problem_set_5_sol

# problem_set_5_sol - Astronomy 160 Frontiers and...

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

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: Astronomy 160: Frontiers and Controversies in Astrophysics Homework Set # 5 Solutions 1) Consider a binary star system consisting of two 1.5M neutron stars. Sup- pose one of the neutron stars is a pulsar with an average observed pulse period of exactly 2 seconds. If the orbit is circular and edge-on to our line of sight, what is the maximum and minimum pulse period observed? (Note: the pulse period P p obeys the same Doppler shift formulas as wave- length does). Total Mass: M = M NS + M P = 1 . 5 M + 1 . 5 M = 3 M = 6 × 10 30 kg, ORBITAL period: P = 8 hr = 8 hr × 3600 s hr = 3 × 10 4 s. Now use this information to solve for the orbital velocity of the pulsar. In order to do that, we must first find the total orbital velocity of the system, where V = V NS + V P . Because V NS M NS = V P M P and M NS = M P , then V P = V/ 2. Now combine the following equations a 3 = P 2 GM 4 π 2 and a = V P 2 π V P 2 π 3 = P 2 GM 4 π 2 V 3 P 3 8 π 3 = P 2 GM 4 π 2 V 3 = 2 πGM P V = 2 πGM P 1 / 3 V = 2 π × 7 × 10- 11 × 6 × 10 30 3 × 10 4 ! 1 / 3 V = 80 × 10 19 10 4 ! 1 / 3 V = 80 × 10 15 1 / 3 V = 4 × 10 5 m / s therefore V P = V 2 = 4 × 10 5 m / s 2 = 2 × 10 5 m / s This isn’t very fast so we can use the Newtonian Doppler formula. We are total that the average pulse period is P p = 2 s: Δ P p P p = v c 1 Δ P p = v c × P p = 2 × 10 5 m / s 3 × 10 8 m / s × 2 s = 4 3 × 10-...
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

problem_set_5_sol - Astronomy 160 Frontiers and...

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

View Full Document
Ask a homework question - tutors are online