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23PH101 - Properties of Matter and Acoustics
Ms Dhivya R
Assistant Professor
Department of Physics
Sri Ramakrishna College of Arts and Science
Coimbatore - 641 006
Tamil Nadu, India
1
Unit I - Gravitation
Newton’s law of gravitation
Mass and density of earth
Kepler’s law of planetary motion
Deduction of Newton’s law of gravitation
from Kepler’s law
Determination of G - Boy’s Experiment
Gravitational field and Gravitational potential
Variation of ‘g’ with latitude, altitude and
depth
2
Sri Ramakrishna College of Arts and Science
Newton’s law of gravitation
Newton’s law of gravitation, statement that any
particle of matter in the universe attracts any
other with a force varying directly as the
product of the masses and inversely as the
square of the distance between them.
𝐹 = 𝐺
𝑚1𝑚2
𝑟2
G = 6.673 x 10-11 N m2/kg2
3
Sri Ramakrishna College of Arts and Science
Mass and Density of Earth
The current best estimate for the mass of Earth is M⊕ =
5.9722×1024 kg.
M = Mass of earth; m = mass of a body
Force acting in the mass(m) due to acceleration due to
gravity(g) F = mg -------- (1)
According to NL = F = 𝐺
𝑀𝑚
𝑅2 ------ (2)
Both the forces are equal mg = 𝐺
𝑀𝑚
𝑅2 ------- (3)
From 3 M =
𝑅2
𝑔
𝐺
-------(4)
It is equivalent to an average density of 5515 kg/m3.
4
Sri Ramakrishna College of Arts and Science
Mass and Density of Earth
From 3 M =
𝑅2
𝑔
𝐺
-------(4)
Volume of Earth V=
4
3
𝜋𝑅3
Density of Earth 𝜌 =
𝑀
𝑉
=
𝑅2𝑔 𝐺
4
3
𝜋𝑅3
=
3𝑔
4𝜋𝑅𝐺
Inertial Mass – Mass obtained from F=mg and
Gravitational Mass – Mass obtained from
F = 𝐺
𝑀𝑚
𝑅2
5
Sri Ramakrishna College of Arts and Science
Mass and Density of Earth
From 3 M =
𝑅2
𝑔
𝐺
-------(4)
Volume of Earth V=
4
3
𝜋𝑅3
Density of Earth 𝜌 =
𝑀
𝑉
=
𝑅2𝑔 𝐺
4
3
𝜋𝑅3
=
3𝑔
4𝜋𝑅𝐺
Inertial Mass – Mass obtained from F=mg and
Gravitational Mass – Mass obtained from
F = 𝐺
𝑀𝑚
𝑅2
6
Sri Ramakrishna College of Arts and Science
Kepler’s law of planetary motion
In astronomy, Kepler’s laws of planetary motion
are three scientific laws describing the motion
of planets around the sun.
Kepler’s first law – The law of orbits
Kepler’s second law – The law of equal areas
Kepler’s third law – The law of periods
7
Sri Ramakrishna College of Arts and Science
Kepler’s First Law – The Law of
Orbits
According to Kepler’s first law, “All the planets
revolve around the sun in elliptical orbits
having the sun at one of the foci”
8
Sri Ramakrishna College of Arts and Science
Kepler’s Second Law – The Law
of Equal Areas
Kepler’s second law states, “The radius vector
drawn from the sun to the planet sweeps out
equal areas in equal intervals of time”.
9
Sri Ramakrishna College of Arts and Science
Kepler’s Third Law – The Law of
Periods
According to Kepler’s law of periods, “The
square of the time period of revolution of a
planet around the sun in an elliptical orbit is
directly proportional to the cube of its semi-
major axis”.
T2 ∝ a3
10
Sri Ramakrishna College of Arts and Science
Derivation of Newton’s Law from
Kepler’s Law
Centrifugal Force on Planet 1 = 𝐹1 = 𝑚1𝑟1𝜔2
𝜔 =
2𝜋
𝑇1
𝐹1 = 𝑚1𝑟1(
2𝜋
𝑇1
)2
11
Sri Ramakrishna College of Arts and Science
Derivation of Newton’s Law from
Kepler’s Law
Centrifugal Force on Planet 2 = 𝐹2 = 𝑚2𝑟2𝜔2
𝜔 =
2𝜋
𝑇2
𝐹2 = 𝑚2𝑟2(
2𝜋
𝑇2
)2
12
Sri Ramakrishna College of Arts and Science
Derivation of Newton’s Law from
Kepler’s Law
Now
𝐹1
𝐹2
=
𝑚1𝑟1
𝑚2𝑟2
𝑇2
𝑇1
2
According to Kepler's Third Law
𝑇2
𝑇1
2
=
𝑟2
𝑟1
3
𝐹1
𝐹2
=
𝑚1𝑟1
𝑚2𝑟2
𝑟2
𝑟1
3
=
𝑚1𝑟22
𝑚2𝑟1
2
Since force is mutual, 𝐹 ∝
𝑚
𝑟2 ; 𝐹 ∝
𝑀𝑚
𝑟2 ;
𝐹 = 𝐺
𝑀𝑚
𝑟2
13
Sri Ramakrishna College of Arts and Science
Determination of G
Boy’s Experiment
• T1 – Fixed &
• T2 – Rotatable
• RS – Small Mirror
• Suspension – QuartzFibre
• A & B – two Gold Spheres
• C & D Lead Balls
• Centre of C level with A
• Centre of D level with B
• P1&P2 Rubber Pads
14
Sri Ramakrishna College of Arts and Science
Determination of G
Force of attraction between spheres A & C
𝐹 =
𝐺𝑀𝑚
𝐴𝐶 2
Force of attraction between spheres B & D
𝐹 =
𝐺𝑀𝑚
𝐵𝐷 2
15
Sri Ramakrishna College of Arts and Science
Determination of G
Since AC = BD, the two forces are equal, Parallel
and act in opposite direction constituting a
couple
∴ 𝑡ℎ𝑒 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑑𝑒𝑓𝑒𝑙𝑒𝑐𝑡𝑖𝑛𝑔 𝑐𝑜𝑢𝑝𝑙𝑒
=
𝐺𝑀𝑚
𝐴𝐶 2
𝑥2𝑙
=
𝐺𝑀𝑚
𝑑2 𝑥2𝑙
Length of the mirror strip RS = 2l & AC = d
16
Sri Ramakrishna College of Arts and Science
Determination of G
𝐺𝑀𝑚
𝑑2
𝑥2𝑙 = 𝑐. θ
From this G can be calculated.
Mirror and gold Spheres can be treated as a
torsional pendulum and T can be found.
𝑇 =
2𝜋 𝐼
𝑐
. From this C can be found.
G found to be 6.6576 X 10-11Nm2Kg-2
17
Sri Ramakrishna College of Arts and Science
Advantages
1. The size of the apparatus is reduced –
disturbances due to convectional current is
reduced.
2. By arranging the masses at different levels,
attraction between the larger mass and
remote small mass is reduced.
3. Lamp and scale arrangement – small
deflection can also be measured
4. Quartz fibre – sensitive and accurate measure
18
Sri Ramakrishna College of Arts and Science
Gravitational Field
• The space or region around a body within
which other bodies experience its
gravitational force of attraction is called a
gravitational field.
• Vector field
19
Sri Ramakrishna College of Arts and Science
Gravitational Potential
• The work done in moving a unit mass from
infinity to a point in a gravitational field is
called the gravitational potential at that point.
• GP is always negative in sign. Highest value
being zero at infinity.
• It’s a scalar quatity.
20
Sri Ramakrishna College of Arts and Science
Intensity of Gravitational field at
a point
It is defined as the space rate of change of
gravitational potential at a point.
𝐹 = −
ⅆ𝑉
ⅆ𝑟
ⅆ𝑉 – small change in gravitational potential for a
small distance ⅆr
21
Sri Ramakrishna College of Arts and Science
Gravitational Potential due to a
point mass
Force of attraction experienced by a unit mass at
A =
𝐺.𝑚
𝑥2
Work done in displacing the unit mass from A to
B through a distance dx =
𝐺.𝑚
𝑥2 𝑑𝑥
22
Sri Ramakrishna College of Arts and Science
Gravitational Potential due to a
point mass
The potential difference between A and B =
𝛿𝑣 =
𝐺⋅𝑚
𝑥2 ⅆ𝑥
Hence total work done in moving the unit mass
from infinity to P or the potential at P =
V= 𝛿 𝑉 =
∞
𝑟
𝐺⋅𝑚
𝑥2 ⅆ𝑥 = −
𝐺⋅𝑚
𝑟
Thus, the gravitational potential has the
maximum value of zero at infinity and
decreases as the distance is decreased.
Sri Ramakrishna College of Arts and Science 23
Equipotential Surface
A surface at all the points of which the
gravitational potential is called an
equipotential surface
Sri Ramakrishna College of Arts and Science 24
Variation of G with Latitude
Centrifugal force along PB = 𝑚 𝑅 cos 𝜆 𝜔2
Net force represented by PC
= 𝑚𝑔 cos 𝜆 − 𝑚𝑅𝜔2 cos 𝜆
PA = 𝑚𝑔 sin 𝜆
PQ2=PA2+QA2
PQ2=PA2+PC2 sin𝑐𝑒 𝑄𝐴 = 𝑃𝐶
mg′ 2 = 𝑚𝑔 sin 𝜆 2 + (𝑚𝑔 cos 𝜆 − 𝑚𝑅𝜔2 cos 𝜆)2
Sri Ramakrishna College of Arts and Science 25
P
Q
O
W E
N
S
A
B
C
𝜆
Variation of G with Latitude
mg′ 2 = 𝑚𝑔 sin 𝜆 2 + (𝑚𝑔 cos 𝜆 − 𝑚𝑅𝜔2 cos 𝜆)2
mg′ 2 = 𝑚𝑔 sin 𝜆 2 + 𝑚𝑔 cos 𝜆 2 + (𝑚𝑅𝜔2 cos 𝜆)2 − 2 𝑚𝑔 cos 𝜆 (𝑚𝑅𝜔2 cos 𝜆)
mg′ 2 = 𝑚𝑔 2(sin 2𝜆 + cos2 𝜆) + (𝑚𝑔)2{(
𝑅2
𝜔4
cos 2
𝜆
𝑔2 ) − 2
𝑅𝜔2
cos2
𝜆
𝑔
}
g′
= 𝑔 {1 + (
𝑅2𝜔4 cos 2𝜆
𝑔2
) − 2
𝑅𝜔2 cos2 𝜆
𝑔
}
1 2
𝒈′ = 𝒈 {𝟏 − 𝟐
𝑹𝝎𝟐
𝐜𝐨𝐬𝟐
𝝀
𝒈
}
Sri Ramakrishna College of Arts and Science 26
Variation of G with Altitude
𝐹 = 𝑚𝑔 =
𝐺𝑀𝑚
𝑅2
𝐹′ = 𝑚𝑔′ =
𝐺𝑀𝑚
(𝑅 + ℎ)2
𝑚𝑔′
𝑚𝑔
=
𝐺𝑀𝑚
(𝑅+ℎ)2
𝐺𝑀𝑚
𝑅2
=
𝐺𝑀𝑚
(𝑅+ℎ)2 X
𝑅2
𝐺𝑀𝑚
𝑔′
𝑔
=
𝑅2
(𝑅+ℎ)2 = 𝑅2
𝑅2(1+ℎ/𝑅)2 =
1
(1+ℎ/𝑅)2 = (1 +
ℎ
𝑅
)−2
Sri Ramakrishna College of Arts and Science 27
h
O
R
P
Q m
M
Variation of G with Altitude
𝑔′
𝑔
= (1 +
ℎ
𝑅
)−2
= (1 −
2ℎ
𝑅
)
(1+x)-n form – neglecting high powers
𝒈′ = 𝒈(𝟏 −
𝟐𝒉
𝑹
)
Value of g decreases with increases in altitude
Sri Ramakrishna College of Arts and Science 28
𝐹 = 𝑚𝑔 =
𝐺𝑀𝑚
𝑅2
𝐹′ = 𝑚𝑔′ =
𝐺𝑀′𝑚
(𝑅 − ℎ)2
M =
4
3
𝜋𝑅3
𝜌
M′ =
4
3
𝜋(𝑅 − ℎ)3
𝜌
Variation of G with Depth
Sri Ramakrishna College of Arts and Science 29
P
Q
h
O
R
𝐹 =
𝑚𝑔′
𝑚𝑔
=
𝐺𝑀′𝑚
(𝑅 − ℎ)2
𝐺𝑀𝑚
𝑅2
=
𝐺𝑀′𝑚
(𝑅 − ℎ)2
𝑋
𝑅2
𝐺𝑀𝑚
𝑔′
𝑔
=
𝑀′
𝑀
𝑋
𝑅2
(𝑅 − ℎ)2
=
4
3
𝜋(𝑅 − ℎ)3
𝜌
4
3
𝜋𝑅3𝜌
𝑋
𝑅2
(𝑅 − ℎ)2
𝑔′
𝑔
=
(𝑅 − ℎ)
𝑅
= (1 −
ℎ
𝑅
)
𝒈′ = 𝒈(𝟏 −
𝒉
𝑹
)
Value of g decreases with increases in Depth
Variation of G with Depth
Sri Ramakrishna College of Arts and Science 30
Q&A
Calculate the gravitational force of attraction
between the Earth and a 70 kg man standing at a sea
level, a distance of 6.38 x 106 m from the earth’s
centre.
Solution:
Given:
m1 =5.98 x 1024 kg; m2 = 70 kg; d = 6.38 x 106 m
The value of G = 6.673 x 10-11 N m2/kg2
Now substituting the values in the Gravitational force formula, we get
F=685 N
Q&A
• Is the force of Gravity the same all over the
Earth?
Gravity isn’t the same everywhere on earth. Gravity is slightly stronger
over the places with more underground mass than places with less mass.
NASA uses two spacecraft to measure the variation in the Earth’s gravity.
These spacecraft are a part of the Gravity Recovery and Climate
Experiment (GRACE) mission.
• Is there gravity in space?
Gravity is everywhere. It gives shape to the orbits of the planets, the solar
system, and even galaxies. Gravity from the Sun reaches throughout the
solar system and beyond, keeping the planets in their orbits. Gravity from
Earth keeps the Moon and human-made satellites in orbit.
Q&A
• How is Einstein’s theory of gravity different from
Newton’s?
According to Einstein, objects move toward one another because of the
curves in space-time, not because of the force of attraction between them.
• Why are the orbits of the planets not circular?
For the orbits to be circular, it requires the planets to travel with a certain
velocity, which is extremely unlikely. If there is any change in the velocity
of the planet, the orbit will be elliptical.
Text Books
Properties of matter (Unit I – IV)
-R. Murugeshan
A Text book of Sound (Unit V)
- N. Subramanyam &
Brij Lal
34
Sri Ramakrishna College of Arts and Science

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Properties of Matter - U1 - Gravitation -PPT 1.pptx

  • 1. 23PH101 - Properties of Matter and Acoustics Ms Dhivya R Assistant Professor Department of Physics Sri Ramakrishna College of Arts and Science Coimbatore - 641 006 Tamil Nadu, India 1
  • 2. Unit I - Gravitation Newton’s law of gravitation Mass and density of earth Kepler’s law of planetary motion Deduction of Newton’s law of gravitation from Kepler’s law Determination of G - Boy’s Experiment Gravitational field and Gravitational potential Variation of ‘g’ with latitude, altitude and depth 2 Sri Ramakrishna College of Arts and Science
  • 3. Newton’s law of gravitation Newton’s law of gravitation, statement that any particle of matter in the universe attracts any other with a force varying directly as the product of the masses and inversely as the square of the distance between them. 𝐹 = 𝐺 𝑚1𝑚2 𝑟2 G = 6.673 x 10-11 N m2/kg2 3 Sri Ramakrishna College of Arts and Science
  • 4. Mass and Density of Earth The current best estimate for the mass of Earth is M⊕ = 5.9722×1024 kg. M = Mass of earth; m = mass of a body Force acting in the mass(m) due to acceleration due to gravity(g) F = mg -------- (1) According to NL = F = 𝐺 𝑀𝑚 𝑅2 ------ (2) Both the forces are equal mg = 𝐺 𝑀𝑚 𝑅2 ------- (3) From 3 M = 𝑅2 𝑔 𝐺 -------(4) It is equivalent to an average density of 5515 kg/m3. 4 Sri Ramakrishna College of Arts and Science
  • 5. Mass and Density of Earth From 3 M = 𝑅2 𝑔 𝐺 -------(4) Volume of Earth V= 4 3 𝜋𝑅3 Density of Earth 𝜌 = 𝑀 𝑉 = 𝑅2𝑔 𝐺 4 3 𝜋𝑅3 = 3𝑔 4𝜋𝑅𝐺 Inertial Mass – Mass obtained from F=mg and Gravitational Mass – Mass obtained from F = 𝐺 𝑀𝑚 𝑅2 5 Sri Ramakrishna College of Arts and Science
  • 6. Mass and Density of Earth From 3 M = 𝑅2 𝑔 𝐺 -------(4) Volume of Earth V= 4 3 𝜋𝑅3 Density of Earth 𝜌 = 𝑀 𝑉 = 𝑅2𝑔 𝐺 4 3 𝜋𝑅3 = 3𝑔 4𝜋𝑅𝐺 Inertial Mass – Mass obtained from F=mg and Gravitational Mass – Mass obtained from F = 𝐺 𝑀𝑚 𝑅2 6 Sri Ramakrishna College of Arts and Science
  • 7. Kepler’s law of planetary motion In astronomy, Kepler’s laws of planetary motion are three scientific laws describing the motion of planets around the sun. Kepler’s first law – The law of orbits Kepler’s second law – The law of equal areas Kepler’s third law – The law of periods 7 Sri Ramakrishna College of Arts and Science
  • 8. Kepler’s First Law – The Law of Orbits According to Kepler’s first law, “All the planets revolve around the sun in elliptical orbits having the sun at one of the foci” 8 Sri Ramakrishna College of Arts and Science
  • 9. Kepler’s Second Law – The Law of Equal Areas Kepler’s second law states, “The radius vector drawn from the sun to the planet sweeps out equal areas in equal intervals of time”. 9 Sri Ramakrishna College of Arts and Science
  • 10. Kepler’s Third Law – The Law of Periods According to Kepler’s law of periods, “The square of the time period of revolution of a planet around the sun in an elliptical orbit is directly proportional to the cube of its semi- major axis”. T2 ∝ a3 10 Sri Ramakrishna College of Arts and Science
  • 11. Derivation of Newton’s Law from Kepler’s Law Centrifugal Force on Planet 1 = 𝐹1 = 𝑚1𝑟1𝜔2 𝜔 = 2𝜋 𝑇1 𝐹1 = 𝑚1𝑟1( 2𝜋 𝑇1 )2 11 Sri Ramakrishna College of Arts and Science
  • 12. Derivation of Newton’s Law from Kepler’s Law Centrifugal Force on Planet 2 = 𝐹2 = 𝑚2𝑟2𝜔2 𝜔 = 2𝜋 𝑇2 𝐹2 = 𝑚2𝑟2( 2𝜋 𝑇2 )2 12 Sri Ramakrishna College of Arts and Science
  • 13. Derivation of Newton’s Law from Kepler’s Law Now 𝐹1 𝐹2 = 𝑚1𝑟1 𝑚2𝑟2 𝑇2 𝑇1 2 According to Kepler's Third Law 𝑇2 𝑇1 2 = 𝑟2 𝑟1 3 𝐹1 𝐹2 = 𝑚1𝑟1 𝑚2𝑟2 𝑟2 𝑟1 3 = 𝑚1𝑟22 𝑚2𝑟1 2 Since force is mutual, 𝐹 ∝ 𝑚 𝑟2 ; 𝐹 ∝ 𝑀𝑚 𝑟2 ; 𝐹 = 𝐺 𝑀𝑚 𝑟2 13 Sri Ramakrishna College of Arts and Science
  • 14. Determination of G Boy’s Experiment • T1 – Fixed & • T2 – Rotatable • RS – Small Mirror • Suspension – QuartzFibre • A & B – two Gold Spheres • C & D Lead Balls • Centre of C level with A • Centre of D level with B • P1&P2 Rubber Pads 14 Sri Ramakrishna College of Arts and Science
  • 15. Determination of G Force of attraction between spheres A & C 𝐹 = 𝐺𝑀𝑚 𝐴𝐶 2 Force of attraction between spheres B & D 𝐹 = 𝐺𝑀𝑚 𝐵𝐷 2 15 Sri Ramakrishna College of Arts and Science
  • 16. Determination of G Since AC = BD, the two forces are equal, Parallel and act in opposite direction constituting a couple ∴ 𝑡ℎ𝑒 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑑𝑒𝑓𝑒𝑙𝑒𝑐𝑡𝑖𝑛𝑔 𝑐𝑜𝑢𝑝𝑙𝑒 = 𝐺𝑀𝑚 𝐴𝐶 2 𝑥2𝑙 = 𝐺𝑀𝑚 𝑑2 𝑥2𝑙 Length of the mirror strip RS = 2l & AC = d 16 Sri Ramakrishna College of Arts and Science
  • 17. Determination of G 𝐺𝑀𝑚 𝑑2 𝑥2𝑙 = 𝑐. θ From this G can be calculated. Mirror and gold Spheres can be treated as a torsional pendulum and T can be found. 𝑇 = 2𝜋 𝐼 𝑐 . From this C can be found. G found to be 6.6576 X 10-11Nm2Kg-2 17 Sri Ramakrishna College of Arts and Science
  • 18. Advantages 1. The size of the apparatus is reduced – disturbances due to convectional current is reduced. 2. By arranging the masses at different levels, attraction between the larger mass and remote small mass is reduced. 3. Lamp and scale arrangement – small deflection can also be measured 4. Quartz fibre – sensitive and accurate measure 18 Sri Ramakrishna College of Arts and Science
  • 19. Gravitational Field • The space or region around a body within which other bodies experience its gravitational force of attraction is called a gravitational field. • Vector field 19 Sri Ramakrishna College of Arts and Science
  • 20. Gravitational Potential • The work done in moving a unit mass from infinity to a point in a gravitational field is called the gravitational potential at that point. • GP is always negative in sign. Highest value being zero at infinity. • It’s a scalar quatity. 20 Sri Ramakrishna College of Arts and Science
  • 21. Intensity of Gravitational field at a point It is defined as the space rate of change of gravitational potential at a point. 𝐹 = − ⅆ𝑉 ⅆ𝑟 ⅆ𝑉 – small change in gravitational potential for a small distance ⅆr 21 Sri Ramakrishna College of Arts and Science
  • 22. Gravitational Potential due to a point mass Force of attraction experienced by a unit mass at A = 𝐺.𝑚 𝑥2 Work done in displacing the unit mass from A to B through a distance dx = 𝐺.𝑚 𝑥2 𝑑𝑥 22 Sri Ramakrishna College of Arts and Science
  • 23. Gravitational Potential due to a point mass The potential difference between A and B = 𝛿𝑣 = 𝐺⋅𝑚 𝑥2 ⅆ𝑥 Hence total work done in moving the unit mass from infinity to P or the potential at P = V= 𝛿 𝑉 = ∞ 𝑟 𝐺⋅𝑚 𝑥2 ⅆ𝑥 = − 𝐺⋅𝑚 𝑟 Thus, the gravitational potential has the maximum value of zero at infinity and decreases as the distance is decreased. Sri Ramakrishna College of Arts and Science 23
  • 24. Equipotential Surface A surface at all the points of which the gravitational potential is called an equipotential surface Sri Ramakrishna College of Arts and Science 24
  • 25. Variation of G with Latitude Centrifugal force along PB = 𝑚 𝑅 cos 𝜆 𝜔2 Net force represented by PC = 𝑚𝑔 cos 𝜆 − 𝑚𝑅𝜔2 cos 𝜆 PA = 𝑚𝑔 sin 𝜆 PQ2=PA2+QA2 PQ2=PA2+PC2 sin𝑐𝑒 𝑄𝐴 = 𝑃𝐶 mg′ 2 = 𝑚𝑔 sin 𝜆 2 + (𝑚𝑔 cos 𝜆 − 𝑚𝑅𝜔2 cos 𝜆)2 Sri Ramakrishna College of Arts and Science 25 P Q O W E N S A B C 𝜆
  • 26. Variation of G with Latitude mg′ 2 = 𝑚𝑔 sin 𝜆 2 + (𝑚𝑔 cos 𝜆 − 𝑚𝑅𝜔2 cos 𝜆)2 mg′ 2 = 𝑚𝑔 sin 𝜆 2 + 𝑚𝑔 cos 𝜆 2 + (𝑚𝑅𝜔2 cos 𝜆)2 − 2 𝑚𝑔 cos 𝜆 (𝑚𝑅𝜔2 cos 𝜆) mg′ 2 = 𝑚𝑔 2(sin 2𝜆 + cos2 𝜆) + (𝑚𝑔)2{( 𝑅2 𝜔4 cos 2 𝜆 𝑔2 ) − 2 𝑅𝜔2 cos2 𝜆 𝑔 } g′ = 𝑔 {1 + ( 𝑅2𝜔4 cos 2𝜆 𝑔2 ) − 2 𝑅𝜔2 cos2 𝜆 𝑔 } 1 2 𝒈′ = 𝒈 {𝟏 − 𝟐 𝑹𝝎𝟐 𝐜𝐨𝐬𝟐 𝝀 𝒈 } Sri Ramakrishna College of Arts and Science 26
  • 27. Variation of G with Altitude 𝐹 = 𝑚𝑔 = 𝐺𝑀𝑚 𝑅2 𝐹′ = 𝑚𝑔′ = 𝐺𝑀𝑚 (𝑅 + ℎ)2 𝑚𝑔′ 𝑚𝑔 = 𝐺𝑀𝑚 (𝑅+ℎ)2 𝐺𝑀𝑚 𝑅2 = 𝐺𝑀𝑚 (𝑅+ℎ)2 X 𝑅2 𝐺𝑀𝑚 𝑔′ 𝑔 = 𝑅2 (𝑅+ℎ)2 = 𝑅2 𝑅2(1+ℎ/𝑅)2 = 1 (1+ℎ/𝑅)2 = (1 + ℎ 𝑅 )−2 Sri Ramakrishna College of Arts and Science 27 h O R P Q m M
  • 28. Variation of G with Altitude 𝑔′ 𝑔 = (1 + ℎ 𝑅 )−2 = (1 − 2ℎ 𝑅 ) (1+x)-n form – neglecting high powers 𝒈′ = 𝒈(𝟏 − 𝟐𝒉 𝑹 ) Value of g decreases with increases in altitude Sri Ramakrishna College of Arts and Science 28
  • 29. 𝐹 = 𝑚𝑔 = 𝐺𝑀𝑚 𝑅2 𝐹′ = 𝑚𝑔′ = 𝐺𝑀′𝑚 (𝑅 − ℎ)2 M = 4 3 𝜋𝑅3 𝜌 M′ = 4 3 𝜋(𝑅 − ℎ)3 𝜌 Variation of G with Depth Sri Ramakrishna College of Arts and Science 29 P Q h O R
  • 30. 𝐹 = 𝑚𝑔′ 𝑚𝑔 = 𝐺𝑀′𝑚 (𝑅 − ℎ)2 𝐺𝑀𝑚 𝑅2 = 𝐺𝑀′𝑚 (𝑅 − ℎ)2 𝑋 𝑅2 𝐺𝑀𝑚 𝑔′ 𝑔 = 𝑀′ 𝑀 𝑋 𝑅2 (𝑅 − ℎ)2 = 4 3 𝜋(𝑅 − ℎ)3 𝜌 4 3 𝜋𝑅3𝜌 𝑋 𝑅2 (𝑅 − ℎ)2 𝑔′ 𝑔 = (𝑅 − ℎ) 𝑅 = (1 − ℎ 𝑅 ) 𝒈′ = 𝒈(𝟏 − 𝒉 𝑹 ) Value of g decreases with increases in Depth Variation of G with Depth Sri Ramakrishna College of Arts and Science 30
  • 31. Q&A Calculate the gravitational force of attraction between the Earth and a 70 kg man standing at a sea level, a distance of 6.38 x 106 m from the earth’s centre. Solution: Given: m1 =5.98 x 1024 kg; m2 = 70 kg; d = 6.38 x 106 m The value of G = 6.673 x 10-11 N m2/kg2 Now substituting the values in the Gravitational force formula, we get F=685 N
  • 32. Q&A • Is the force of Gravity the same all over the Earth? Gravity isn’t the same everywhere on earth. Gravity is slightly stronger over the places with more underground mass than places with less mass. NASA uses two spacecraft to measure the variation in the Earth’s gravity. These spacecraft are a part of the Gravity Recovery and Climate Experiment (GRACE) mission. • Is there gravity in space? Gravity is everywhere. It gives shape to the orbits of the planets, the solar system, and even galaxies. Gravity from the Sun reaches throughout the solar system and beyond, keeping the planets in their orbits. Gravity from Earth keeps the Moon and human-made satellites in orbit.
  • 33. Q&A • How is Einstein’s theory of gravity different from Newton’s? According to Einstein, objects move toward one another because of the curves in space-time, not because of the force of attraction between them. • Why are the orbits of the planets not circular? For the orbits to be circular, it requires the planets to travel with a certain velocity, which is extremely unlikely. If there is any change in the velocity of the planet, the orbit will be elliptical.
  • 34. Text Books Properties of matter (Unit I – IV) -R. Murugeshan A Text book of Sound (Unit V) - N. Subramanyam & Brij Lal 34 Sri Ramakrishna College of Arts and Science