CHAPTER 45 - 4

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

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ,...
View Full Document

## This note was uploaded on 09/13/2010 for the course PHYSICS 7 taught by Professor ? during the Spring '08 term at Berkeley.

### Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online