ is the mass of inner solid sphere of radius (R - d)


Variation of g due to depthMass = volume x density

……………………………………………(2)


Variation of g due to depth

This is the expression for the acceleration due to gravity at a depth d below the surface of the earth.

Therefore, acceleration due to gravity decreases with depth.

Dividing (2) by (1), we get,


Variation of g due to depthAt the centre of the earth, d = R, therefore

gd = 0So if a body of mass m is taken to the centre of the earth, its weight will be equal to zero (since w= mg). But its mass will not become 0.


Variation of g due to depth

g’

depth

altitude

Inside

the

earth

Outside the

earth


What is latitudeThe latitude of a location on the Earth is the angular distance of that location south or north of the Equator.

Latitude of equator is 0o

Latitude of North pole is : 90o north (+90o)Latitude of South pole is : 90o south (-90o)


Variation of g due to latitudeThe Earth is rotating from west to east

and the axis of rotation passes through the poles.

Let angular velocity of earth be ω.Every point on the surface of the

earth is moving in a circle, i.e. every point is in an accelerated motion


Variation of g due to latitudeNP-North PoleSP-South Pole


Variation of g due to latitude

Consider a body at point P on the surface of the earth.

Let the latitude of point P be

The body at point P moves in a circular path whose center is at Q and radius is PQ.

PQ = r

XOP =



Therefore, OPQ = (alternate angles – transversal cutting two parallel lines)

Centripetal acceleration needed for a body at point P is ar = rω2

Consider OPQ

cos =



cos =

cos =

r = R cos

Therefore, ar = R cos x ω2

ar = R ω2 cos ………………. (1)

a



a

From OPQ, the radial component of centripetal acceleration is

a = ar cos\a = R ω2 cos cos

\ a = R ω2 cos2 ……………… (2)

\The effective acceleration g’ due to gravity at point P is directed towards the centre of the earth and is given by:

g’ = g - a



a

\ g’ = g - R ω2 cos2

\As latitude increases, cos decreases, so g’ will increase.

\The value of g’ increases as we move from the equator to the pole due to rotation of the earth.


Variation of g due to latitudeAssignment 1: Obtain an expression / value for the acceleration due to gravity at the equator.

Assignment 2: Obtain an expression / value for the acceleration due to gravity at the poles.

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Syjc chap2-gravitation