Questions on the HR diagram and the life of Sun-like stars

Question 1

State the quantities plotted on the two axes of a Hertzsprung-Russell diagram, including the direction each axis runs and its typical range.

Accepted Answer

Absolute magnitude on the vertical axis, from about +15 at the bottom to −10 at the top, brighter upward on a reversed scale (1); temperature on the horizontal axis from about 50 000 K down to 2500 K left to right (1), often labelled instead by spectral class O B A F G K M. (If luminosity is plotted instead of magnitude, it increases upward normally on a log scale; only the temperature and magnitude scales run backwards.) (1)

Question 2

Name the three populated regions of the HR diagram and state where each sits.

Accepted Answer

The main sequence, a diagonal band from hot-bright (top left) to cool-dim (bottom right) (1); the red giants, above it on the cool side (1); and the white dwarfs, below it on the hot side (1).

Question 3

State what a star on the main sequence is doing, and where the Sun sits on the diagram.

Accepted Answer

It is fusing hydrogen in its core, the long stable phase of a star's life (1). The Sun sits partway down the main sequence, class G, absolute magnitude about +4.8 (1).

Question 4

State the Chandrasekhar limit, and what happens to a stellar core more massive than it.

Accepted Answer

About 1.4 solar masses, the greatest mass electron degeneracy pressure can support as a white dwarf (1). A heavier core collapses further, to a neutron star or a black hole (1).

Question 5

State two properties of a neutron star.

Accepted Answer

Any two of the following, one mark each (2): composed almost entirely of neutrons; mass of order one to two solar masses; radius of order 10 km; density comparable with that of an atomic nucleus.

Question 6

State what supports a white dwarf against gravitational collapse, and what energy source it has left.

Accepted Answer

Electron degeneracy pressure, the quantum refusal of electrons to be squeezed into the same state (1). None: fusion has ended, so it only radiates its stored thermal energy, cooling and fading (1).

Question 7

Describe the future evolution of the Sun as a path on the HR diagram, naming each stage and the physical change that drives it.

Accepted Answer

When core hydrogen runs low, fusion moves to a shell (1), the outer layers swell and the surface cools: the Sun leaves the main sequence and moves up and to the right, becoming a red giant (1). It then sheds its outer layers; the exposed core drops down and to the left as a white dwarf, no longer fusing (1), and slowly cools and fades (1).

Question 8

A red giant is cooler than the Sun yet far more luminous. Explain, using Stefan's law, how both can be true.

Accepted Answer

P = σAT⁴: its lower temperature cuts the power per unit area (1), but its enormously larger surface area more than compensates (1). High luminosity at low temperature forces a huge radius (1).

Question 9

A type Ia supernova peaks at apparent magnitude 18.2. Taking the peak absolute magnitude of a type Ia to be −19.3, calculate the distance to its host galaxy.

Accepted Answer

m − M = 18.2 − (−19.3) (1) m − M = 37.5 (1) 5 log(d/10) = 37.5, so d = 10 × 10^7.5 (1) d = 3.2 × 10⁸ pc (1)

Question 10

A neutron star has mass 3.0 × 10³⁰ kg and radius 12 km. Calculate its density.

Accepted Answer

V = (4/3)πr³ = 7.24 × 10¹² m³ (1) ρ = M/V = 3.0 × 10³⁰/(7.24 × 10¹²) (1) ρ = 4.1 × 10¹⁷ kg m⁻³ (1)

Question 11

Describe the formation of a star, from a cloud of gas and dust to its arrival on the main sequence.

Accepted Answer

The nebula collapses under its own gravity (1); the collapse heats the core, forming a protostar (1); when the core becomes hot enough, hydrogen fusion begins, and the star settles where the outward push of radiation and gas pressure balances gravity: it has joined the main sequence (1).

Question 12

An X-ray source is identified as a black hole with a Schwarzschild radius of 30 km. Determine its mass in solar masses. G = 6.67 × 10⁻¹¹ N m² kg⁻², c = 3.0 × 10⁸ m s⁻¹, solar mass = 1.99 × 10³⁰ kg.

Accepted Answer

M = R_sc²/2G (1) = 3.0 × 10⁴ × (3.0 × 10⁸)²/(2 × 6.67 × 10⁻¹¹) = 2.0 × 10³¹ kg (1), which is 2.0 × 10³¹/(1.99 × 10³⁰) ≈ 10 solar masses (1).

Question 13

The escape velocity from a spherical mass M of radius r is √(2GM/r). Show that setting the escape velocity equal to the speed of light leads to the expression R_s ≈ 2GM/c² for the Schwarzschild radius.

Accepted Answer

The event horizon is the radius at which the escape velocity reaches c (1). √(2GM/r) = c gives c² = 2GM/r (1), so r = R_s = 2GM/c² (1).

Question 14

Calculate the Schwarzschild radius of a black hole of six solar masses (solar mass 1.99 × 10³⁰ kg), and state what the radius represents.

Accepted Answer

R_s ≈ 2GM/c² (1)
= 2 × 6.67 × 10⁻¹¹ × 1.19 × 10³¹/(3.0 × 10⁸)² (1)
R_s = 18 km (1)
It is the radius of the event horizon, the boundary from inside which nothing, light included, can escape (1)

Question 15

Explain why type Ia supernovae can be used as standard candles, and what surprising conclusion their use led to in the late 1990s.

Accepted Answer

Each is a white dwarf detonating in closely similar conditions, near the Chandrasekhar limit, so every one peaks at nearly the same absolute magnitude, about −19.3 (1); comparing with the observed peak apparent magnitude gives the distance (1). Distant ones proved fainter than their red shifts predicted, which within the standard cosmological model means the expansion of the universe is accelerating, driven by what is now called dark energy (1). The peaks are not truly identical: they scatter by a few tenths of a magnitude, and the measured rate of decline of each light curve is used to correct the peak, so a type Ia is a standardisable candle.

Question 16

Describe the shape of the light curve of a type Ia supernova, and name a transient event that instead accompanies the death of a massive star.

Accepted Answer

A steep rise over about two weeks to the peak near absolute magnitude −19.3 (1), then a decline lasting months, slower than the rise, the rate of that decline being what standardises the peak (1). A type Ia is a detonating white dwarf, not a massive star; some collapsing massive stars fire gamma-ray bursts, brief flashes of gamma radiation detectable across most of the universe (1).

Question 17

A type Ia supernova detonates in a galaxy 40 Mpc away. An astronomy society's telescope can record objects brighter than apparent magnitude +15. Taking the peak absolute magnitude of a type Ia to be −19.3, deduce whether the society could record the supernova at its peak.

Accepted Answer

d = 40 Mpc = 4.0 × 10⁷ pc (1). m = M + 5 log(d/10) = −19.3 + 5 log(4.0 × 10⁶) (1) = +13.7 (1). Since +13.7 is brighter (smaller) than the +15 limit, the society could record the supernova (1).

Question 18

A star of twenty solar masses leaves the main sequence. Describe fully its subsequent evolution and its possible final states, explaining why its fate differs from the Sun's.

Accepted Answer

With core hydrogen exhausted, the star swells to a giant, and its greater mass drives fusion on beyond helium (1). Its core ends more massive than the Chandrasekhar limit of about 1.4 solar masses (1), so electron degeneracy pressure cannot hold it up as a white dwarf, and the core collapses (1). The outer layers detonate as a supernova, briefly outshining the galaxy (a gamma-ray burst may accompany the collapse) (1). The collapsed core survives as a neutron star, its protons and electrons crushed together into neutrons (1). If the remnant is massive enough, it falls inside its own Schwarzschild radius and becomes a black hole; the Sun, far below these masses, ends instead as a white dwarf (1).

Question 19

The Schwarzschild radius of one solar mass is about 3.0 km. A supermassive black hole at the centre of a galaxy has a mass of 2 × 10⁹ solar masses. Calculate its Schwarzschild radius and compare it with the Earth-Sun distance, 1.5 × 10¹¹ m.

Accepted Answer

R_s is directly proportional to M (1), so R_s = 3.0 km × 2 × 10⁹ ≈ 6.0 × 10¹² m (1). That is about 40 times the Earth-Sun distance: the event horizon alone would swallow the whole planetary system (1).

Question 20

A gamma-ray burst radiates 1.0 × 10⁴⁵ J in total, in a flash lasting only seconds. Calculate the total energy the Sun radiates over its whole ten-billion-year main-sequence lifetime, taking its luminosity as 3.9 × 10²⁶ W, and hence compare the two energy outputs.

Accepted Answer

Lifetime in seconds: t = 1.0 × 10¹⁰ × 3.156 × 10⁷ = 3.16 × 10¹⁷ s (1)
E_Sun = Lt = 3.9 × 10²⁶ × 3.16 × 10¹⁷ = 1.23 × 10⁴⁴ J (1)
ratio = E_burst/E_Sun = (1.0 × 10⁴⁵)/(1.23 × 10⁴⁴) ≈ 8.1 (1)
A single gamma-ray burst therefore radiates several times as much energy in a few seconds as the Sun manages across its entire ten-billion-year main-sequence lifetime (1)