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Newton’s Universal Law of Gravitation

Newton’s Universal Law of Gravitation. Newton believed that every object ______________ every other object. The force of the attraction depends on the __________ and ___________ of the two objects. The Universal Law of Gravitation. F g = G (m 1 m 2 )

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Newton’s Universal Law of Gravitation

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  1. Newton’s Universal Law of Gravitation Newton believed that every object ______________ every other object. The force of the attraction depends on the __________ and ___________ of the two objects.

  2. The Universal Law of Gravitation Fg = G (m1m2) r2 Fg = gravitational force in (N) m1 = mass object 1 (kg) m2 = mass object 2 (kg) r = the distance between the masses (m) (measured from their centers) G = 6.67 x 10-11 N m2/kg2 (a constant everywhere!)

  3. Ex. 1 Are you attracted to the person sitting next to you? Calculate the gravitational force between you (mass 70kg) and the person next to you (mass 65kg) if you are 1.2 m apart. Given: m1 = 70 kg m2 = 65 kg r = 1.2 m G = 6.67 x 10-11 N m2/kg2 Unknown: Fg Equation: Fg = G (m1m2) r2 = (6.67 x 10-11 N m2/kg2)(70 kg)(65 kg) (1.2 m)2 = 2.11 x 10-7 N = 0.000000211 N(tiny!)

  4. We can use the Universal Law of Gravitation to find the acceleration due to gravity at various distances from earth. • As you go further away from the earth’s surface the acceleration due to gravity _____________. • Two equations for gravitational force F = ma becomes Fg = mg and Fg = G (m1m2) r2 We can set them equal to each other: Fg = Fg mg = G m1 m2 one of the masses cancels r2 g = G m r2

  5. g = G m r2 • g = the acceleration due to gravity (m/s2) • G = 6.67 x 10-11 N m2/kg2 • m = the mass of the planet in kg • r = the distance from the center of the planet in (m) • Remember- r measures from CENTER of planet, not the surface.

  6. Ex. 2 Find the gravity if you are 2.1 x 105 m above the earth’s surface. Given: r = 2.1 x 105 m + 6.37 x 106 m = 6.58 x 106 m mearth = 5.98 x 1024 kg Unknown: Fg Equation: F = ma, now Fg = mg (We need g at that altitude.) Fg = Fg mg = G m1 m2 r2 g = G m r2 = (6.67 x 10-11 N m2/kg2)(5.98 x 1024 kg) (6.58 x 106 m)2 = 9.21 m/s2

  7. If you have a mass of 60 kg, what is your weight? So you would weigh Fg = mg = (60kg) (9.21 m/s2) Fg = 553 N

  8. Gravity Review

  9. Geosynchronous Orbit A geosynchronous satellite orbits above the same point on the equator of the earth at all times. Examples: GPS cell phone satellites TV satellites

  10. Satellites are _________________________. In order to not fall back to earth, they need to maintain a certain velocity… In order for a satellite to stay in a consistent orbit __________________ = ________________ Fg= Fc mg= m v2 r (The mass of the satellite cancels out!) g = v2 r g = the acceleration due to gravity in m/s2 v = the speed of the satellite in m/s r = the distance from the center of the planet in meters

  11. Ex. 3 Calculate the speed needed for one of the Direct TV satellites to orbit at an altitude of 320,000 m above the earth’s surface. Given: r = 320,000 m + 6.37 x 106 m = 6,690,000 m or 6.69 x 106 m Unknown: v Equation: Fg = Fc mg = mv2 r g = v2 r We need g at that altitude: Fg = Fg mg = G m1 m2 r2 g = G m r2 = (6.67 x 10-11 N m2/kg2)(5.98 x 1024 kg) (6.69 x 106 m)2 g = 8.91 m/s2 8.91 m/s2 = ___v2__ (6.69 x 106 m) v = 7,721 m/s (that’s about 17,000 mph!)

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