Chapter 7 gravitation

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Chapter 7 gravitation

  1. 1. CHAPTER 7:GRAVITATION WONG LIHERNG TAN SI YOUNG TAN SZE MING YEE SIEW QI
  2. 2. 7.1NEWTON’S LAW OFGRAVITA TION
  3. 3. Newton ‘s law of gravitation states that theforce of attraction between particles isdirectly proportional to their mass andinversely proportional to the square ofdistance apart.
  4. 4. whereThe negative sign shows that the force is anattractive force.
  5. 5. Case 1: Determine the force of gravitationalattraction between the earth 5.98 x 1024 kg and a 70kg boy who is standing at sea level, a distance of6.38 x 106 m from earths center.m1 = 5.98 x 1024 kg, m2 = 70 kg, r = 6.38 x 106 m, G= 6.6726 x 10-11N-m2/kg2
  6. 6. 7.2GRAVITATIONALFIELDSTRENGTH
  7. 7. • A gravitational field is a region where gravitational force acts on massive bodies. Eg . gravitaional field of the Earth.• The gravitational field strength tells us how strong a gravitational field is. The gravitational field strength of the Earth near its surface is 9.81m / s2. This means an object that is near the surface of the earth will accelerate towards it at 9.81ms2. We could then define the gravitational field strength as the acceleration an object will experience within that gravitational field.• A better definition, however, can be derived from the equation, F = ma.• The gravitational field strength , E at a point is the force of gravity per unit mass exerted on a mass placed
  8. 8. • Gravitational field strength is a vector quantity which measured in , though it is perfectly acceptable to use ms − 2 for situations where it is treated as an acceleration (such as the acceleration of an object in free fall).• (r > R)
  9. 9. • The –ve sign shows that the direction of E is in the direction of decreasing r, that is, towards the centre of the Earth.• (by substituting F for mg)• (by cancelling the lower case ms)• On the surface of the Earth, r=R and
  10. 10. Questions:• 1. A point mass produces a gravitational field strength of 3.4 x 10-2 N/kg at a distance of 4.9 x 102 m away. What is the mass of the point?
  11. 11. Ans:g = Gm/r2Thus, m = gr2/G = 1.2 x 1014 kg
  12. 12. • 2. Estimate the gravitational field strength at the surface of an interstellar body whose density is 5.5 x 103kg/m3 and radius is 6.4 x 106 m.
  13. 13. Ans : g = Gm/r2and m = density x volume = D x V= D x 4/3 x p x r3Substitute for m in the field equation above.Thus, g = 4pGDr/3= 4 x p x 6.67 x 10-11 x 5.5 x 103 x 6.4 x 106/3 = 9.8 N/kg
  14. 14. Variation of Acceleration Due to Gravity g with Distance r from the centre of the Earth• Weight = mg = gravitational force• Acceleration due to gravity, g=gravitational force/m =gravitational field strength,E # The variation of the acceleration due to gravity g with distance r from the centre of the Earth equals the variation of the gravitational field strength.
  15. 15. • For a mass m on the surface of the Earth, weight =mg = Acceleration due to gravity on the surface of the Earth ,Where g = accleration due to gravity G = universal gravitational constant M = mass of the earth R = radius of the earth
  16. 16. • The variation of the acceleration due to gravity g’ with distance r from the centre of the Earth is illustrated by the graph above.
  17. 17. • Figure above shows a mass m at a distance of r < R from the centre of the Earth.
  18. 18. Variation of g with latitude• The acceleration due to gravity, and the gravitational field strength on the surface of the Earth is not the same at different points on the Earth’s surface because :-• The Earth is not a perfect sphere, but it is an ellipsoid as shown in below.• The rotation of the Earth about its axis.• The Earth’s rotation on its axis causes it to bulge at the equator and flatten at the poles, forming an ellipsoidal shape.
  19. 19. • The figure below shows that the radius of the Earth at the equator R1 is greater than the distance of the poles from the centre of the Earth R2.• g α ⅟r²• The value of g at the equator is less than the value of g at the poles.
  20. 20. -The centripetal forcem(R cos Ѳ)ω² is providedby the component of moIn the direction to the axisof the rotation.-At the N-pole, Ѳ=90° g’=g-At the equator, Ѳ=0° g’=g-Rω²
  21. 21. Gravitational Potential• The strength of the gravitational force at a point in a gravitational field is described by the gravitational field strength E or g is a vector quantity.• Another quantity associated with the point in the gravitational field is the gravitational potential. It is a scalar quantity.
  22. 22. • The gravitational potential V at a point P in a gravitational field is defined as the work done per unit mass to bring a body from infinity to P. The unit for gravitational potential is Jkgˉ¹.• The gravitational potential energy U of a body at a point P in a gravitational field is defined as the work done to bring the body from infinity to P. The unit for gravitational potential energy is J.• Hence the gravitational potential energy U of a body of mass m at a point where the gravitational potential, V is given by U = mV Conversely, U V= m
  23. 23. • Figure below shows a body of mass m at a distance of x from the centre of the Earth. Mm The gravitational force on the body is F = - G x² P -dx m M R Mm F=-G x² r x• The work done to move the body a small distance dx towards the Earth, that is, a displacement ( dx ) is dU = F (-dx) = (-G Mm )( -dx) x² Mm = G x² dx
  24. 24. • The work done to bring the body from infinity to the point P which is at a distance r from the centre of the Earth is r ∞Gravitational potential energy, U = - GMm r• Gravitational potential at a distance r from the centre of the Earth is U V= m GM Gravitational potential, V = - r• The negative sign in the expression for U and V shows that the work is done by the Earth to bring the mass from infinity.
  25. 25. • On the surface on the Earth, r = R GM – Gravitational potential, V = - R GMm – Gravitational potential energy, U = - R• The graph illustrates the variation of the gravitational potential V with distance r from the centre of the Earth. V 1 R r² 0 Eα r 1 - GM r R Vα
  26. 26. Changes in Gravitational Potential Energy• A body of mass m on the surface of the Earth has Mm gravitational potential energy, U = - G R• When the body is raised through a small height dR, the change in its gravitational potential energy dU is obtained by differentiating U with respect to R. Mm dU = G R² dR = mg (dR)• If dR is written as h, then the change in gravitational potential energy of a body of mass m when it raised through a height h is mgh. ∆ U = mgh
  27. 27.  G is the Universal Gravitational Constant. It is a scalar quantity with dimensiong is the acceleration due to gravity . It is a vector quantity with dimension
  28. 28. Where g = accerelation due to gravity R = constant radius of earth G = universal gravitational constant M = mass of Earth
  29. 29. EXAMPLE :A planet of mass M and radius r rotates about its axiswith an angular velocity large enough to substances onits equator just able to stay on its surface . Find interms of M , r and G the period of rotation of theplanet.
  30. 30. 7.5 Satellite Motion in Circular Orbits
  31. 31. Satellite is a body that revolves round aplanet. Satellites can be categorized asnatural satellites or man-made satellites.The moon, the planets and comets areexamples of natural satellites. Examples ofman-made satellites are Sputnik I , MeasatI ,II and III which are communicationsatellites.
  32. 32. In order to launch satellite into orbit, rockets are used. When rocket thatcarries the satellite reaches the requiredheight , the satellite is launched intocircular orbit with a certain velocity v thatis tangential to intended orbit.
  33. 33. Since the centripetal force required forthe circular motion of the satellite isprovide by the gravitational attractionof the Earth on the satellite, (mv2) / r = G(Mm) / r2Velocity of satellite, v = [ (GM) / r]1/2since GM= gR2, v = [ (gR2) / r]1/2
  34. 34. For a satellite orbiting close to theEarth’s surface , the radius of thesatellite’s orbit is approximate to theradius of the surface. ( r = R) V = 7.92 x 103 ms-1 Period of satellite’s orbital , T = 85 minutes
  35. 35. Synchronous Satellite
  36. 36. Synchronous Satellite is the communicationsatellite in which the orbit of it is synchronisedwith the rotation of the Earth about its axis .It should have the characteristics of :Period of its orbit = 1.0 day (which is equals tothe period of the earth’s rotation about itsaxis)
  37. 37. The satellite orbits the Earth in the samedirection as the direction of the Earth.The satellite’s orbit is in the same plane as theequator because the centripetal force isprovided by the gravitational attraction of theEarth.
  38. 38. As Qravitational force = Centripetal force G[ (Mm) / r2] = mrw2 r3 = GM * T / (2π) + 2*Radius of satellite’s orbit , r = 4.24 x 10 7 m (6.6 times the radius of the Earth)
  39. 39. Planetary System
  40. 40. 1. Planet is a body in orbit round the Sun which sastifies the condition of •Has enough mass to form itself into a spherical shape •Has cleared its immediate neighbourhood of all smaller objects
  41. 41. 2. Due to the above definition of ‘planet”, Pluto was excluded from planethood and reclassified as a ‘dwarf planet”.3.Planets orbit the Sun in elliptical orbits. As an approximation, we assume that the orbits of the planets are circular.
  42. 42. If a planet of mass m is in a circular orbit of radius rround the Sun (mass Ms), the centripetal force isprovided by the gravitational attraction between theplanet and the Sun.
  43. 43. G[ (Mm) / r2] = mrw2 r3 = GM * T / (2π) + 2 T2/ r3 = (4 π2) / GMs = constant(which sastifies Kepler’s Third Law)
  44. 44. Energy of a SatelliteTotal energy of the satellite , E = U+ K E = -(GMm)/r + (GMm)/2rAs an object approaches the Earth , *Its kinectic energy K increases as the gravitational pull of the Earth on it increase. *Its gravitational potential energy U decrease , becoming more and more negative. *Its total energy E also increase.
  45. 45. Weightlessness
  46. 46. 1.The apparent loss of weght is knownas weightlessness.For example , an astronaut in aspacecraft feels that he has no mass asall objects in the spacecraft experiencean apparent loss of weght.
  47. 47. 2. A spacecraft orbiting the Earth has an acceleration that equals the acceleration due to gravity g at the position of th spacecraft.
  48. 48. 3. If N = normal reaction of the floorboard of the spacecraft on an object of mass m , Using F = ma, Mg – N = ma N=0 *means that the floorboard does not exert any force on the object.
  49. 49. 4. Condition for a person to experienceweightlessness :The person must fall towards the Earthwith an acceleration that equals theacceleration due to gravity.
  50. 50. 7.6 EscapeVelocity
  51. 51. When a rocket is launchedfrom the earth’s surface , it canescape from gravitational pullof the earth if it has sufficientenergy to travel to infinity.The energy required comefrom the kinetic energy whenthe rocket is launched .The minimum velocity is alsoknown as escape velocity .
  52. 52. Expression for escape velocityUsing the principle of conservation of energy,Kinetic energy required by the body on theEarth’s surface = Gain in gravitational potential energy of thebody when it moves from the Earth ‘s surfaceto infinity.
  53. 53. Case 1: What is theescape velocity onearth with respect toEarths gravity?Mass of the earth =5.98x1024kg,R = 6378100 m,G = 6.6726 x 10-11N-m2/kg2.

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