Download presentation

1
Chapter 4 Gravitation Physics Beyond 2000

2
Gravity Newton

3
**Gravity The moon is performing circular motion round the earth.**

The centripetal force comes from the gravity. v Fc moon earth

4
Gravity Newton found that the gravity on the moon is the same force making an apple fall. W Ground

5
**Newton’s Law of Gravitation**

Objects attract each other with gravitational force. In the diagram, m1 and m2 are the masses of the objects and r is the distance between them. m1 m2 F r

6
**Newton’s Law of Gravitation**

Every particle of matter attracts every other particle with a force whose magnitude is G is a universal constant G = 6.67 m3kg-1s-2 m1 m2 F r Note that the law applies to particles only.

7
Example 1 Find how small the gravitation is.

8
**Shell Theorem Extends the formula**

to spherical objects like a ball, the earth, the sun and all planets.

9
**Theorem 1a. Outside a uniform spherical shell.**

The shell attracts the external particle as if all the shell’s mass were concentrated at its centre. F r m1 m2 O

10
**Theorem 1b. Outside a uniform sphere.**

The sphere attracts the external particle as if all the sphere’s mass were concentrated at its centre. m1 F F m2 O r

11
**Example 2 Outside a uniform sphere.**

The earth is almost a uniform sphere. F r m1 m2 O earth

12
**Theorem 2a. Inside a uniform spherical shell.**

The net gravitational force is zero on an object inside a uniform shell. m2 m1 The two forces on m2 cancel.

13
**Theorem 2b. Inside a uniform sphere.**

where m1 is the mass of the core with r the distance from the centre to the mass m2 m2 m1 r F

14
Example 3 Inside a uniform sphere. m2 m1 r F

15
Gravitational Field A gravitational field is a region in which any mass will experience a gravitational force. A uniform gravitational field is a field in which the gravitational force in independent of the position.

16
Field strength, g The gravitational field strength, g, is the gravitational force per unit mass on a test mass. F is the gravitational force m is the mass of the test mass g is a vector, in the same direction of F. SI unit of g is Nkg-1. test mass F m

17
**Field strength, g SI unit of g is Nkg-1.**

The gravitational field strength, g, is the gravitational force per unit mass on a test mass. F is the gravitational force m is the mass of the test mass test mass F m SI unit of g is Nkg-1.

18
**Field strength, g, outside an isolated sphere of mass M**

The gravitational field strength, g, outside an isolated sphere of mass M is M r field strength at X X O Prove it by placing a test mass m at a point X with distance r from the centre of the isolated sphere M.

19
Example 4 The field strength of the earth at the position of the moon.

20
**Field strength, g Unit of g is Nkg-1.**

g is also a measure of the acceleration of the test mass. g is also the acceleration due to gravity, unit is ms-2.

21
**Field strength, g Field strength, g. Unit Nkg-1.**

A measure of the strength of the gravitational field. Acceleration due to gravity, g. Unit ms-2. A description of the motion of a test mass in free fall.

22
Field lines We can represent the field strength by drawing field lines. The field lines for a planet are radially inward. planet Radial field

23
Field lines We can represent the field strength by drawing field lines. The field lines for a uniform field are parallel. Uniform field earth’s surface

24
Field lines The density of the field lines indicates the relative field strength. g1= 10 Nkg-1 g2= 5 Nkg-1

25
Field lines The arrow and the tangent to the field lines indicates the direction of the force acting on the test mass. direction of the force test mass

26
**The earth’s gravitational field**

Mass of the earth Me 5.98 1024 kg Radius of the earth Re 6.37 106 m O Re

27
**Gravity on the earth’s surface, go**

The gravitational field go near the earth’s surface is uniform and The value of go 9.8 Nkg-1

28
Example 5 The gravity on the earth’s surface, go.

29
Apparent Weight Use a spring-balance to measure the weight of a body. Depending on the case, the measured weight R (the apparent weight) is not equal to the gravitational force mgo. R mgo

30
Apparent Weight The reading on the spring-balance is affected by the following factors: The density of the earth crust is not uniform. The earth is not a perfect sphere. The earth is rotating.

31
**Apparent Weight The density of the earth crust is not uniform.**

Places have different density underneath. Thus the gravitational force is not uniform.

32
**Apparent Weight 2. The earth is not a perfect sphere.**

Points at the poles are closer to the centre than points on the equators. rpole < requator gpole > gequator N-pole Equator S-pole

33
**Apparent Weight X Y 3. The earth is rotating.**

Except at the pole, all points on earth are performing circular motion with the same angular velocity . However the radii of the circles may be different. X Y

34
**Apparent Weight 3. The earth is rotating.**

Consider a mass m is at point X with latitude . The radius of the circle is r = Re.cos . m Y X r Re O

35
**Apparent Weight 3. The earth is rotating.**

The net force on the mass m must be equal to the centripetal force. Fc m X Y r Re O Note that Fnet points to Y.

36
**Apparent Weight R R Fc X Y r m mgo O 3. The earth is rotating.**

The net force on the mass m must be equal to the centripetal force. So the apparent weight (normal reaction) R does not cancel the gravitational force mgo. Fc X Y r m mgo O

37
**Apparent Weight R R 3. The earth is rotating.**

The apparent weight R is not equal to the gravitational force mgo in magnitude. Fc X Y r m mgo O

38
**Apparent weight R on the equator**

mgo R The apparent field strength on the equator is

39
**Apparent weight R at the poles**

mgo The apparent field strength at the poles is

40
Example 6 Compare the apparent weights.

41
**Apparent weight at latitude **

Fc X Y r m mgo O Note that the apparent weight R is not exactly along the line through the centre of the earth.

42
**Variation of g with height and depth**

Outside the earth at height h. r m h Re Me g O h = height of the mass m from the earth’s surface

43
**Variation of g with height and depth**

Outside the earth at height h. r m h Re Me g O where go is the field strength on the earth’s surface.

44
**Variation of g with height and depth**

Outside the earth at height h. r m h Re Me g O where go is the field strength on the earth’s surface.

45
**Variation of g with height and depth**

Outside the earth at height h close to the earth’s surface. h<<Re. r m h Re Me g O where go is the field strength on the earth’s surface.

46
**Variation of g with height and depth**

Below the earth’s surface. Re r O d Me Only the core with colour gives the gravitational force. g r = Re-d

47
**Variation of g with height and depth**

Below the earth’s surface. Re r O d Me Find the mass Mr of g r = Re-d

48
**Variation of g with height and depth**

Below the earth’s surface. Re r O d Me g r = Re-d

49
**Variation of g with height and depth**

Below the earth’s surface. Re r O d Me g g r r = Re-d

50
**Variation of g with height and depth**

r < Re , g r. r > Re , earth g go r distance from the centre of the earth Re

51
**Gravitational potential energy Up**

Object inside a gravitational field has gravitational potential energy. When object falls towards the earth, it gains kinetic energy and loses gravitational potential energy. This object possesses Up earth

52
Zero potential energy By convention, the gravitational potential energy of the object is zero when its separation x from the centre of the earth is . Up = 0 earth O x

53
**Negative potential energy**

For separation less than r, the gravitational potential energy of the object is less than zero. So it is negative. Up < 0 earth O r

54
**Gravitational potential energy Up**

Definition 1 It is the negative of the work done by the gravitational force FG as the object moves from infinity to that point. earth FG O r dx

55
**Gravitational potential energy Up**

Definition 1 earth FG O r dx

56
**Gravitational potential energy Up**

Definition 2 It is the negative of the work done by the external force F to bring the object from that point to infinity. Me earth F O r dx m

57
**Gravitational potential energy Up**

Definition 2 Me earth F O r dx m

58
**Gravitational potential energy Up**

earth O r Me m

59
Example 7 Conservation of kinetic and gravitational potential energy.

60
Example 8 - Work done = gravitational potential energy

61
**Example 9 Two particles are each in the other’s gravitational field.**

Thus each particle possesses gravitational energy.

62
**System of three particles**

Each particle is in another two particles’ gravitational fields. Each particle possesses gravitational potential energy due to the other two particles. m M1 M2 r1 r2 Up of

63
**System of three particles**

Up of m M1 M2 r1 r2 r3 Up of

64
**Example 10 Up of the moon due to the earth’s gravitational field. r**

What is the Up of the earth due to the moon’s gravitational field?

65
Escape speed ve Escape speed ve is the minimum projection speed required for any object to escape from the surface of a planet without return. ve

66
Escape speed ve Escape speed ve is the minimum projection speed required for any object to escape from the surface of a planet without return. ve

67
Escape speed ve On the surface of the planet, the body possesses both kinetic energy Uk and gravitational potential energy Up. Uk ve R m M UP

68
Escape speed ve If the body is able to escape away, it means the body still possesses kinetic energy at infinity. Note that the gravitational energy of the body at infinity is zero. ve R m M

69
Escape speed ve If there is not any loss of energy on projection, the total energy of the body at lift-off = the total energy of the body at infinity ve R m M

70
Escape speed ve = kinetic energy at infinity ≧0 ve R m M

71
Escape speed ve where go is the gravitational acceleration on the surface of the earth. ve R m M

72
Escape speed ve So the escape speed from earth is ve R m M

73
Escape speed ve Example: Find ve ve R m M

74
**Gravitational potential V**

Definition: The gravitational potential at a point is the gravitational potential energy per unit test mass. where U is the gravitational potential energy of a mass m at the point

75
**Gravitational potential V**

Definition: The gravitational potential at a point is the gravitational potential energy per unit test mass. unit of V is J kg-1

76
**Gravitational potential V**

Example 12 – to find the change in gravitational potential energy. ΔU = U – Uo If ΔU >0, there is a gain in U. If ΔU <0, there is a loss in U.

77
Equipotentials Equipotentials are lines or surfaces on which all points have the same potential. The equipotentials are always perpendicular to the field lines.

78
Equipotentials The equipotentials around the earth are imaginary spherical shells centered at the earth’s centre.

79
Equipotentials The field is radial.

80
Equipotentials The equipotentials near the earth’s surface are parallel and evenly spaced surface. The field is uniform. surface

81
Equipotentials Example 13 – Earth’s equipotential.

82
**Potential V and field strength g**

83
**Potential V and field strength g**

If we consider the magnitude of g only, r

84
**Earth-moon system Me Mm r P D-r moon D earth**

The potential is the sum of the potentials due to the earth and the moon. Me earth moon Mm r P D-r D

85
Earth-moon system earth moon P D r D-r Me Mm

86
Earth-moon system V r earth moon

87
Earth-moon system V r earth moon

88
Earth-moon system V g r earth moon

89
**Earth-moon system V g=0 r X moon g = 0 at a point X between the earth**

r earth moon X g = 0 at a point X between the earth and the moon. X is a neutral point.

90
**Earth-moon system V g>0 r X moon**

r earth moon X g points to the centre of the earth if it is positive.

91
**Earth-moon system V g<0 r X moon**

r earth moon X g points to the centre of the moon if it is negative.

92
**Earth-moon system Given: Me = 5.98 × 1024 kg Mm = 7.35 × 1022 kg**

D = 3.84 × 108 m G = 6.67 × Nm2kg-2 Find: the position X at which g = 0. earth moon X x Hint:

93
**Earth-moon system Given: Me = 5.98 × 1024 kg Mm = 7.35 × 1022 kg**

D = 3.84 × 108 m G = 6.67 × Nm2kg-2 Find: the position X at which g = 0. earth moon X x Answer: x = 3.46 × 108 m

94
Earth-moon system Example 14 – potential difference near the earth’s surface.

95
Orbital motion The description of the motion of a planet round the sun. sun

96
**Orbital motion Kepler’s law: The law of orbits.**

All planets move in elliptical orbits, with the sun at one focus. sun

97
**Orbital motion Kepler’s law: 2. The law of areas.**

The area swept out in a given time by the line joining any planet to the sun is always the same. sun

98
**Orbital motion Kepler’s law: 3. The law of periods.**

The square of the period T of any planet about the sun is proportional to the cube of their mean distance r from the sun. sun

99
Orbital motion Basically, we only study the simple case of circular orbit. r

100
Orbital motion A satellite of mass m performs circular motion round the earth with speed vc . The radius of the orbit is r. r satellite earth vc

101
**Orbital motion The centripetal force is provided by the**

gravitational force. r satellite earth vc Fc

102
**Orbital motion Show that where Me is the mass of the earth r satellite**

vc Fc

103
**Orbital motion Example 15 – find the speed of a satellite. vc**

earth vc

104
**Proof of Kepler’s 3rd law in a circular orbit**

satellite 1 earth vc1 satellite 2 r2 vc2

105
**Proof of Kepler’s 3rd law in a circular orbit**

Note that the proof is true for satellites round the same planet. r1 satellite 1 earth vc1 satellite 2 r2 vc2

106
Kepler’s 3rd law Example 16 – apply Kepler’s 3rd law.

107
**Satellites Natural satellites – e.g. moon. Artificial satellites –**

e.g. communication satellites, weather satellites.

108
**Geosynchronous satellites**

A geosynchronous satellite is above the earth’s equator. It rotates about the earth with the same angular speed as the earth and in the same direction. It seems stationary by observers on earth.

109
**Geosynchronous satellites**

ω equator axis satellite h Re

110
**Geosynchronous satellites**

Find the radius of the orbit of a geosynchornous satellite. ω equator axis satellite h Re rs h + Re = rs

111
**Geosynchronous satellites**

rs = 4.23×107 m ω equator axis satellite h Re r h + Re = rs

112
**Geosynchronous satellites**

h = 3.59×107 m ω equator axis satellite h Re r h + Re = rs

113
**Parking Orbit Note that there is only one such orbit.**

It is called a parking orbit. ω equator axis satellite h Re r h + Re = rs

114
**Satellites Near the Earth’s surface**

Assume that the orbit is circular with radius r Re , the radius of the earth. The gravitational field strength go is almost a constant (9.8 N kg-1). The gravitational force provides the required centripetal force.

115
**Satellites Near the Earth’s surface**

Find vr vr satellite r Re earth

116
**Energy and Satellite Motion**

Find v and the kinetic energy Uk of the satellite. r satellite earth Me v m

117
**Energy and Satellite Motion**

The satellite in the orbit possesses both kinetic energy and gravitational energy. r satellite earth Me v m

118
**Energy and Satellite Motion**

earth Me v m Note that Uk > 0

119
**Energy and Satellite Motion**

Find Up the gravitational potential of the satellite. r satellite earth Me v m

120
**Energy and Satellite Motion**

earth Me v m Note that Up < 0

121
**Energy and Satellite Motion**

Find U, the total energy of the satellite. r satellite earth Me v m

122
**Energy and Satellite Motion**

earth Me v m Note that U < 0

123
**Energy and Satellite Motion**

U : Up : Uk = -1 : -2 : 1 r satellite earth Me v m

124
**Falling to the earth The satellite may lose energy due to air**

resistance. The total energy becomes more negative and r becomes less. r satellite earth Me v m

125
**Falling to the earth The satellite follows a spiral path towards**

earth Me v m

126
Falling to the earth As r decreases, the kinetic energy of the satellite increases and the satellite moves faster. r satellite earth Me v m

127
Falling to the earth Example 17 – Loss of energy

128
**Weightlessness in spacecraft**

v v mg The astronaut is weightless.

129
**Weightlessness in spacecraft**

We fell our weight because there is normal reaction on us. Normal reaction ground mg

130
**Weightlessness in spacecraft**

If there is not any normal reaction on us, we feel weightless. e.g. free falling mg

131
**Weightlessness in spacecraft**

v The gravitational force mg on the astronaut is the required centripetal force. He does not require any normal reaction to act on him. mg

132
**Weightlessness in spacecraft**

The astronaut is weightless. v mg

Similar presentations

OK

Gravitational Field Historical facts Geocentric Theory Heliocentric Theory – Nicholas Copernicus (1473 – 1543) Nicholas Copernicus – All planets, including.

Gravitational Field Historical facts Geocentric Theory Heliocentric Theory – Nicholas Copernicus (1473 – 1543) Nicholas Copernicus – All planets, including.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on boilers operations research Download ppt on transportation in human beings Ppt on class ab power amplifier Ppt on odisha culture of india Ppt on ready to serve beverages and more store Ppt on principles of peace building initiatives Ppt on next generation 2-stroke engine upgrade Ppt on data collection methods examples Ppt on hotel industry in india Ppt on power generation using footsteps prayer