Unassigned%20Problems%204

# Unassigned%20Problems%204 - Chapter 21: Stellar Explosions...

This preview shows pages 1–3. Sign up to view the full content.

Chapter 21: Stellar Explosions 1 Chapter 20 Problems 3. Refer to More Precisely 17-2 : (a) 1 A.U. = 215 solar radii. L = 215 2 (3000 / 5800) 4 , L = 3300 solar luminosities. (b) The radius is five time bigger than in (a) and so L is 25 times bigger or 82,500 solar luminosities. 4. 0.0004 = R 2 (12000 / 6000) 4 , R = 0.005 solar radii = 3480 km. 5. It looks like about a factor of 90 ± 10. If C, N, and O were 10 times less abundant, the CNO energy production would decline by a factor of 10. 7. 2.5log 10 (100) = 5, so a luminosity change of 100 corresponds to an absolute magnitude change of 5 over a given the time span of 10 5 years. This is a change in magnitude of 5x10 -5 per year, which would not be noticeable to the naked eye, but possibly with modern instrumentation. 8. Density is mass divided by volume. For the core: () 30 3 7 0.25 2 10 Density 4 1.5 10 3 π ×× = × Density = 3.5 × 10 7 kg/m 3 The core density of the Sun, as stated in Chapter 16, is 1.5 × 10 5 kg/m 3 so this is 230 times higher. For the giant’s envelope: () 30 3 10 0.5 2 10 Density 4 7.5 10 3 ×× = × Density = 5.6 × 10 -4 kg/m 3 This 3.7 × 10 -9 times lower than the core density of the Sun. 9. Neptune is at 30.1 A.U. This distance is 4.5 × 10 9 km. At 50 km/s, the time it takes to travel this distance will be 9 × 10 7 s or about 2.9 years. Proxima Centauri, the nearest star, is 1.30 pc distant, which is 4x10 13 km. This will take 8.1x10 11 seconds or 26,000 years. 12. L 1 /1 2 = L 2 /1.6 2 , L 2 / L 1 = 2.56. This is almost exactly a factor of one magnitude. 13. In solar units L = M 4 . We also know the main sequence lifetime of a star ( t ) is M/L times the lifetime of the Sun. That is t = 10 10 M/L . Substituting for L gives t = 10 10 / M 3 . (a) For a value of t = 400 million years gives 4 × 10 8 = 10 10 / M 3 , M = 2.9 solar masses.

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

View Full Document
Chapter 21: Stellar Explosions 2 (b) 2 × 10 9 = 10 10 / M 3 , M = 1.7 solar masses. 14. (a) Angular momentum is mvr but v = 2 GM/r . Here m is Jupiter’s mass, M the Sun’s mass and r is the radius of the orbit. Substituting for v gives angular momentum of m 2 GMr . If angular momentum is conserved then M 1 r 1 = M 2 r 2
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 11/01/2009 for the course PHYS 229 taught by Professor Mehmetyali during the Fall '09 term at Sabancı University.

### Page1 / 4

Unassigned%20Problems%204 - Chapter 21: Stellar Explosions...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online