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LAW OF UNIVERSAL GRAVITATION F G gravitational force (in two directions) G universal gravitation constant 6.67x10 -11 Nm 2 kg -2 r distance between the.

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Presentation on theme: "LAW OF UNIVERSAL GRAVITATION F G gravitational force (in two directions) G universal gravitation constant 6.67x10 -11 Nm 2 kg -2 r distance between the."— Presentation transcript:

1 LAW OF UNIVERSAL GRAVITATION F G gravitational force (in two directions) G universal gravitation constant 6.67x10 -11 Nm 2 kg -2 r distance between the objects m 1 mass of the larger object

2 near the earth’s surface... both of these equations could be applied to the surface of any planet

3 Planet X has a radius that is 3.5 times the radius of the earth and a mass that is 2.0 times the earth’s. Compare the acceleration due to gravity at the surface of each planet.

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5 What happens to the gravitational attraction between two particles if one mass is doubled, the other tripled and the distance between them cut in half?

6 read p. 139-142 p. 141 1-6extra p. 143 8-13 p. 144 1-6

7 SATELLITES A satellite is an object or a body that revolves around another object, which is usually larger in mass. Planets, moons, space shuttles, space stations, comets, and “satellites” are satellites. Satellites remain in a constant orbit because they are acted upon by a centripetal force and display centripetal acceleration.

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10 remember m 1 is the larger mass and the central object

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12 What is the period of rotation of the moon about the earth?

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14 read 145-146 p. 151 1, 3-6 extra p. 147 2-4, 6 p. 160 14-20

15 GRAVITATIONAL FIELDS A force field exists in the space surrounding an object in which a force is exerted on objects (e.g. gravitational, electric, magnetic). The strength of gravitational force fields is deter- mined by the Law of Universal Gravitation. If two or more gravitational fields are acting on an object then the net field is the sum of all the individual fields. read 274-275p.276 2-6p.277 1-8

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17 KEPLER’S LAWS In 1543 Copernicus proposes the heliocentric model of the solar system in which planets revolve around the sun in circular orbits. Slight irregularities show up over long periods of study. Tycho Brahe takes painstaking observations for 20 years with large precision instruments but dies (1600) before he can analyze them properly. A young mathematician continues Brahe’s work.

18 From his analysis the kinematics of the planets is fully understood. Kepler’s First Law of Planetary Motion Each planet moves around the Sun in an orbit that is an ellipse, with the Sun at one focus of the ellipse. Kepler’s Second Law of Planetary Motion The straight line joining a planet and the Sun sweeps out equal areas in space in equal intervals of time.

19 Planets move faster when they are closer to the Sun (centripetal force is stronger). equal areas equal times orbits are elliptical but are not very elongated

20 Kepler’s Third Law of Planetary Motion The cube of the average radius of a planet is directly proportional to the square of the period of the planet’s orbit. We have already proved this a few slides back. Recall

21 For our solar system m 1 is the mass of the sun. constant

22 Mars’ average distance from the sun is 2.28 x10 11 m while its period of rotation is 5.94 x 10 7 s. What is Jupiter’s average distance from the sun if its period of rotation is 3.75 x 10 8 s ? this equation holds for objects orbiting the same mass

23 read 278-283 p. 283 10-12 p. 284 4-7, 9

24 GRAVITATIONAL POTENTIAL ENERGY, AGAIN Recall the Law of Universal Gravitation for constant masses, a graph of force vs. radius would be...

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26 The graph above is a F vs. d graph which means the shaded area is the work required to move an object from r 1 to r 2. The shaded area is not easy to calculate but can be done with a geometric mean. In this case the work done by the lifter is equal to  E p. Another method involves calculus and integration over a range from r 1 to r 2.

27 geometric mean of force

28 Know these two equations, you are not required to know the previous development. Which preceding equation can be simplified to mg  h, the potential energy change near the earth’s surface?

29 Potential energy is a negative function! It increases until it is zero. PE stops here because the objects come into contact and cannot get closer.

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31 Recall

32 so... read p. 285-287p. 287 1-5

33 Escape from a Gravitational Field To escape a gravitational field an object must have at least a total mechanical energy of zero!! for escape

34 Escape energy - the minimum E K needed to pro- ject a mass (m 2 ) from the surface of another mass (m 1 ) to escape the gravitational force of m 1 Escape speed - the minimum speed needed to project a mass (m 2 ) from the surface of another mass (m 1 ) to escape the gravitational force of m 1 Binding energy - the additional E K needed by a mass (m 2 ) to escape the gravitational force of m 1 (similar to escape energy but applies to objects that possess E k i.e. satellites).

35 To calculate the escape energy or the escape speed of a mass (m 2 ): To calculate the binding energy of a mass (m 2 ): binding energy

36 Calculate the escape velocity of any object on the Earth’s surface.

37 The escape velocity is the same for all objects on the Earth’s surface while the escape energy is different for different massed object.

38 What is E k and E M of an orbiting body (satellite)? this is always true of satellites

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40 for an orbiting satellite !! Note that the total energy is negative since the satellite is “bound” to the central body.

41 read p.288-293 p. 293 6-11 #12 is interesting! extra p. 294 1-8 p. 300 1-17 25,26 look fun

42 a) What is the speed of Earth in orbit about the Sun? b) What is the total energy of Earth? c) What is the binding energy of Earth? d) If Earth was launched from the surface of the Sun to its present orbit then what velocity must it be launched with (Ignore the radius of Earth.)? e) If Earth came to rest and fell to the Sun then what velocity would it have when it hit the Sun (Ignore radius of Earth.)?

43 m e = 5.98x10 24 kg m s = 1.99x10 30 kg r e = 1.49x10 11 m (of orbit) r s = 6.96x10 8 m (of the body) G= 6.67x10 -11 Nm 2 kg -2

44 a) or

45 b) c) The binding energy is 2.664 x 10 33 J

46 d)

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48 e)


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