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Universal Gravitation. Space Station link  Nasa Nasa.

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Presentation on theme: "Universal Gravitation. Space Station link  Nasa Nasa."— Presentation transcript:

1 Universal Gravitation

2 Space Station link  Nasa Nasa

3 Gravitational Force  An attractive force that exists between all objects  The weakest of the four basic kinds of forces (EM, strong nuclear and weak nuclear are the others)  We know how things fall but not why

4 Gravity  Galileo and Newton gave the name gravity to the force that exists between Earth and objects.

5  The force of the earth on the moon holds the moon in its orbit

6 Tides  The periodic change in the surface level of the oceans due to the gravitational force of the sun and moon on the Earth

7 Spring tide: greater than normal, since the moon,Earth and sun are aligned Neap tide:lower than normal due to the sun, Earth and moon at right angles

8 Inverse Square Law  The net force between objects is inversely proportional to the square of the distance between their centers  The net force is directly proportional to the product of the masses.

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10 Cavendish Balance

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12 Henry Cavendish 1798  Used his torsion balance to measure the forces needed to rotate the rods through given angles  Calculated the attractive forces  Calculated G = 6.67 x 10 -11 Nm 2 /kg 2

13 Gravitation Formula  F g = Gmm/d 2  d is the distance between the center of the objects or the average radius of orbit in meters  m is the mass in kg  G is 6.67 x 10 -11 Nm 2 /kg 2  Problems P. 191 # 53 to 59  (64, 75, 78 to 80, 82, 83)

14 Using gravitation  F = ma  Weight = F g = mg  Force of gravitational attraction = Gmm e /d 2

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16 Calculate acceleration due to gravity.  F gravitation = F g  Substitute in gravitation formula  Gmm = mg d 2 d 2 Cancel mass of object being accelerated g = Gm/d 2 p. 192 # 72 to 74, 81, 84

17 Gravitational Field  Anything that has mass has a gravitational field  Acts on a body resulting in attraction  g is the field strength  g is F/m  Field is numerically equal to gravity

18 Mass calculations  Use F g = F c  Gravitation = Centripetal  Gm e m s = m s 4  2 R R 2 T 2 R 2 T 2 Cancel the mass of the “satellite” M e = 4  2 R 3 GT 2 GT 2

19  You can also rearrange the formula to find the mass of a planet  Ex:  M e = gr 2 /G where r is the radius of the earth  Substitute in the values for G, d and g  M e = 9.8(6.37 x 10 6 )/6.67 x 10 -11 p. 193 # 80

20 Kepler’s Laws of Planetary Motion  Brahe – the Earth s the center of the universe  Kepler – Brahe’s assistant, performed careful mathematical analysis of Brahe’s data

21 Kepler’s Laws  Describe the behavior of planets and satellites.  His explanations are not considered correct today. http://animate

22 Law #1  The paths of the planets are ellipses with the center of the sun at one focus.

23 Law #2  An imaginary line from the sun to a planet sweeps out equal areas in equal time intervals.

24 Speeds at different points

25 Law #3 The square of the period of revolution of a planet about the sun is proportional to the cube of its mean distance from the sun. (T a /T b ) 2 = (r a /r b ) 3

26 Law #3 con’t. R 3 /T 2 = K where K is Kepler’s constant R 3 /T 2 = K where K is Kepler’s constant  K = 3.35 x 10 18 m 3 /s 2  Can also be used to determine distances and periods of two objects around earth  P. 174 # 1 to 5 Practice  P. 192 # 61 to 63, 67 to 69, 71, 85, 88

27 Review of Kepler’s Laws

28  The squares of the orbital periods of the planets around the Sun are proportional to the cubes of the orbital semi-major axes.  This means that if you know either how much time a planet's orbit around the Sun takes you can easily know it's average distance from the Sun, or vice- versa!

29 Major Axis defined  Kepler's 1st Law: An ellipse has a long axis (called the major axis). It also has two foci (focuses). One focus is where the Sun is located, the other focus is empty. As the two foci are brought together, the ellipse looks more and more like a circle. In fact, a circle is just a special case of an ellipse with the two foci at the same place (the center of the circle), in which case the major axis is the diameter of the circle. Half the major axis is called the semi-major axis (a in the upper figure, r in the lower figure).

30 Notes:  R = d (radius is the distance between the center of the objects)  Satellite can mean any orbiting object  “planet” is the center of motion  If an object is at some distance above a planet, add the radius and the altitude of orbit  Mass of the “satellite” cancels out

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32 Modeling the Orbits of Planets and Satellites  P. 186  A – aphelion is the furthest distance from the sun along the major axis  P – perihelion is the closest distance along the major axis  e is the eccentricity – ratio of the distance between the foci to the length of the major orbit

33 Lab

34  P 2 =a 3  Where P is the orbital period in Earth years and a is the length of the semimajor axis (average distance from the Sun) in Astronomical Units.

35 Calculations  See procedure 4. Use the data table to find e  D = 2e(10 cm) / e + 1  d/2 C d/2  Use a different color for each planet

36 Motions of Satellites  F c = F g  F c is directed towards the center of the earth  The weight (f g ) of the object keeps it in its path

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38 Critical speed  The speed at which an object is perpetually falling towards the earth but never landing.  The rate of fall is equal to the curvature of the earth.  F g = F c  mg = mv 2 /r so v = √gr

39 Period of a Satellite  Gm e m s = m s 4  2 R R 2 T 2 R 2 T 2  m s cancels  Cross multiply Gm e T 2 = 4  2 R 3  Divide to solve for T  T 2 = 4  2 R 3 Gm e Gm e Simplify to T = 2  √R 3 /Gm e (R = radius of the planet + altitude of the satellite) (m p = mass of planet) Practice p. 178 # 6

40 Velocity of a satellite  Gm e m s = m s v 2 R 2 R R 2 R Cancel mass of the satellite  Gm e = v 2 R 2 R R 2 R Simplify and solve for v V = √Gm/R p. 181 # 12 to 14

41 Example : T = 2  √R 3 /Gm e V = √Gm/R  Engineers are planning to place the International Space Station into orbit at an altitude of 450 km above Earth’s surface. What would be the orbital speed and period of the ISS?  Re = 6.38 x 10 6 m  Me = 5.97 x 10 24 kg  G = 6.67 x 10 -11 N. m 2 /kg 2  P. 191 # 71, 85, 88, 90 to 92, 94, 96

42 Geosynchronous satellite From Earth, a satellite in geosynchronous orbit appears to "hover" over one spot on the Equator. That means a receiving dish on the Earth can point at the satellite at one spot in the sky and not have to "track" its motion.geosynchronous orbit The satellite isn't motionless, though. It's in a very high orbit where it circles the Earth once a day, matching the Earth's rotation on its axis. Weather and Cable TV are examples.

43 Escape speed  The minimum speed an object must possess in order to escape the gravitational pull of a body.

44 Calculate escape speed  V = √2Gm/d  M = mass of the celestial body  D = radius  Mass of the escaping object does not matter

45 Voyager Satellites  Update link

46 Einstein’s Theory of Gravity

47  Gravity is not a force but an effect of space itself.  Mass causes space to be curved.  Bodies accelerate as they move in curved space.

48 Answers  R = 6.83 x 10 3 m  V = 7.63 x 10 3 m/s  T = 5.62 x 10 3 s


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