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Chapter 4 Gravitation Physics Beyond 2000 Gravity Newton ry/newtongrav.htmlhttp://csep10.phys.utk.edu/astr161/lect/histo.

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Presentation on theme: "Chapter 4 Gravitation Physics Beyond 2000 Gravity Newton ry/newtongrav.htmlhttp://csep10.phys.utk.edu/astr161/lect/histo."— Presentation transcript:

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2 Chapter 4 Gravitation Physics Beyond 2000

3 Gravity Newton ry/newtongrav.htmlhttp://csep10.phys.utk.edu/astr161/lect/histo ry/newtongrav.html 9/0,5716, ,00.htmlhttp://www.britannica.com/bcom/eb/article/ 9/0,5716, ,00.html ges/gravity/g1.htmlhttp://www.nelsonitp.com/physics/guide/pa ges/gravity/g1.html

4 Gravity The moon is performing circular motion round the earth. The centripetal force comes from the gravity. FcFc earth moon v

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

6 Newtons Law of Gravitation Objects attract each other with gravitational force. In the diagram, m 1 and m 2 are the masses of the objects and r is the distance between them. m1m1 m2m2 F F r

7 Newtons Law of Gravitation Every particle of matter attracts every other particle with a force whose magnitude is m1m1 m2m2 F F r G is a universal constant G = m 3 kg -1 s -2 Note that the law applies to particles only.

8 Example 1 Find how small the gravitation is.

9 Shell Theorem Extends the formula to spherical objects like a ball, the earth, the sun and all planets.

10 Theorem 1a. Outside a uniform spherical shell. The shell attracts the external particle as if all the shells mass were concentrated at its centre. FF r m1m1 m2m2 O

11 Theorem 1b. Outside a uniform sphere. The sphere attracts the external particle as if all the spheres mass were concentrated at its centre. FF r m1m1 m2m2 O

12 Example 2 Outside a uniform sphere. The earth is almost a uniform sphere. F r m1m1 m2m2 O earth F

13 Theorem 2a. Inside a uniform spherical shell. The net gravitational force is zero on an object inside a uniform shell. m2m2 m1m1 The two forces on m 2 cancel.

14 Theorem 2b. Inside a uniform sphere. m2m2 m1m1 r F where m 1 is the mass of the core with r the distance from the centre to the mass m 2

15 Example 3 Inside a uniform sphere. m2m2 m1m1 r F

16 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. mhttp://saturn.vcu.edu/~rgowdy/mod/g33/s.ht m

17 Field strength, g The gravitational field strength, g, is the gravitational force per unit mass on a test mass. test mass F m 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.

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

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

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

21 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.

22 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.

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

24 Field lines We can represent the field strength by drawing field lines. The field lines for a uniform field are parallel. Uniform field earths surface

25 Field lines The density of the field lines indicates the relative field strength. g 1 = 10 Nkg -1 g 2 = 5 Nkg -1

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

27 The earths gravitational field Mass of the earth M e kg Radius of the earth R e m ReRe O

28 Gravity on the earths surface, g o The gravitational field g o near the earths surface is uniform and The value of g o 9.8 Nkg -1

29 Example 5 The gravity on the earths surface, g o.

30 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 mg o. R mg o

31 Apparent Weight The reading on the spring-balance is affected by the following factors: 1.The density of the earth crust is not uniform. 2.The earth is not a perfect sphere. 3.The earth is rotating.

32 Apparent Weight 1.The density of the earth crust is not uniform. Places have different density underneath. Thus the gravitational force is not uniform.

33 Apparent Weight 2. The earth is not a perfect sphere. Points at the poles are closer to the centre than points on the equators. r pole < r equator g pole > g equator N-pole S-pole Equator

34 Apparent Weight 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

35 Apparent Weight 3. The earth is rotating. Consider a mass m is at point X with latitude. The radius of the circle is r = R e.cos. X r ReRe m O Y

36 Apparent Weight 3. The earth is rotating. The net force on the mass m must be equal to the centripetal force. X r ReRe m O Y FcFc Note that F net points to Y.

37 Apparent Weight R 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 mg o. X r m O Y FcFc R mg o

38 Apparent Weight R 3. The earth is rotating. The apparent weight R is not equal to the gravitational force mg o in magnitude. X r m O Y FcFc R mg o

39 Apparent weight R on the equator The apparent field strength on the equator is mg o R

40 Apparent weight R at the poles The apparent field strength at the poles is mg o R

41 Example 6 Compare the apparent weights.

42 Apparent weight at latitude X r m O Y FcFc R mg o Note that the apparent weight R is not exactly along the line through the centre of the earth.

43 Variation of g with height and depth Outside the earth at height h. h = height of the mass m from the earths surface rm h ReRe MeMe g O

44 Variation of g with height and depth Outside the earth at height h. r m h ReRe MeMe g O where g o is the field strength on the earths surface.

45 Variation of g with height and depth Outside the earth at height h. r m h ReRe MeMe g O where g o is the field strength on the earths surface.

46 Variation of g with height and depth Outside the earth at height h close to the earths surface. h<

47 Variation of g with height and depth Below the earths surface. ReRe r Od MeMe g Only the core with colour gives the gravitational force. r = R e -d

48 Variation of g with height and depth Below the earths surface. ReRe r Od MeMe g Find the mass M r of r = R e -d

49 Variation of g with height and depth Below the earths surface. ReRe r Od MeMe g r = R e -d

50 Variation of g with height and depth Below the earths surface. ReRe r Od MeMe g r = R e -d g r

51 Variation of g with height and depth earth g gogo 0 r distance from the centre of the earth ReRe 1.r < R e, g r. 2.r > R e,

52 Gravitational potential energy U p 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 U p earth

53 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. U p = 0 earth x O

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

55 Gravitational potential energy U p Definition 1 It is the negative of the work done by the gravitational force F G as the object moves from infinity to that point. earth O r FGFG dx

56 Gravitational potential energy U p Definition 1 earth O r FGFG dx

57 Gravitational potential energy U p Definition 2 It is the negative of the work done by the external force F to bring the object from that point to infinity. earth O r F dx MeMe m

58 Gravitational potential energy U p Definition 2 earth O r F dx MeMe m

59 Gravitational potential energy U p earth O r MeMe m

60 Example 7 Conservation of kinetic and gravitational potential energy.

61 Example 8 - Work done = gravitational potential energy

62 Example 9 Two particles are each in the others gravitational field. Thus each particle possesses gravitational energy.

63 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 M1M1 M2M2 r1r1 r2r2 U p of

64 System of three particles U p of m M1M1 M2M2 r1r1 r2r2 r3r3

65 Example 10 U p of the moon due to the earths gravitational field. r earth moon What is the U p of the earth due to the moons gravitational field?

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

67 Escape speed v e Escape speed v e is the minimum projection speed required for any object to escape from the surface of a planet without return. veve

68 Escape speed v e On the surface of the planet, the body possesses both kinetic energy U k and gravitational potential energy U p. veve R m M UkUk UPUP

69 Escape speed v e 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. veve R m M

70 Escape speed v e veve R m M 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

71 Escape speed v e veve R m M = kinetic energy at infinity 0

72 Escape speed v e veve R m M where g o is the gravitational acceleration on the surface of the earth.

73 Escape speed v e veve R m M So the escape speed from earth is

74 Escape speed v e veve R m M Example: Find v e

75 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

76 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

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

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

79 Equipotentials The equipotentials around the earth are imaginary spherical shells centered at the earths centre.

80 Equipotentials The field is radial.

81 Equipotentials The equipotentials near the earths surface are parallel and evenly spaced surface. The field is uniform. surface

82 Equipotentials Example 13 – Earths equipotential.

83 Potential V and field strength g r

84 r If we consider the magnitude of g only,

85 Earth-moon system The potential is the sum of the potentials due to the earth and the moon. earth moon P D r D-r MeMe MmMm

86 Earth-moon system earth moon P D r D-r MeMe MmMm

87 Earth-moon system earth moon r V 0

88 Earth-moon system earth moon r V 0

89 Earth-moon system earth moon r V 0 g

90 Earth-moon system earth moon r V 0 g=0 g = 0 at a point X between the earth and the moon. X is a neutral point. X

91 Earth-moon system earth moon r V 0 g>0 g points to the centre of the earth if it is positive. X

92 Earth-moon system earth moon r V 0 g<0 g points to the centre of the moon if it is negative. X

93 Earth-moon system Given: M e = 5.98 × kg M m = 7.35 × kg D = 3.84 × 10 8 m G = 6.67 × Nm 2 kg -2 Find: the position X at which g = 0. earth moon X x Hint:

94 Earth-moon system Given: M e = 5.98 × kg M m = 7.35 × kg D = 3.84 × 10 8 m G = 6.67 × Nm 2 kg -2 Find: the position X at which g = 0. earth moon X x Answer: x = 3.46 × 10 8 m

95 Earth-moon system Example 14 – potential difference near the earths surface.

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

97 Orbital motion Keplers law: 1.The law of orbits. All planets move in elliptical orbits, with the sun at one focus. sun

98 Orbital motion Keplers 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

99 Orbital motion Keplers 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

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

101 Orbital motion A satellite of mass m performs circular motion round the earth with speed v c. The radius of the orbit is r. r satellite earth vcvc

102 Orbital motion The centripetal force is provided by the gravitational force. r satellite earth vcvc FcFc

103 Orbital motion Show that r satellite earth vcvc FcFc where M e is the mass of the earth

104 Orbital motion Example 15 – find the speed of a satellite. r satellite earth vcvc

105 Proof of Keplers 3 rd law in a circular orbit r1r1 satellite 1 earth v c1 r2r2 satellite 2 v c2

106 Proof of Keplers 3 rd law in a circular orbit r1r1 satellite 1 earth v c1 r2r2 satellite 2 v c2 Note that the proof is true for satellites round the same planet.

107 Keplers 3 rd law Example 16 – apply Keplers 3 rd law.

108 Satellites Natural satellites – e.g. moon. Artificial satellites – e.g. communication satellites, weather satellites.

109 Geosynchronous satellites A geosynchronous satellite is above the earths 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.

110 Geosynchronous satellites ω equator axis satellite h ReRe

111 Geosynchronous satellites Find the radius of the orbit of a geosynchornous satellite. ω equator axis satellite hReRe rsrs h + R e = r s

112 Geosynchronous satellites r s = 4.23×10 7 m ω equator axis satellite hReRe r h + R e = r s

113 Geosynchronous satellites h = 3.59×10 7 m ω equator axis satellite hReRe r h + R e = r s

114 Parking Orbit Note that there is only one such orbit. It is called a parking orbit. ω equator axis satellite hReRe r h + R e = r s

115 Satellites Near the Earths surface Assume that the orbit is circular with radius r R e, the radius of the earth. The gravitational field strength g o is almost a constant (9.8 N kg -1 ). The gravitational force provides the required centripetal force.

116 Satellites Near the Earths surface r R e satellite earth vrvr Find v r

117 Energy and Satellite Motion Find v and the kinetic energy U k of the satellite. r satellite earth M e v m

118 Energy and Satellite Motion The satellite in the orbit possesses both kinetic energy and gravitational energy. r satellite earth M e v m

119 Energy and Satellite Motion r satellite earth M e v m Note that U k > 0

120 Energy and Satellite Motion Find U p the gravitational potential of the satellite. r satellite earth M e v m

121 Energy and Satellite Motion r satellite earth M e v m Note that U p < 0

122 Energy and Satellite Motion r satellite earth M e v m Find U, the total energy of the satellite.

123 Energy and Satellite Motion r satellite earth M e v m Note that U < 0

124 Energy and Satellite Motion r satellite earth M e v m U : U p : U k = -1 : -2 : 1

125 Falling to the earth r satellite earth M e v m The satellite may lose energy due to air resistance. The total energy becomes more negative and r becomes less.

126 Falling to the earth r satellite earth M e v m The satellite follows a spiral path towards the earth.

127 Falling to the earth r satellite earth M e v m As r decreases, the kinetic energy of the satellite increases and the satellite moves faster.

128 Falling to the earth Example 17 – Loss of energy

129 Weightlessness in spacecraft mg v v The astronaut is weightless.

130 Weightlessness in spacecraft We fell our weight because there is normal reaction on us. mg Normal reaction ground

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

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

133 Weightlessness in spacecraft v mg The astronaut is weightless.


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