CH 8. Universal Gravitation Planetary and Satellite Motion

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1 CH 8 Universal Gravitation Planetary and Satellite Motion

2 Sir Isaac Newton UNIVERSAL GRAVITATION

3 Newton: Universal Gravitation Newton concluded that earthly objects and heavenly objects obey the same physical laws. When an object is placed above the earth s surface and dropped, the earth exerts a force on the object. This is called gravity

4 Sitting under the old apple tree one day, Newton realized that the moon s motion is caused by gravity (naturally) He used his own three laws, and Kepler s laws to formulate the law of Universal Gravitation.

5 Newton discovered that the force of gravity is inversely proportional to the square of the distance from the sun to the planet. This is called the inverse square law

6 Newton also theorized that the force of gravity was directly proportional to the product of the masses of two objects.

7 In the end, we get

directly proportional to the product of the masses of the

8 Newton s Laws of Universal Gravitation Every body in the universe attracts every other body with a force that is directly proportional to the product of the masses of the bodies and inversely proportional to the square of the distance between the bodies.

9 G is called Newton s Gravitational Constant, and has been determined to be Nm 2 /kg 2

example calculation Calculate the gravitational force between two objects when they are 7.50 10-1 m apart. Each object has a mass of 50 kg.

10 example calculation Calculate the gravitational force between two objects when they are m apart. Each object has a mass of 50 kg. F g = G (m 1 xm 2 ) d 2 G = 6.67 x Nm 2 /kg 2. F = (6.67 x ) x (50 kg) 2 (.75 m) 2 = (6.67 x )(4.44 x 10 3 ) = 2.96 X 10-7 N

11 Weight and Gravitational Attraction. 1. Find the weight of a 75 kg person on earth. W = 75 X 9.8 = 735 N 2. Find the gravitational force of attraction between a 75 kg person and the Earth m earth = 5.98 x kg r earth = 6.37 x 10 6 m Fg = G(Mm/r 2 ) = 6.67E-11(5.98E24)(75) / (6.37E6) 2 Fg = N

example calculation Calculate the gravitational force on a 6.0 10 2 kg spaceship that is 1.6 10 4 km above the surface of the Earth. M e = 5.98 10 24 kg r E = 6.38 x 10 6 m G = 6.

12 example calculation Calculate the gravitational force on a kg spaceship that is km above the surface of the Earth. M e = kg r E = 6.38 x 10 6 m G = 6.67 x Nm 2 /kg 2. F g = G (m 1xm 2 ) d 2 F g = 6.67 x Nm 2 /kg 2 x ( kg)( kg) (2.238 X 10 4 km) 2 = (6.67 x Nm 2 /kg 2 )(3.58 x ) 5 x 10 8 = 4.8 x 10 5 N

13 Johannes Kepler s Laws SATELLITE AND PLANETARY MOTION

14 In the early 1600s, Johannes Kepler proposed three laws of planetary motion: The path of the planets about the sun is elliptical in shape, with the center of the sun being located at one focus. (The Law of Ellipses) An imaginary line drawn from the center of the sun to the center of the planet will sweep out equal areas in equal intervals of time. (The Law of Equal Areas) The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of their average distances from the sun. (The Law of Harmonies)

15 Kepler's first law - the law of ellipses - planets orbit the sun in an elliptical path, with the sun located at one foci of that ellipse. And now to remember some geometry: An ellipse is a special curve in which the sum of the distances from every point on the curve to two other points is a constant. The two other points are known as the foci of the ellipse. The closer together that these points are, the more closely that the ellipse resembles the shape of a circle. a circle is the special case of an ellipse in which the two foci are at the same location.

17 Kepler's second law - the law of equal areas describes the speed at which any given planet will move while orbiting the sun. A planet moves fastest when it is closest to the sun and slowest when it is furthest away.

Kepler s Second Law An equal area is swept out (A 1 = A 2 ), by a line drawn between the two bodies, during the same time period ( t) anywhere along the orbital

18 Kepler s Second Law An equal area is swept out (A 1 = A 2 ), by a line drawn between the two bodies, during the same time period ( t) anywhere along the orbital path. It takes as much time to go from C to D as from A to B: t apogee perigee t This implies that the satellite velocity is greater when the radial distance is smaller.

Kepler's third law - the law of harmonies compares the orbital period and radius of orbit of

the ratio of the squares of the periods to the cubes of their average distances from the sun

This happens because as the distance from the sun increases, the average orbital speed of the

19 Kepler's third law - the law of harmonies compares the orbital period and radius of orbit of a planet to those of other planets. the ratio of the squares of the periods to the cubes of their average distances from the sun is the same for every one of the planets T 2 /r 3 Is pluto a planet? This happens because as the distance from the sun increases, the average orbital speed of the planet decreases, while the distance around the orbit is also increasing. For example Mercury completes one orbit every 88 days and Pluto completes one orbit every 249 years!

20 example problem Suppose a small planet is discovered that is 14 times as far from the sun as the Earth's distance is from the sun (1.5 x m). Use Kepler's law of harmonies to predict the orbital period of such a planet. T e2 /r e 3 = T p2 /r p 3 Rearranging to solve for T p : T p 2 = (T e 2 )(r p3 )/r e 3 (r p ) / (r e ) = 14 so T p 2 = T e where T e = 1 yr T p 2 = 1 yr 2 x 14 3 = 2744 yr 2 T p = 2744 yr 2 = 52.4 yrs

21 example problem The average orbital distance of Mars is 1.52 times the average orbital distance of the Earth. Knowing that the Earth orbits the sun in approximately 365 days, predict the time for Mars to orbit the sun. Given: r mars = 1.52 r earth and T earth = 365 days Kepler's third law relates the ratio of the period squared to the ratio of radius cubed (T mars / T earth ) 2 = (r mars / r earth ) 3 T 2 mars = T 2 earth x (r mars / r earth ) 3 r mars / r earth = 1.52 T 2 mars = (365 days) 2 x (1.52) 3 T mars = 684 days

25 Cannonball A cannonball is fired horizontally from a tall mountain to the ground below. Because of gravity, it strikes the ground with increased speed. A second cannonball is fired fast enough to go into circular orbit but gravity does not increase its speed. Why Refer to #19 & #23 in conceptual

27 MOON MOTION Draw a FBD and sum the forces acting on the moon as it orbits the earth. If the moon orbits the earth every 27.3 days at a radius of 3.8 x 10 8 meters. Determine the velocity and the acceleration due to gravity that the moon experiences.

28 Speed of a Satellite: F = G M m e = g r² mv 2 r

29 Speed of a Satellite: v = GM e r

30 v = velocity (m/s) G = gravitational consta M = mass of Earth (kg) e r = distance to center of the Earth (m)

32 Satellites What is the minimum velocity needed for a satellite to stay in orbit at 1500 km above the earth s surface. Given M earth = 5.98 x kg R earth = 6.37 x 10 6 m And the answer is m/s

33 TREETOP ORBIT If the earth had no air (atmosphere) or mountains to interfere, could a satellite given adequate initial velocity orbit arbitrarily close to the earth s surface provided it did not touch? a) Yes, it could. b) No, orbits are only possible at a sufficient distance above the earth s surface where gravitation is reduced. Explain.

PHYSICS. Chapter 13 Lecture FOR SCIENTISTS AND ENGINEERS A STRATEGIC APPROACH 4/E RANDALL D. KNIGHT Pearson Education, Inc.

PHYSICS. Chapter 13 Lecture FOR SCIENTISTS AND ENGINEERS A STRATEGIC APPROACH 4/E RANDALL D. KNIGHT Pearson Education, Inc. PHYSICS FOR SCIENTISTS AND ENGINEERS A STRATEGIC APPROACH 4/E Chapter 13 Lecture RANDALL D. KNIGHT Chapter 13 Newton s Theory of Gravity IN THIS CHAPTER, you will learn to understand the motion of satellites