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

91. Earth-moon system V g<0 r X moonr
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 satellitevc 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 satellitesω
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 spacecraftv 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 spacecraftv
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