1 Gravitation © David Hoult 2009
2 © David Hoult 2009
3 © David Hoult 2009
4 © David Hoult 2009
5 © David Hoult 2009
6 © David Hoult 2009
7 © David Hoult 2009
8 F a m1m2 © David Hoult 2009
9 © David Hoult 2009
10 © David Hoult 2009
11 © David Hoult 2009
12 © David Hoult 2009
13 © David Hoult 2009
14 © David Hoult 2009
15 1 F a r2 © David Hoult 2009
16 m1m2 F = G r2 where G is the universal gravitation constant© David Hoult 2009
17 m1m2 F = G r2 N m2 kg-2 where G is the universal gravitation constant© David Hoult 2009
18 Testing the Inverse Square Law of Gravitation© David Hoult 2009
19 © David Hoult 2009
20 9.8 = × 10-3 ms-2 602 © David Hoult 2009
21 9.8 = × 10-3 ms-2 602 v2 a = r © David Hoult 2009
22 9.8 = 2.72 × 10-3 ms-2 602 v2 a = r r = 3.84 × 108 m T = 27.3 days© David Hoult 2009
23 Centripetal acceleration of the moon (caused by the force of gravity)2.72 × 10-3 ms-2 © David Hoult 2009
24 The inverse square law is a good theoryConclusion The inverse square law is a good theory © David Hoult 2009
25 Relation between g and G© David Hoult 2009
26 Relation between g and G© David Hoult 2009
27 Relation between g and G© David Hoult 2009
28 Relation between g and G© David Hoult 2009
29 we have assumed the equivalence of inertial and gravitational mass© David Hoult 2009
30 Gravitational Field Strength© David Hoult 2009
31 The g.f.s. at a point in a gravitational field is the force per unit mass acting on point mass© David Hoult 2009
32 The g.f.s. at a point in a gravitational field is the force per unit mass acting on point massUnits Nkg-1 © David Hoult 2009
33 “Force per unit mass” is equivalent to acceleration© David Hoult 2009
34 G.f.s. is another name for acceleration due to gravity© David Hoult 2009
35 © David Hoult 2009
36 © David Hoult 2009
37 1 g a r2 © David Hoult 2009
38 1 g a r2 © David Hoult 2009
39 © David Hoult 2009
40 1 outside the sphere g a r2
41 1 outside the sphere g a r2 © David Hoult 2009
42 1 outside the sphere g a r2 inside the sphere g a r © David Hoult 2009
43 1 outside the sphere g a r2 inside the sphere g a r © David Hoult 2009
44 © David Hoult 2009
45
46 © David Hoult 2009
47 World High Jump Record... © David Hoult 2009
48 World High Jump Record... on Mars ? © David Hoult 2009
49 © David Hoult 2009
50 © David Hoult 2009
51 maximum height, s depends on:© David Hoult 2009
52 maximum height, s depends on:initial velocity, u © David Hoult 2009
53 maximum height, s depends on:initial velocity, u acceleration due to gravity, g © David Hoult 2009
54 so, for a given initial velocityu2 = -2gs so, for a given initial velocity © David Hoult 2009
55 so, for a given initial velocityu2 = -2gs so, for a given initial velocity gs = a constant © David Hoult 2009
56 For a given initial velocity, the maximum height reached by the body is inversely proportional to the acceleration due to gravity © David Hoult 2009
57 1 s a g © David Hoult 2009
58 1 s a g sg = a constant © David Hoult 2009
59 1 s a g gs = a constant g1s1 = g2s2 or s1 g2 = s2 g1© David Hoult 2009
60 Gravitational Potential© David Hoult 2009
61 The potential at a point in a gravitational field is the work done per unit mass moving point mass from infinity to that point © David Hoult 2009
62 units of potential J kg-1The potential at a point in a gravitational field is the work done per unit mass moving point mass from infinity to that point units of potential J kg-1 © David Hoult 2009
63 © David Hoult 2009
64 © David Hoult 2009
65 © David Hoult 2009
66 w = Fs but in this situation the force is not of constant magnitude© David Hoult 2009
67 w = Fs but in this situation the force is not of constant magnitude© David Hoult 2009
68 It is clear that the work done will depend on:© David Hoult 2009
69 It is clear that the work done will depend on:the mass of the planet, M © David Hoult 2009
70 It is clear that the work done will depend on:the mass of the planet, M the distance, r of point p from the planet © David Hoult 2009
71 It is clear that the work done will depend on:the mass of the planet, M guess: w a M the distance, r of point p from the planet © David Hoult 2009
72 It is clear that the work done will depend on:the mass of the planet, M guess: w a M the distance, r of point p from the planet guess: w a 1/r © David Hoult 2009
73 ...it can be shown that... © David Hoult 2009
74 GM w = r © David Hoult 2009
75 “at infinity” means that the body is out of the gravitational fieldA body at infinity, has zero gravitational potential “at infinity” means that the body is out of the gravitational field © David Hoult 2009
76 All bodies fall to their lowest state of potential (energy)A body at infinity, has zero gravitational potential “at infinity” means that the body is out of the gravitational field All bodies fall to their lowest state of potential (energy) © David Hoult 2009
77 A body at infinity, has zero gravitational potential“at infinity” means that the body is out of the gravitational field All bodies fall to their lowest state of potential (energy) All gravitational potentials are therefore negative quantities © David Hoult 2009
78 GM V = r © David Hoult 2009
79 GM V = r Therefore the gravitational potential energy possessed by a body of mass m placed at point p is given by © David Hoult 2009
80 GM V = r The gravitational potential energy possessed by a body of mass m placed at point p is given by G P E = V m © David Hoult 2009
81 Escape Velocity © David Hoult 2009
82 © David Hoult 2009
83 © David Hoult 2009
84 © David Hoult 2009
85 © David Hoult 2009
86 G P E = zero © David Hoult 2009
87 G P E = zero To find the minimum velocity, ve which will cause the rocket to escape the Earth’s gravity, assume K E of distant rocket is also equal to zero. © David Hoult 2009
88 G P E = zero To find the minimum velocity, ve which will cause the rocket to escape the Earth’s gravity, assume K E of distant rocket is also equal to zero. As the body is moving away from the planet it is losing K E and gaining G P E © David Hoult 2009
89 G P E = zero To find the minimum velocity, ve which will cause the rocket to escape the Earth’s gravity, assume K E of distant rocket is also equal to zero. As the body is moving away from the planet it is losing K E and gaining G P E D K E = D G P E © David Hoult 2009
90 If the mass of the rocket is m, then the G P E it possesses at the surface of the planet isGMm G P E = R © David Hoult 2009
91 If the mass of the rocket is m, then the G P E it possesses at the surface of the planet isGMm G P E = R GMm D G P E = r © David Hoult 2009
92 GMm G P E = R GMm D G P E = R ½mve2If the mass of the rocket is m, then the G P E it possesses at the surface of the planet is GMm G P E = R GMm D G P E = R D K E = ½mve2 © David Hoult 2009
93 GMm ½mve2 = R © David Hoult 2009
94 Also, as g = GM/R2 © David Hoult 2009
95 Also, as g = GM/R2 © David Hoult 2009