CHAPTER 45 - 4

CHAPTER 45 - 4 - that the sum of the angles is 540° 24 To...

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180° 90° 90° - 0° 90° 90° 180° 180° 180° 21. ( a ) The Schwarzschild radius for a star with mass equal to that of our Sun is R   = 2 GM S / c 2   = 2(6.67 × 10 –11  N     m 2 /kg 2 )(1.99 × 10 30  kg)/(3.00 × 10 8  m/s) 2  = 2.95 × 10 3  m = 2.95 km. ( b ) The Schwarzschild radius for a star with mass equal to that of the Earth is R   = 2 GM E / c 2   = 2(6.67 × 10 –11  N     m 2 /kg 2 )(5.97 × 10 24  kg)/(3.00 × 10 8  m/s) 2  = 8.9 × 10 –3  m = 8.9 mm. 22. If we use the data for our galaxy, we have R   = 2 GM / c 2   = 2(6.67 × 10 –11  N     m 2 /kg 2 )(3 × 10 41  kg)/(3.00 × 10 8  m/s) 2  =       4 × 10 14  m . 23. If we consider a triangle with its three vertices  on a great circle, such as one through the North  and South poles as shown in the diagram, we see 
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Unformatted text preview: that the sum of the angles is 540°. 24. To escape from an object with mass M and radius r , the kinetic energy of the body must be greater than the gravitational potential energy at the surface: ! mv 2 = GMm / r , or v esc 2 = 2 GM / r . At the Schwarzschild radius we have v esc 2 = 2 GM /(2 GM / c 2 ) = c 2 , or v esc = c . 25. We find the distance from Hubble’s law: v = Hd ; (0.010)(3.00 × 10 8 m/s) = [(70 × 10 3 m/s/Mpc)/(10 6 pc/Mpc)(3.26 ly/pc)] d ,...
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This note was uploaded on 09/13/2010 for the course PHYSICS 7 taught by Professor ? during the Spring '08 term at Berkeley.

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CHAPTER 45 - 4 - that the sum of the angles is 540° 24 To...

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