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Chapter IX Gravitation A. Newtons Law of Gravitation B. Weight C. Gravitational Potential Energy D. Orbits E. Blach Hole.

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Presentation on theme: "Chapter IX Gravitation A. Newtons Law of Gravitation B. Weight C. Gravitational Potential Energy D. Orbits E. Blach Hole."— Presentation transcript:

1 Chapter IX Gravitation A. Newtons Law of Gravitation B. Weight C. Gravitational Potential Energy D. Orbits E. Blach Hole

2 Every particle or matter in the universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them. A. Newtons Law of Gravitation Gavitational Constant =

3 B. Weight w = mg

4 C. Gravitational Potential Energy work is given by, where F r is the radial component of the gravitational force F that is, the component in the direction outward from the center of the earth. Because F points directly inward toward the center of the earth, F, is negative. The magnitude of the gravitational force, by a minus sign,

5 W grav = - (U 2 – U 1 ) Gravitational Potential Energy

6 D. Orbits

7

8 E. BLACK HOLE The concept of a black hole is one of the most interesting and startling products of modern gravitational theory, yet the basic idea can be understood on the basis of Newtonian principles.

9 The Escape Speed from a Star Think first about the properties of our own sun. Its mass M = 1.99 X kg and radius R = 6.96 X 10 8 m are much larger than those of any planet, but compared to other stars, our sun is not exceptionally massive. The sun's average density,

10 The sun's temperatures range from 5800 K at the surface up to 1.5 X 10 7 K in the interior, so it surely contains no solids or liquids. Yet gravitational attraction pulls the sun's gas atoms together until the sun is, on average, 41 % denser than water and about 1200 times as dense as the air we breathe.

11 From the earth to the moon. Three men were sent to the moon in a shell fired from a giant cannon sunk in the earth in Florida. Find the escape speed-that is, the muzzle speed that would allow the shell to escape from the earth completely. Neglect air resistance, the earth's rotation, and the gravitational pull of the moon. The earth's radius is R E = 6380 km = 6.38 X 10 6 m, and its mass is m E = 5.97 X kg.

12 Solution: This result does not depend on the mass of the shell, nor does it depend on the direction in which the shell is launched

13 To generalize our result, the initial speed v, needed for a body to escape from the surface of a spherical mass M with radius R (ignoring air resistance) is Substituting M = V = (4/3 R 3 ) into Using either form of this equation, you can show that the escape speed for a body at the surface of our sun is v = 6.18 X 10 5 m/ s. This value, roughly 1/500 the speed of light.

14 Noted that if a body with the same average density as the sun had about 500 times the radius of the sun, its escape speed would be greater than the speed of light c. With his statement that "all light emitted from such a body would be made to return toward it," Mitchell became the first person to suggest the existence of what we now call a black hole.

15 The first expression for escape speed suggests that a body of mass M will act as a black hole if its radius R is less than or equal to a certain critical radius. by simply setting v = c For sun, radius of sun R = 6.96 X 10 8 m


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