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Newton’s Law of Universal Gravitation: Two objects of mass m 1 and m 2 separated by

Newton’s Law of Universal Gravitation: Two objects of mass m 1 and m 2 separated by a center-to-center distance r ___________ each other with a gravitational force:. attract. r. m 1. m 2. F g. F g. F g =. …where G = ____________________________

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Newton’s Law of Universal Gravitation: Two objects of mass m 1 and m 2 separated by

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  1. Newton’s Law of Universal Gravitation: Two objects of mass m1 and m2 separated by a center-to-center distance r ___________ each other with a gravitational force: attract r m1 m2 Fg Fg Fg = …where G = ____________________________ is called the _______________ gravitational constant.

  2. PhysRT, page 6:

  3. G m1 m2 r2 Newton’s Law of Universal Gravitation: Two objects of mass m1 and m2 separated by a center-to-center distance r ___________ each other with a gravitational force: attract r m1 Fg m2 Fg Fg = …where G = ____________________________ is called the _______________ gravitational constant.

  4. PhysRT, Page 1: top

  5. G m1 m2 r2 Newton’s Law of Universal Gravitation: Two objects of mass m1 and m2 separated by a center-to-center distance r ___________ each other with a gravitational force: attract r m1 Fg m2 Fg Fg = 6.67 x 10-11 N·m2/kg2 …where G = ____________________________ is called the _______________ gravitational constant. universal

  6. Notes: • Fg is an _________ range, _______________ force. • Fg is stronger when the objects are__________ . • 3. The constant G is very __________  Fg is the • ________________ of the fundamental forces. • 4. Fg is always ___________________. • 5. Both masses pull each other with ____________ • magnitude forces, but in _____________ directions. • 6. Equation is only true for ____________ masses. •  for spheres, you must assume mass is • concentrated at __________________ •  for complicated shapes, _____________ is needed, • but equation works ________________ anyway. infinite "at a distance" closer small weakest attractive equal opposite point its center. calculus approximately

  7. G m1 m2 r2 Ex. A mass of 1.8 x 103 kg (F-150) is 0.50 meter from a mass of 6.0 x 101 kg (student). Find the magnitude of the force of gravitational attraction between the two masses. Show all work. Fg = (6.0 x 101 kg) (6.67 x 10-11 Nm2/kg2) (1.8 x 103 kg) Fg = (0.50 m)2 (7.2 x 10-6 Nm2) (0.25 m2) Fg = / Fg = 2.9 x 10-5 N ≈ 3/1000 of one dollar bill neither Which mass pulls with a greater force?

  8. Fg Fg Fg m1 r m2 G m1 m2 r2 Fg = inverse squared direct direct Double m1Fg ________________ Triple m2  Fg ________________ Double both m1 and m2 Fg ________________ Triple m1 and double m2 Fg ________________ Double r  Fg ________________ Halve r  Fg ________________ Triple r  Fg ________________ Double m1 and r  Fg ________________ Double m1, m2 and r  Fg ________________ 2x greater 3x greater 4x greater 6x greater 4x weaker 4x greater 9x weaker 2x weaker same

  9. Fg = Gm1m2 r2 Ex: If the Fg is between an object of mass m and a planet, then Fg is called the _________: Fg = ___ w weight Ex: Earth m 1 Re r = GMem Fg = Re2 Me GMe m Fg = Re2 g w = m 1 Earth radius Re = ________________ Me = ________________ g = mass of Earth GMe/Re2 9.81 m/s2 =

  10. Ex: Are you weightless in the space shuttle (mass = ms)? The space shuttle orbits at ≈ _______ = __________ above Earth's surface. Its __________ distance from Earth's center is r = _____ + _____ Mm = _______ Mm. = ________ Re So the ___________ (Fg) of the shuttle and all its contents in orbit, compared to its weight on land, is: Earth Re = ___________ ≈ ___________ ms GMems GMems = Fg = = 

  11. PhysRT, Page 1: top

  12. Ex: Are you weightless in the space shuttle (mass = ms)? The space shuttle orbits at ≈ _______ = __________ above Earth's surface. Its __________ distance from Earth's center is r = _____ + _____ Mm = _______ Mm. = ________ Re So the ___________ (Fg) of the shuttle and all its contents in orbit, compared to its weight on land, is: 300 km 0.3 Mm total Earth 0.3 6.4 6.7 1.05 weight 6.37 x106 m Re = ___________ ≈ ___________ 6.4 Mm ms GMems GMems GMems 0.91 = Fg = 91% =  Re2 Re2 (1.05Re)2 (1.05)2

  13. The Vomit Comet: The airplane produces about 25 s of apparent weightless-ness by following a parabolic vertical flight path. A parabolic flight path is the same path that would be taken by an object in free fall, such as a cannonball fired into the air.

  14. Olympic Weightless Events- Scores for Team USA: A 9.5 in the One- Finger Hold event. A 5.2 on the Wrist-Apple Bump event. But a 9.9 out of 10 in the Weightless Hair event.

  15. center-to-center Ex: r is the ____________________ distance r = __ Re 4 C 3 __ Re above surface r = __ Re 3 B 2 __ Re above surface r = __ Re 2 A 1 __ Re above surface r = 1 Re Earth If Fg at surface = 200 N, what is the weight (Fg) at A? = 200/4 = 50 N 200/22 = 22 N 200/32 At B? = 12.5 N At C? 200/42

  16. Ex: A 20-N box on a table is lifted from 1 m to 2 m above the floor. Since the height was doubled, the new weight should be w = 20/22 = 5 N ?????? This _________ ______________ because these heights are ______________ from ________ _____________ . does not 5 N ? happen not measured 20 N 2 m Earth's table center 1 m

  17. Ex: A 600-N volleyball player jumps in the air. What is the force of gravity acting on her… 1/ …while in the air? 2/ …as she is landing? 3/ …when she is again at rest on the ground? 4/ What is her weight in all three cases above? 600-N 600-N 600-N 600-N

  18. Same rock at rest on a table: A rock in freefall: Ex: Fg = 1.33 N Fg = 1.33 N a/ What is the weight of the rock in each case? 1.33 N b/ What is the net force acting on the rock in each case? -1.33 N free fall: Fnet = ______ on table: Fnet = ______ 0 c/ What is the acceleration of the rock in each case? free fall: a = ______ on table: a = ______ -9.81 m/s2 0 d/ What is the reaction force to the weight of the rock in each case? The rock pulls up on Earth with a 1.33 N gravitational force.

  19. G m1 m2 r2 Ex: Cavendish "Weighing the Earth" Experiment: Pb When a ____ sphere (m2) was brought close to the barbells, the _______________ attraction caused the thin wire to _________ . thin wire From the wire's properties, the ______________ needed to make the wire twist that much could be _____________ force, Fg gravitational r m2 m1 Pb barbells twist estimated Then Fg, r, m2 and m1 were substituted into: Fg = G and this was solved to find ______ . G Once _____ was known, an object of known massmand weight w were used to find ___________ unknown mass Me using Me G m Re2 w= Earth's

  20. One last note: g = Fg/m In PhysRT: mg same Solve this for: Fg = ________ equation mg w = Not in PhysRT: acceleration ____________ due to gravity Even though g appears in the equation for w, an object does NOT have to be ________________ to use this equation. Think of g as simply a ________________________ between ____ and ____ . In fact, g can have ___________________ in different locations, which is why ____________ may change even though _________ remains the same. accelerating w conversion factor m different values weight mass

  21. Open your 3-ring binder to the Worksheet Table of Contents. Record the title of the worksheet: Force of Gravity WS

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