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Gravitation

Gravitation. Notes HRW p240. Law of Universal Gravitation (LUG). Isaac Newton (1642-1727) observed: The gravitational force between 2 objects is proportional to the product of their masses and inversely proportional to the square of their separation distance. F  M*m. F 1/d 2.

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Gravitation

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  1. Gravitation Notes HRW p240

  2. Law of Universal Gravitation (LUG) • Isaac Newton (1642-1727) observed: • The gravitational force between 2 objects is • proportional to the product of their masses and • inversely proportional to the square of their separation distance. F  M*m F 1/d2

  3. Gravitational Force Formula • where • G = universal gravitational constant, • G = 6.67 x 10-11 N.m2/kg2 • M, m = the masses of any 2 objects of interest • d = distance between the CENTERS of the objects

  4. Example • What is the gravitational force between 2 students sitting 3 meters apart, if one has a mass of 80 kg, and the other a mass of 120 kg? • Solve using LUG • F = GMm/d2 • F = 6.67e-11(80)(120)/32 • F = 7.11e-8 N

  5. Gravitational Acceleration • Gravitational force experienced by an object of mass m on Earth (or any celestial body of mass M) is Fg • Thus, gravitational acceleration g can be calculated for any (M)assive body of radius d

  6. Example • What is the magnitude of the gravitational acceleration (g) on Neptune if it has a mass of 1.02e26 kg and a radius of 2.48e7 m? • Solve: g = GM/d2 • g = (6.67e-11)(1.02e26)/(2.48e7)2 • g = 11.06 m/s2

  7. Satellite Background • Per the video, an object (satellite) needs a specific speed to go into orbit

  8. Escape Speed (vesc). • Threshold speed required for an object to escape the gravitational pull of a body of mass M and radius d:

  9. Example • What is the escape speed for Neptune? MN = 1.02e26 kg, dN = 2.48e7 m • Solve vesc = √(2GM/d) • vesc = √(2*6.67e-11*1.02e26/2.48e7) • vesc = 23,423 m/s

  10. Example • What is the escape speed for Earth, where ME = 5.97e24 kg, dE = 6.38e6 m? • Solve vesc = √(2GM/d) • vesc = √(2*6.67e-11*5.97e24/6.38e6) • vesc = 11,172.6 m/s

  11. Satellite Orbits

  12. Satellite Orbits

  13. Satellite Orbits • Low Earth Orbit (LEO) • 100 -1000 km • Photo, weather • Orbit ~ 90 mins • Includes Shuttle • Geostationary/synchronous orbit (GEO) • 36,000 km above equator • TV, Weather • Orbit 1 day

  14. Satellite Orbital Radius (do) • Orbital radius • do = R + h

  15. Satellite Orbital Speed (vo) • The speed vo required to keep a satellite in orbit at a distance h above a planet with radius R and mass M • do = orbital radius (R + h) • R = radius of the planet • h = height above the planet (altitude) • G = 6.67 x 10-11

  16. Example • At what speed must a satellite orbit Neptune if its altitude is 100 km? MN = 1.02e26 kg, dN = 2.48e7 m • Solve vo = √(GM/do) • vo = √(6.67e-11*1.02e26/(2.48e7+100,000)) • (note: km > m conversion!) • vo = 16,530 m/s

  17. Practice - LEO • What is the orbital speed and period of a LEO satellite @ 200 km above Earth, where Me = 6e24 kg, Re = 6.4e6 m • Solve vo = √(GM/do) • vo = √(6.67e-11*6e24/(6.4e6+200,000)) • vo = 7,787 m/s • Solve vo = 2πdo/T for T • T = 2πdo/ vo = 5374.2 sec = 89.56 minutes

  18. Practice - GEO • What is the orbital speed and period of a GEO satellite @ 36,000 km above Earth (Me = 6e24 kg, Re = 6.4e6 m) • Solve vo = √(GM/do) • vo = √(6.67e-11*6e24/(6.4e6+36000000)) • vo = 3,072.2 m/s • Solve vo = 2πdo/T for T • T = 2πdo/ vo = 86,714 sec = 24.08 hours • GEOSTATIONARY!

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