Gravitation

DESIGNED

By

Edustudy point
Gravitation: It is a force due to which all things are attracted towards the earth. This force is known
as Gravitation.

• Gravitation is the force of attraction between all masses in the universe, especially the force of
attraction exerted by the earth on all the bodies near its surface.

Kepler’s Laws:
➢ Kepler’s 1st Law: Law of Orbits

Statement: - The orbit of every planet is an ellipse around the sun with sun at one of the two foci
of ellipse.

Whenever a planet revolves around sun it traces an

ellipse around the sun. The closest point is P and the
farthest point is A, P is called the perihelion and A the
aphelion. The semi major axis is half the distance AP.

➢ Kepler’s 2nd law: Law of Areas

Statement: -The line that joins a planet to the sun sweeps out equal areas in equal intervals of time.

• Area covered by the planet while revolving around the

sun will be equal in equal intervals of time. This means
the rate of change of area with time is constant.

• ΔA / Δt =1/2(rxv)

• ΔA / Δt = constant (This means equal areas are

covered in equal intervals of time).

➢ Kepler’s 3rdLaw: Law of periods

Statement: -According to this law the square of time period of a planet is directly proportional to
the cube of the semi-major axis of its orbit.

• T2=r3 (In ellipse semi-major axis is same as radius of the circle)

Universal Law of Gravitational:

Universal law of Gravitation states – Every single body in this universe attracts each other with a force
which is directly proportional to the product of their masses and inversely proportional to the square of the
distance between them.

• This law holds good for all the bodies in the universe.
 • If the product of mass of the bodies increase the force of attraction also increases between them
and if the square of the distance between the bodies increases, force decreases.

Mathematically:-

• Consider 2 boxes having mass m1 and m2. The distance between them is r.

• F ∝ m1m2

• F ∝ 1/r2

• Combining above equations:- F ∝ m1m2/r2

• F= G m1m2/r2

Where G = universal Gravitational constant. Its value is constant and it never changes.

Gravitational Constant: Cavendish experiment

• To calculate the value of G, Cavendish took a wooden plank and attached two 2 balls on either side
of the plank and hung this with a thin thread from the top.

• He introduced 2 very big balls and those balls are near the smaller balls.

Schematic drawing of Cavendish’s experiment:-

• S1 and S2 are large spheres that are kept on either side of the ellipse.

• When the big spheres are taken to the other side of the
ellipse (shown by dotted circles), the bar AB rotates a
little since the torque reverses direction.

• The angle of rotation can be measured experimentally.

• He observed that: - Plank rotates till twisting force

becomes equal to the gravitational force between the
balls. This happened because of force of attraction between the small balls and bigger balls.

𝛕 = G m1m2/r2
LƟ = G m1m2/r2

• By using the above equation, he calculated the value of G,

• G= 6.67x10-11Nm2/kg2

Acceleration due to gravity of the earth

• All the objects fall towards the earth because of gravitational pull of the earth. And when a body is
falling freely, it will have some velocity and therefore it will attain some acceleration. This
acceleration is known as acceleration due to gravity.

• It is a vector quantity.

• Denoted by ‘g’.

• Its value is 9.8m/s2.

Expression for Acceleration due to gravity

• Consider any object of mass ‘m’ at a point A on the surface of the earth. The force of gravity
between the body and earth can be calculated as

• F =G m Me /Re2 (1) where

o m=mass of the body

o Me = mass of the Earth

o Re= distance between the body and the earth is same as the radius of the earth

• Newton’s Second law states that

F = m a (2)

Comparing the equations (1) and (2)

• F = m (G m Me/Re2)

• (G m Me/Re2) is same as g (acceleration due to gravity)

• Therefore, the expression for Acceleration due to gravity.

• g=G Me/Re2

Acceleration due to gravity above the surface of earth

• To calculate the value of acceleration due to gravity of a point mass m at a height h above the
surface of the earth.

given figure shows the value of acceleration due to gravity g at

a height h above the surface of the earth.

• Force of gravitation between the object and the earth will

be

• F= G mMe/(Re+h)2

Where, m = mass of the object, Re = radius of the earth

• g(h) = F/m = GMe/(Re+h)2 = GMe/[Re2(1+h/Re)2]

• h << Re (as radius of the earth is very large)

By calculating we will get,

g(h) = g(1-2h/Re)

Acceleration due to gravity below the surface of earth

• Density of the earth is constant throughout. Therefore,

ρ = Me / (4/3π Re3) equation (1)

where, Me = mass of the earth, Volume of sphere= 4/3π Re3, Re = radius of the earth.
 • As entire mass is concentrated at the centre of the earth. Therefore density can be written as

ρ = Ms / (4/3π Rs3) equation (2)

• Comparing equation (1) and (2)

• Me/Ms = Re3/Rs3 where Rs = (Re-d) 3

• d= distance of the body from the centre to the surface of

the earth.

• Therefore, Me / Ms = Re3/(Re-d)3

• Ms = Me (Re- d)3/ Re3 from equation (3)

• To calculate Gravitational force (F) between earth and point mass m at a depth d below the surface
of the earth.

Above figure shows the value of g at a depth d. In this case only the smaller sphere of radius (Re–d)
contributes to g.

• F = G m Ms/(Re-d)2

• g = F / m where g= acceleration due to gravity at point d below the surface of the earth.

• g = G Ms/(Re-d)2

• Putting the value of Ms from equation (3) we have, GMe (Re-d)3/Re3(Re-d)2

=GMe (Re-d)/Re3

• We know g = G Me/Re2 equation (4)

• g(d) = G Me / Re2 (1-d/Re)

• From equation (4) g (d) = g (1-d/Re)

Inertial and Gravitational Mass

Inertial Mass: - Inertial mass is defined as the mass of body by virtue of inertia of mass.

• By Newton’s Law F=ma

• m=F/a where m= inertial mass (as it is because of inertia of a body)

Gravitational Mass:-Gravitational mass is defined as the mass of the body by virtue of the gravitational
force exerted by the earth.

• By Gravitation Force of attraction –

o F = G m M / r2

▪ M = F r 2/ G M

Gravitational Potential Energy: Gravitational Potential Energy is due to the potential energy
of a body arising out of the force of gravity.

• It depends on the height above the ground and mass of the body.
Expression for Gravitational Potential Energy

Case1:- ‘g’ is constant.

• Consider an object of mass ‘m’ at point A on the surface of earth. Work done will be given as :

WBA = Force X displacement (where F = gravitational force exerted towards the earth)
= m g (h2-h1) (body is brought from position A to B)
= m g h2-mgh1

WAB = VA-VB (where, VA=potential energy at point A, VB= potential energy at point B)

• From above equation we can say that the work done in moving the particle is just the difference of
potential energy between its final and initial positions.

Case2:- ‘g’ is not constant.

• Calculate Work done in lifting a particle from r = r 1 to r = r2 (r2> r1) along a vertical path,

• We will get , W=V (r2) – V (r1)

Conclusion: - In general the gravitational potential energy at a distance ‘r’ is given by :

V(r) = -GMem/r + Vo

• Where, V(r) = potential energy at distance ‘r’

• Vo = At this point gravitational potential energy is zero.
• Gravitational potential energy is ∝ to the mass of the particle.

Gravitational Potential: It is defined as the potential energy of a particle of unit mass at that point
due to the gravitational force exerted by earth.

• Gravitational potential energy of a unit mass is known as gravitational potential.

• Mathematically: Gpotential= -GM/R

Planetary Motion:
• Ptolemy was the first scientist who studied the planetary motion.

• He gave geocentric model. It means all the planets, stars and

sun revolve around the earth and earth is at the centre.

• Heliocentric model was proposed by some Indian

astronomers.

• According to which all planets revolve around the sun.

• Nicholas proposed the Nicholas Copernicus model according

to which all planets move in circles around the sun.

• Finally came Johannes Kepler who used Tycho Brahe observations and he gave Kepler’s 3 laws of
Gravitation.

• These 3 laws became the basis of Newton’s Universal law of Gravitation.

Escape Velocity: Escape velocity is the minimum velocity that a body must attain to escape the
gravitational field of the earth.

Mathematically: -

• Suppose we throw a ball and the initial velocity of the ball is equal to the escape velocity such that
ball never comes back.

➢ At Final Position: At Infinity

Total Energy (∞) = kinetic Energy (∞) + Potential Energy (∞)

• Kinetic Energy (∞) = ½ mvf2 where vf=final velocity

• Potential Energy (∞) = -G M m/r + V0 , (where M=mass of the earth, m= mass of the ball,

V0=potential energy at surface of earth, r=distance from the centre of the earth.)

• Therefore: - Potential Energy (∞) =0

• Total Energy (∞) =½ mvf2

➢ At initial position:-

• E. = 1/2mvi2

• E= - GMm/ (Re+h) + V0 ,(Where h= height of the ball from the surface of the earth.)

• Total Energy (initial) = 1/2mvi2- GMm/ (Re+h)

• According to law of conservation of energy

Total Energy (∞) =Total Energy (initial)

½ mvf2 = 1/2mvi2 - GMm/ (Re+h)

• As L.H.S = positive

1/2mvi2 - GMm/ (Re+h) ≥ 0

1/2mvi2 = GMm/ (Re+h)

• By calculating, vi2 = 2GM/ (Re+h)

• Assume Ball is thrown from earth surface h<<Re

• This implies Re+h is same as Re as we can neglect h.

• Therefore, vi2 = 2GM/ (Re) OR vi = √(2GM/Re)

• This is the initial velocity with which if the ball is thrown it will never fall back on the earth
surface.

In terms of ‘g’

g = GM/Re2

• Escape velocity can be written as Ve = √2gRe

Earth Satellites: Any object revolving around the earth.
➢ Natural Satellite: Satellite created by nature.
Example: - Moon is the only natural satellite of earth.

➢ Artificial Satellites: Human built objects orbiting the earth for practical uses. There are several
purposes which these satellites serve. Example:- Practical Uses of Artificial satellites
• Communication
• Television broadcasts
• Weather observation
• Military support
• Navigation
• Scientific research

Determining the Time Period of Earth Satellite

• As satellites move in circular orbits there will be centripetal force acting on it.

Fc = mv2/Re+h It is towards the centre.

Where, h = distance of satellite form the earth, Fc= centripetal force

FG = GmMe/(Re+h)2

where, Fg= Gravitation force, m= mass of the satellite, Me = mass of the earth

Fc=FG

mv2/Re+h = GmMe/(Re+h)2

v2=GMe/Re+h

v=√ GMe/Re+h (1)

• This is the velocity with which satellite revolve around the earth.

• The satellite covers distance = 2 π(Re+h) with velocity v.

T=2 π(Re+h)/v

= 2 π(Re+h)/ √GMe/Re+h

T=2 π(Re+h)3/2/ √ GMe)

Special Case: - h<< Re (satellite is very near to the surface of the earth)

• Then T=2 π√ Re3/GMe OR T=2 π√ Re/g

Energy of an orbiting satellite:

• m= mass of the satellite, v=velocity of the satellite

• K E = 1/2mv2

= 1/2 m (GMe/Re+h)
 • K E. =1/2 GMe/(Re+h)

• P E.= -GMem/(Re+h)

• Total Energy = K.E. + P.E.

= 1 /2 GMe/(Re+h) + -GMem/(Re+h)

• E.= GMem/2(Re+h)

• Conclusion:- P.E. = 2 x K.E.

• Total energy is negative. This means the satellite cannot escape from the earth’s gravity.

Geostationary Satellite:-Geo means earth and stationary means at rest.This means something
stationary.

• Satellites orbiting around the Earth in equatorial plane with time period equal to 24 hours.Appear to
be stationary with respect to earth.

• These satellites can receive telecommunication signals and broadcast them back to a wide area on
earth.

• Example: INSAT group of satellites.

Polar Satellites: These are low altitude satellites. This means they orbit around earth at lower heights.

• They orbit around the earth in North-South direction. Whereas earth is moving from East to West.

• A camera is fixed above this type of satellite so they can view small strips of earth.

• As earth also moves, so at each instance different types of stripes of earth can be viewed.

• They are useful in remote sensing, meteorology and environmental studies of the earth.

Weightlessness: It is a condition of free fall, in which the effect of gravity is cancelled by the inertial
(e.g., centrifugal) force resulting from orbital flight. There is no force of gravity acting on the objects.

• It is the condition in which body does not feel its weight at all.

• Example: When an apple falls from a tree it won’t feel its weight. This condition experienced by
anybody while in free-fall is known as weightlessness.

Weightlessness in the orbital motion of satellites

• In case of a satellite there is an acceleration which is acting towards the centre of the Earth.

• This acceleration is known as centripetal acceleration (a c).

• There is also earth’s acceleration which is balancing this centripetal acceleration.

 o g = ac they are equal in magnitude and they are balancing each other.

• Inside the satellites there is no acceleration which means everything is moving with uniform
velocity.

• Inside an orbiting satellite weightlessness is experienced.

KHATAM