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Newton’s Law of Universal Gravitation

Newton’s Law of Universal Gravitation Every object in the universe is attracted to every other object. F= Gm 1 m 2 r 2 G = 6.67 X 10 -11 N-m 2 /kg 2 m 1 = mass of one object m 2 = mass of second object r = distance from center of objects. Cavendish proves the law in 1798.

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Newton’s Law of Universal Gravitation

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  1. Newton’s Law of Universal Gravitation Every object in the universe is attracted to every other object. F= Gm1m2 r2 G = 6.67 X 10-11 N-m2/kg2 m1 = mass of one object m2 = mass of second object r = distance from center of objects

  2. Cavendish proves the law in 1798

  3. Everything in the solar system pulls on everything else. Sun pulls on Earth All the other planets also pull on the Earth

  4. Calculate the force of gravity between two 60.0 kg (132 lbs) people who standing 2.00 m apart. (6.00 X 10-8 N) • Calculate the force of gravity between a 60 kg person and the earth? Assume the earth has a mass of 5.98 X 1024 kg and a radius of 6,400,000 m (4,000 miles). (584 N)

  5. A planet has a mass four times that of the earth, but the acceleration of gravity is the same as on the earth’s surface. Relate the planet’s radius to that of the earth (2Re).

  6. A 2000-kg satellite orbits the earth at an altitude of 6380 km (the radius of the earth)above the earth’s surface. What is the force of gravity on the satellite? F= Gm1m2 = (6.67 X 10-11 N m2/kg2)(2000kg)(5.98 X 1024 kg) r2 (6,380,00 m + 6,380,00 m)2 F = 4900 N

  7. A simpler solution: At the earth’s surface: FG = mg At twice that distance: FG = (1/2)2mg = ¼ mg FG = (¼ )(2000 kg)(9.80 m/s2) FG = 4900 N

  8. What is the net force on the moon when it is at a right angle with the sun and the earth? Relevant Data: MM = 7.35 X 1022 kg ME = 5.98 X 1024 m MS = 1.99 X 1030 m rMS = 1.50 X 1011 m rME = 3.84 X 108 m

  9. FME Earth Calculate each force separately: FME = 1.99 X 1020 N FMS = 4.34 X 1020 N FR2 = FME2 + FMS2 FR = 4.77 X 1020 N tan q = opp = FME adj FMS q = 24.6o q FR FMS Sun

  10. Consider an object (any object) Weight = mg (Force due to gravity) mg = GmmE rE2 g = GmE rE2 mE =g rE2 = (9.8 m/s2)(6.38 X 106 m)2 = 5.98 X 1024 kg G 6.67 X 10-11 N-m2/kg2

  11. Calculating “g” g = GmE rE2 g = (6.67 X 10-11 N-m2/kg2)(5.98 X 1024 kg) (6.38 X 106 m)2 g = 9.80 m/s2

  12. Calculate the value of g at the top of Mt. Everest, 8848 m above the earth’s surface. g = GmE r2 g = (6.67 X 10-11 N-m2/kg2)(5.98 X 1024 kg) (6.38 X 106 m + 8848 m)2 g = 9.77 m/s2

  13. g varies with: • Altitude • Location • Earth is not a perfect sphere • Different mineral deposits can change density • “salt domes” are low density salt regions near petroleum deposits

  14. Gravitational Potential Energy • mgh is not valid except near the surface of the earth • Zero point is an infinite distance from the earth • Need calculus since the force varies with distance U = - ∫F dx U = - ∫ Gm1m2 dr r2 U = -Gm1m2 r

  15. A 1000 kg rocket is launched from the surface of the earth. Calculate the escape velocity. Hint: set the final K and U equal to zero. (11,200 m/s)

  16. A 1000 kg rocket is launched from the surface of the moon. Calculate the escape velocity. The mass of the moon is 7.35 X 1022 kg and the radius is 1.74 X 106 m. (2383 m/s)

  17. A tile of a rocket falls off at an altitude of 1500 km above the surface of the earth. The rocket is travelling upwards at 2000 m/s. • Calculate the speed of the tile as it hits the ground. (5276 m/s) • In reality. Could the tile actually reach this speed?

  18. Suppose the earth stops orbiting and falls into the sun (see page 395) • Use conservation of energy to calculate the speed of the earth as it just crashes into the sun. (6.13 X 105 m/s) • Why doesn’t our earth get pulled into the sun?

  19. What speed should a satellite be launched if it needs to have a speed of 500 m/s at 400 km. Remember to include the radius of the earth in your calculations. (2755 m/s)

  20. Why don’t satellites fall back onto the earth? • Speed • They are “falling” due to the pull of gravity • Can feel “weightless” if you are on board (just like in the elevator)

  21. Speed of a Satellite F= GMm = mv2 r2 r GM = v2 r v = \/GM/r

  22. The Starship Enterprise wishes to orbit the earth at a 300 km height. • Calculate the total distance of the Enterprise from the center of the earth. (6.68 x 106 m) • Calculate the proper orbital speed for the Enterprise. (7730 m/s) • Would a heavier starship require a greater orbital speed?

  23. Kepler’s Laws (1571-1630) • The orbit of each planet is an ellipse, with the sun at one focus • Each planet sweeps out equal areas in equal time • T1 2 = r13 T2 r2

  24. 1. The orbit of each planet is an ellipse, with the sun at one focus http://csep10.phys.utk.edu/astr161/lect/history/kepler.html

  25. 2. Each planet sweeps out equal areas in equal time • Suppose the travel time in both cases is three days. • Shaded areas are exactly the same area

  26. Mars has a year that is about 1.88 earth years. What is the distance from Mars to the Sun, using the Earth as a reference (rES = 1.496 X 108 m) T12 = r13 T22 r23 TM2 = rM3 TE2 rE3

  27. rM3 = TM2rE3 TE2 rM3 = (1.88y)2(1.496 X 108 m)3 (1 y)2 rM3 = 1.18 X 1025 m3 rM = 2.28 X 108 m

  28. How long is a year on Jupiter if Jupiter is 5.2 times farther from the Sun than the earth? TJ2 = rJ3 TE2 rE3 TJ2 = rJ3 TE2 rE3

  29. TJ2 = rJ3 TE2 = (5.2)3(1 y)2 rE3 (1)3 TJ2 = 141 y2 TJ= 11.8 y

  30. How high should a geosynchronous satellite be placed above the earth? Assume the satellite’s period is 1 day, and compare it to the moon, whose period is 27 days. Ts2 = rs3 TM2 rM3 rs3 = rM3 Ts2 TM2

  31. rs3 = rM3 Ts2 = rM3 (1 day)2 TM2 (27 day)2 rS3 = rM3 Take the 3rd root of 729 both sides rs = rM 9 The satellite must orbit 1/9 the distance to the moon.

  32. Deriving the Third Law To derive Kepler’s Law, we will need two formulas. F= Gm1mJ r2 F=m1v2 r mJ m1

  33. GMm = mv2 r2 r GM = v2 Substitute v=2pr r T GM = 4p2r2 r T2 T2= 4p2 r3 GM

  34. T2= 4p2We can do this for two r3 GMdifferent moons T12= 4p2T22= 4p2 r13 GMr23 GM T12 = T22 r13 r23

  35. A Useful Form This form of the equation: T2= 4p2 r3 GM Is useful for determining the mass of the central planet, using only the period and distance of one of the satellites. S could be the Sun, Earth, or other body with satellites.

  36. Calculate the mass of the sun, knowing that the earth is 1.496 X 108 m from the sun. (2.0 X 1030 kg)

  37. A student is given the following data and asked to calculate the mass of Saturn. The data describes the orbital periods and radii of several of Saturn’s moons. Orbital Period, T Orbital Radius, R (seconds) (m) 8.14 X 104 1.85 X 108 1.18 X 105 2.38 X 108 1.63 X 105 2.95 X 108 2.37 X 105 3.77 X 108 4p2

  38. Let’s use this equation: T2= 4p2 r3 GmS And rearrange it: GmS= 1 4p2 r3 T2 Once more: • = GmS T2 4p2 r3

  39. Calculate the following values and graph them. 1 G T2 4p2 r3

  40. Calculate the slope of the graph y = m x + b • = mS G T2 4p2 r3 ms = 5.9 X 1026 kg

  41. Orbital Energies GMm = mv2 r2 r GMm = mv2 r 1 mv2 = GMm = K • 2r K = -½ Ug

  42. E = K + Ug E = - ½ Ug + Ug E = ½ Ug

  43. A 1500 kg satellite orbits at an altitude of 4,000 km above the earth’s surface. Calculate the work required to move that satellite to an orbit of 12,000 km above the earth’s surface. (1.25 X 1010 J)

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