Upcoming SlideShare
×

# Gravitational Fields

4,117 views

Published on

• Full Name
Comment goes here.

Are you sure you want to Yes No
Your message goes here
• Be the first to comment

### Gravitational Fields

1. 1. Gravitational Fields and Forces
2. 2. Gravitational Force and Field <ul><li>Newton proposed that a force of attraction exists between any two masses. </li></ul><ul><li>This force law applies to point masses not extended masses </li></ul><ul><li>However the interaction between two spherical masses is the same as if the masses were concentrated at the centres of the spheres. </li></ul>
3. 3. Newton´s Law of Universal Gravitation <ul><li>Newton proposed that </li></ul><ul><li>“ every particle of matter in the universe attracts every other particle with a force which is directly proportional to the product of their masses, and inversely proportional to the square of their distance apart ” </li></ul>
4. 4. <ul><li>This can be written as </li></ul><ul><li>F = G m 1 m 2 r 2 </li></ul><ul><li>Where G is Newton´s constant of Universal Gravitation </li></ul><ul><li>It has a value of 6.67 x 10 -11 Nm 2 kg -2 </li></ul>
5. 5. Calculate the gravitational pull of the Earth on the moon <ul><li>M m =0.01230*M e </li></ul><ul><li>M e =5.976 x 10 24 kg </li></ul><ul><li>R = 384400 km </li></ul>
6. 6. Gravitational Field Strength <ul><li>A mass M creates a gravitational field in space around it. </li></ul><ul><li>If a mass m is placed at some point in space around the mass M it will experience the existance of the field in the form of a gravitational force </li></ul>
7. 7. <ul><li>We define the gravitational field strength as the ratio of the force the mass m would experience to the mass, M </li></ul><ul><li>That is the gravitational field strength at a point, is the force exerted per unit mass on a particle of small mass placed at that point </li></ul>
8. 8. <ul><li>The force experienced by a mass m placed a distance r from a mass M is </li></ul><ul><li>F = G Mm r 2 </li></ul><ul><li>And so the gravitational field strength of the mass M is given by dividing both sides by m </li></ul><ul><li>g = G M r 2 </li></ul>
9. 9. <ul><li>The units of gravitational field strength are N kg -1 </li></ul><ul><li>The gravitational field strength is a vector quantity whose direction is given by the direction of the force a mass would experience if placed at the point of interest </li></ul>
10. 10. Field Strength at the Surface of a Planet <ul><li>If we replace the particle M with a sphere of mass M and radius R then relying on the fact that the sphere behaves as a point mass situated at its centre the field strength at the surface of the sphere will be given by </li></ul><ul><li>g = G M R 2 </li></ul>
11. 11. <ul><li>If the sphere is the Earth then we have </li></ul><ul><li>g = G M e R e 2 </li></ul><ul><li>But the field strength is equal to the acceleration that is produced on the mass, hence we have that the acceleration of free fall at the surface of the Earth, g </li></ul><ul><li>g = G M e R e 2 </li></ul>
12. 12. Is g = 9.81 ? <ul><li>Use the data from before and verify whether the acceleration due to gravity is 9.81 ms -1 </li></ul>
13. 13. Gravitational Energy and Potential <ul><li>We know that the graviational potential energy increases as a mass is raised above the Earth </li></ul><ul><li>The work done in moving a mass between two points is positive when moving away from the Earth </li></ul><ul><li>By definition the gravitational potential energy is taken as being zero at infinity </li></ul><ul><li>It is a scalar quantity </li></ul>
14. 14. <ul><li>The gravitational potential at any point in the Earth´s field is given by the formula </li></ul><ul><li>V = - G M e r </li></ul><ul><li>Where r is the distance from the centre of the Earth (providing r >R) </li></ul><ul><li>The negative sign allows for the fact that all the potentials are negative as they have to increase to zero </li></ul>
15. 15. Definition <ul><li>The potential is therefore a measure of the amount of work that has to be done to move particles between points in a gravitational field and its units are J kg –1 </li></ul><ul><li>The work done is independent of the path taken between the two points in the field, as it is the difference between the initial and final potentials that give the value </li></ul>
16. 16. Gravitational potential and strength <ul><li>The gravitational field strength is a vector quantity whose direction is given by the direction of the force a mass would experience if placed at the point of interest </li></ul><ul><li>g = G M r 2 </li></ul><ul><li>Compare this to gravitational potential </li></ul><ul><li>V = - G M </li></ul><ul><li>r </li></ul><ul><li>And we get the relationship that </li></ul><ul><li> g = - V </li></ul><ul><li> r </li></ul>
17. 17. Equipotentials <ul><li>Equipotentials join points of </li></ul><ul><li>equal potential together </li></ul><ul><li>They are always perpendicular </li></ul><ul><li>to field lines </li></ul><ul><li>They are very simple for radial and uniform fields </li></ul>
18. 18. <ul><li>In this image the lines are equally spaced…it is a uniform field </li></ul><ul><li>In the real world the lines are surfaces, but we cant show that on paper very well </li></ul>
19. 19. Equipotentials for 2 point masses is like two positive charges
20. 20. Escape Speed <ul><li>The escape speed is the speed required for a projectile to leave the Earth´s gravitational attraction. </li></ul><ul><li>i.e. To get to infinity! </li></ul>
21. 21. <ul><li>If the potential at the Earth´s surface is </li></ul><ul><li>V = - G M e R e </li></ul><ul><li>Then the E p change to get to infinity is </li></ul><ul><li>G M e x m R e </li></ul><ul><li>Where m is the mass of the projectile </li></ul>
22. 22. <ul><li>For this amount of energy to be gained the projectile must have had an equal amount of E k </li></ul><ul><li>Therefore ½mv 2 = G M e x m R e </li></ul><ul><li>v =  ( 2GM e R e ) </li></ul><ul><li>But using the fact that g = G M e R e 2 </li></ul><ul><li>Then v =  (2gR e ) </li></ul>
23. 23. <ul><li>Find the gravitational potential 1000km above the Earth’s surface? </li></ul>
24. 24. <ul><li>A satellite of mass 50kg moves from a point where the potential is –20 MJkg -1 to another point where the potential is –60 MJkg -1 </li></ul><ul><ul><li>What is the change in potential </li></ul></ul><ul><ul><li>What is the speed of the satellite </li></ul></ul>
25. 25. <ul><li>A 2000kg spacecraft in orbit at R above the Earth of radius R. The potential at the Earth's surface is –60 MJ kg -1 . What is the change in potential energy if the spacecraft returns to Earth </li></ul>
26. 26. Graphs Gravitational field strength versus distance g α 1/r 2 Gravitational potential versus distance V α -1/r
27. 27. Components of Motion <ul><li>When a body is in free motion, (moving through the air without any forces apart from gravity and air resistance), it is called a projectile </li></ul><ul><li>Normally air resistance is ignored so the only force acting on the object is the force due to gravity </li></ul><ul><li>This is a uniform force acting downwards </li></ul>
28. 28. <ul><li>Therefore if the motion of the projectile is resolved into the vertical and horizontal components </li></ul><ul><li>The horizontal component will be unaffected as there are no forces acting on it </li></ul><ul><li>The vertical component will be accelerated downwards by the force due to gravity </li></ul>
29. 29. <ul><li>These two components can be considered as independent factors in the motion of a projectile in a uniform field </li></ul><ul><li>In the absence of air resistance the path taken by any projectile is parabolic </li></ul>
30. 30. Solving Problems <ul><li>In solving problems it is necessary to consider the 2 components independently </li></ul>
31. 31. <ul><li>Therefore the horizontal motion it is necessary to use the equation </li></ul><ul><li>speed = distance time </li></ul><ul><li>Where speed is the horizontal component of the velocity </li></ul>
32. 32. <ul><li>Therefore the vertical motion it is necessary to use the kinematic equations for uniform acceleration </li></ul><ul><li>i.e. Using the s.u.v.a.t equations </li></ul><ul><li>Where u and v are the initial and final vertical components of the velocity </li></ul>
33. 33. Example <ul><li>A ball is kicked at an angle of 40.0 o with a velocity of 10.0 ms -1 . Taking g = 10 ms –2 . How far does it travel horizontally? </li></ul>40 o 10ms -1
34. 34. <ul><li>To be able to calculate the horizontal distance we need to know the horizontal speed, and the time. </li></ul><ul><li>The horizontal distance is easy to calculate by resolving the velocity </li></ul>10.0 sin 40.0 o 10.0 cos 40.0 o 40.0 o 10.0ms -1
35. 35. <ul><li>However, to calculate the time we will need to use the vertical component and the s.u.v.a.t. equations </li></ul>
36. 36. <ul><li>s = ? </li></ul><ul><li>u = 10.0 sin 40.0 o ms -1 </li></ul><ul><li>v = ? </li></ul><ul><li>a = -10.0 ms -2 (Up is positive, therefore acceleration here is negative) </li></ul><ul><li>t = ? </li></ul><ul><li>We only have 2 of the values when we need three to find any other </li></ul>
37. 37. <ul><li>However, if we ignore air resistance, then the final vertical component of the velocity will be equal and opposite of the initial component </li></ul><ul><li>i.e. v = -10.0 sin 40.0 o ms -1 </li></ul><ul><li>Looking at the equations for uniform acceleration, we need an equation that links u, v, a and t. </li></ul>
38. 38. <ul><li>v = u + at </li></ul><ul><li>Rearranging to make t the subject </li></ul><ul><li>t = v – u a </li></ul><ul><li>Substitute in </li></ul><ul><li>t = -10.0 sin 40.0 o – 10.0 sin 40.0 o -10 </li></ul><ul><li>t = 1.286 seconds </li></ul>
39. 39. <ul><li>Now returning to the horizontal components </li></ul><ul><li>Using speed = distance time </li></ul><ul><li>Rearranging distance = speed x time </li></ul><ul><li>Distance = 10.0 cos 40.0 o x 1.286 </li></ul><ul><li>Distance = 9.851 = 9.9 metres </li></ul>
40. 40. Using the Conservation of Energy <ul><li>In some situations the use of the conservation of energy can be a much simpler method than using the kinematic equations </li></ul><ul><li>Solving projectile motion problems makes use of the fact that E k + E p = constant at every point in the objects flight (assuming no loss of energy due to friction) </li></ul>
41. 41. Example <ul><li>A ball is projected at 25.0 ms -1 at an angle of 40.0 0 to the horizontal. The ball is released 2.00m above the ground. Taking g = 10.0 ms -2 . Find the maximum height it reaches. </li></ul>
42. 42. Solution 2.0m 25.0 ms -1 A B v = v horizontal H
43. 43. <ul><li>Total energy at A is given by </li></ul><ul><li>E k + E p = ½ m (25.0) 2 + mg x 2.0 </li></ul><ul><li> =312.5m + 20m </li></ul><ul><li> = 332.5m </li></ul>
44. 44. <ul><li>Next, to find the total energy at B we need to know the velocity at B, which is given by the horizontal component of the velocity at A </li></ul><ul><li>Total energy at B is given by </li></ul><ul><li>E k + E p = ½ m (25.0 cos 40 o ) 2 + mg x H </li></ul><ul><li> = 183.38m+ 10mH </li></ul><ul><li>Then using the conservation of energy </li></ul>
45. 45. <ul><li>Equating the 2 equations </li></ul><ul><li>332.5m = 183.38m + 10mH </li></ul><ul><li>332.5 = 183.38 + 10H </li></ul><ul><li>332.5 – 183.38 = 10H </li></ul><ul><li>10H = 149.12 </li></ul><ul><li>H = 14.912 = 14.9m </li></ul>
46. 46. Orbital Motion <ul><li>Gravitation provides the centripetal force for circular orbital motion </li></ul><ul><li>The behaviour of the solar system is summarised by Kepler´s laws </li></ul><ul><li>Kepler´s law state </li></ul><ul><ul><li>1. Each planet moves in an ellipse which has the sun at one focus </li></ul></ul><ul><ul><li>2. The line joining the sun to the moving planet sweeps out equal areas in equal times </li></ul></ul>
47. 47. Deriving the Third Law <ul><li>Suppose a planet of mass m moves with speed v in a circle of radius r round the sun of mass M </li></ul><ul><li>The gravitational attraction of the sun for the planet is = G Mm r 2 </li></ul><ul><li>From Newton’s Law of Universal Gravitation </li></ul>
48. 48. <ul><li>If this is the centripetal force keeping the planet in orbit then </li></ul><ul><li>G Mm = mv 2 ( from centripetal equation) </li></ul><ul><li>r 2 r </li></ul><ul><li> GM = v 2 r </li></ul>
49. 49. <ul><li>If T is the time for the planet to make one orbit </li></ul><ul><li>v = 2  r v 2 = 2 2  2 r 2 T T 2 </li></ul><ul><li> GM = 4  2 r 2 r T 2 </li></ul><ul><li> GM = 4  2 r 3 T 2 </li></ul><ul><li> r 3 = GM T 2 4  2 </li></ul><ul><li>r 3 = a constant </li></ul><ul><li>T 2 </li></ul>
50. 50. <ul><li>The square of the times of revolution of the planets (i.e. Their periodic time T ) about the sun are proportional to the cubes of their mean distance ( r ) from it. </li></ul><ul><li>We have considered a circular orbit but more advanced mathematics gives the same result for an elliptical one. </li></ul>Kepler´s Third Law
51. 51. Energy of Orbiting Satellites <ul><li>Potential Energy, E p </li></ul><ul><li>A satellite of mass m orbiting the Earth at a distance r from its centre has </li></ul><ul><li>Gravitational Potential, V = - G M e r </li></ul><ul><li>Therefore the gravitational potential energy, </li></ul><ul><li>E p = - G M e m r </li></ul>
52. 52. Energy of Orbiting Satellite <ul><li>Kinetic Energy, E k </li></ul><ul><li>By the law of Universal Gravitation and Newton’s Second Law </li></ul><ul><li>G M e m = m v 2 r 2 r </li></ul><ul><li>Therefore the kinetic energy, </li></ul><ul><li>E k = ½mv 2 = G M e m 2r </li></ul>
53. 53. Energy of Orbiting Satellite <ul><li>Total Energy, E p + E k </li></ul><ul><li>Total Energy = - G M e m + G M e m r 2r </li></ul><ul><li>Total Energy = - G M e m 2r </li></ul><ul><li>Total energy is constant for a circular orbit. </li></ul>
54. 54. Graphs of Energy Against Radius <ul><li>Potential Energy </li></ul>E p 1/r E p r <ul><li>i.e. E p  -1/r or E p = -k/r </li></ul><ul><ul><li>where k is the constant of proportionality </li></ul></ul><ul><ul><li>= GM e m </li></ul></ul>
55. 55. Graphs of Energy Against Radius <ul><li>Kinetic Energy </li></ul>E k 1/r E k r <ul><li>i.e. E k  1/r or E k = k/r </li></ul><ul><ul><li>where k is the constant of proportionality </li></ul></ul><ul><ul><li>= GM e m </li></ul></ul><ul><li>2 </li></ul>
56. 56. Graphs of Energy Against Radius <ul><li>Total Energy </li></ul>E total 1/r E total r <ul><li>i.e. E total  -1/r or E total = -k/r </li></ul><ul><ul><li>where k is the constant of proportionality </li></ul></ul><ul><ul><li>= GM e m </li></ul></ul><ul><li>2 </li></ul>
57. 57. Weightlessness <ul><li>Consider an object of mass m hanging from a spring balance which is itself hanging from the roof of a lift </li></ul>mg T a
58. 58. <ul><li>The body is subjected to a downward directed force mg due to the Earth, and an upwards force T , due to the tension in the spring. </li></ul><ul><li>The net down ward force is ( mg – T ), and by Newton´s second law </li></ul><ul><li>mg – T = ma </li></ul><ul><li>Where a is downward directed acceleration of the body </li></ul>
59. 59. <ul><li>If the lift is stationary, or is moving with a constant speed, a = 0 and therefore T = mg </li></ul><ul><li>i.e. The balance registers the weight of the body as mg </li></ul><ul><li>However, if the lift is falling freely under gravity, both it and the body have a downward directed acceleration of g </li></ul><ul><li>i.e. g = a </li></ul>
60. 60. <ul><li>It follows from the equation that T = 0 </li></ul><ul><li>i.e. The balance registers the weight of the body as zero </li></ul><ul><li>It is usual to refer to a body in this situation as being weightless </li></ul><ul><li>The term should be used with care, a gravitational pull of magnitude mg acts on the body whether it is in free fall or not, and therefore, in the strictest sense it has weight even when in free fall. </li></ul>
61. 61. <ul><li>The reason it is said to be weightless is that, whilst falling freely, it exerts no force on its support. </li></ul><ul><li>Similarly, a man standing on the floor of a lift would exert no force on the floor if the lift were in free fall. In accordance with Newton´s third law, the floor of the lift would exert no upwards push on the man and therefore he would not have the sensation of weight. </li></ul>
62. 62. Orbital Weightlessness <ul><li>An astronaut in an orbiting spacecraft has a centripetal acceleration equal to g 1 , where g 1 is the acceleration due to gravity at the height of the orbit </li></ul><ul><li>The spacecraft also has the same centripetal acceleration </li></ul><ul><li>The astronaut therefore has no acceleration relative to his spacecraft, i.e. he is weightless </li></ul>