Astrophysics problem-solving question

The Sun has absolute magnitude +4.8 and surface temperature 5800 K. A distant star has absolute magnitude −4.2 and surface temperature 3500 K. (a) Calculate the difference in absolute magnitude and hence the ratio of the luminosities of the two stars. [2] (b) Use the Stefan-Boltzmann law to calculate the ratio of the radius of the distant star to the radius of the Sun. [3] (c) State where each star would be located on a Hertzsprung–Russell diagram. [2] (d) Describe the likely future evolution of a star similar to the Sun from the main sequence to its final state. [3]

(a) ΔM = −4.2 − (+4.8) = −9.0. L_star / L_Sun = 10^{0.4(M_Sun − M_star)} = 10^{0.4(4.8 − (−4.2))} = 10^{3.6} ≈ 4.0 × 10^3. So the star is about 4000 times more luminous than the Sun. (b) L = 4πr^2 σT^4, so L ∝ r^2 T^4. r_star / r_Sun = √(L_star/L_Sun) × (T_Sun/T_star)^2 = √(4.0 × 10^3) × (5800/3500)^2 = 63 × (1.66)^2 = 63 × 2.75 ≈ 173 ≈ 1.7 × 10^2. (c) The Sun is on the main sequence. The distant star is much cooler but much more luminous, so it is a red giant (upper right of the HR diagram). (d) A Sun-like star spends most of its life on the main sequence fusing hydrogen to helium in its core. When hydrogen in the core is exhausted, the core contracts and heats up while the outer layers expand and cool, making the star a red giant. Later the outer layers are ejected as a planetary nebula and the hot core remains as a white dwarf.

(a) [2] - Correct magnitude difference −9.0 (1) - Correct luminosity ratio ≈ 4 × 10^3 (1) (b) [3] - Correct proportional relationship r ∝ √(L)/T^2 (1) - Correct substitution of values (1) - Correct ratio ≈ 170 (accept 150–180) (1) (c) [2] - Sun on main sequence (1) - Distant star is a red giant / upper right of HR diagram (1) (d) [3] - Main-sequence hydrogen fusion phase (1) - Expansion to red giant with core contraction (1) - Planetary nebula and white-dwarf final state (1)

A lower absolute magnitude means a more luminous star. Luminosity depends on both radius and temperature.