Specification: AQA A-level Physics (7408) 3.9.2.6 Astrophysics option | Paper: 3 Section B | Data and formulae booklet: given in the exam
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📌 KEY DEFINITIONS – learn them word for word
Black hole and event horizon: a black hole is an object whose escape velocity is greater than the speed of light, so not even light can escape. The event horizon is the boundary around a black hole at which the escape velocity is equal to the speed of light; its radius is the Schwarzschild radius, Rs ≈ 2GM ÷ c².
Supernova: a rapid increase in the absolute magnitude of a star, typically by 10 or more, when it explodes. A core-collapse supernova happens when the iron core of a very massive star collapses; type Ia supernovae have a known peak absolute magnitude and are used as standard candles to measure distance.
Neutron star: the collapsed core left after a supernova, made almost entirely of neutrons, with a density similar to that of an atomic nucleus (about 1017 kg m−3). If the remaining core is massive enough, even neutron degeneracy cannot stop the collapse and a black hole forms.
The equation
Schwarzschild radius = (2 × gravitational constant × mass) ÷ (speed of light)² Rs ≈ 2GM ÷ c²
| Quantity | Symbol | Unit |
|---|---|---|
| Schwarzschild radius (radius of the event horizon) | Rs | metres (m) |
| gravitational constant, 6.67 × 10−11 | G | N m² kg−2 |
| mass of the black hole | M | kilograms (kg) |
| speed of light in a vacuum, 3.00 × 108 | c | m s−1 |
Rearranged: M = Rsc² ÷ 2G | Rs ∝ M, so a black hole of 10 solar masses has a Schwarzschild radius 10 times that of a 1 solar mass black hole | Convert solar masses to kg (× 1.99 × 1030) and km to m before you substitute, and remember to square c.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) Calculate the Schwarzschild radius of a black hole with the same mass as the Sun, 1.99 × 1030 kg, to 2 significant figures. (G = 6.67 × 10−11 N m² kg−2, c = 3.00 × 108 m s−1)
Show solution
F: Rs = 2GM ÷ c²
I: Rs = 2 × 6.67 × 10−11 × 1.99 × 1030 ÷ (3.00 × 108)² = 2.654… × 1020 ÷ 9.00 × 1016
F: Rs is already the subject
A: 2949.6… m = 2.9 km (2 s.f.), compared with the Sun’s actual radius of 696 000 km
Watch out: forgetting to square c. Dividing by 3.00 × 108 instead of 9.00 × 1016 gives an answer 3.00 × 108 times too big.
Q2 (Core) A stellar black hole has a mass of 10 solar masses. Calculate the radius of its event horizon in km, to 2 significant figures. (Mass of Sun = 1.99 × 1030 kg, G = 6.67 × 10−11 N m² kg−2, c = 3.00 × 108 m s−1)
Show solution
Convert first: M = 10 × 1.99 × 1030 = 1.99 × 1031 kg
F: Rs = 2GM ÷ c²
I: Rs = 2 × 6.67 × 10−11 × 1.99 × 1031 ÷ (3.00 × 108)²
F: Rs is already the subject
A: 29 496.2… m = 29 km (2 s.f.)
Q3 (Core) Calculate the radius to which the Earth (mass 5.97 × 1024 kg) would have to be compressed to become a black hole. Give your answer in mm, to 2 significant figures. (G = 6.67 × 10−11 N m² kg−2, c = 3.00 × 108 m s−1)
Show solution
F: Rs = 2GM ÷ c²
I: Rs = 2 × 6.67 × 10−11 × 5.97 × 1024 ÷ (3.00 × 108)²
F: Rs is already the subject
A: 8.848… × 10−3 m = 8.8 mm (2 s.f.), about the size of a marble
Q4 (Core) The event horizon of a black hole has a radius of 30 km. Calculate the mass of the black hole in kg, to 2 significant figures, and express it in solar masses. (Mass of Sun = 1.99 × 1030 kg, G = 6.67 × 10−11 N m² kg−2, c = 3.00 × 108 m s−1)
Show solution
Convert first: 30 km = 3.0 × 104 m
F: Rs = 2GM ÷ c²
I: 3.0 × 104 = 2 × 6.67 × 10−11 × M ÷ (3.00 × 108)²
F: M = Rsc² ÷ 2G = 3.0 × 104 × 9.00 × 1016 ÷ 1.334 × 10−10
A: 2.023… × 1031 kg = 2.0 × 1031 kg (2 s.f.), and 2.023… × 1031 ÷ 1.99 × 1030 = 10 solar masses (2 s.f.)
Q5 (Exam) Sagittarius A*, the supermassive black hole at the centre of the Milky Way, has a mass of 4.3 × 106 solar masses. Calculate its Schwarzschild radius, to 2 significant figures, and express it (a) as a multiple of the Sun’s radius and (b) in astronomical units, both to 2 significant figures. (Mass of Sun = 1.99 × 1030 kg, radius of Sun = 6.96 × 108 m, 1 AU = 1.50 × 1011 m, G = 6.67 × 10−11 N m² kg−2, c = 3.00 × 108 m s−1)
Show solution
Convert first: M = 4.3 × 106 × 1.99 × 1030 = 8.557 × 1036 kg
F: Rs = 2GM ÷ c²
I: Rs = 2 × 6.67 × 10−11 × 8.557 × 1036 ÷ (3.00 × 108)²
F: Rs is already the subject
A: 1.268… × 1010 m = 1.3 × 1010 m (2 s.f.). (a) 1.268… × 1010 ÷ 6.96 × 108 = 18 solar radii. (b) 1.268… × 1010 ÷ 1.50 × 1011 = 0.085 AU, well inside the orbit of Mercury.
Q6 (Exam) Cygnus X-1 contains a black hole of about 21 solar masses orbiting a blue supergiant. (a) Calculate the radius of its event horizon in km, to 2 significant figures. (b) Without repeating the full calculation, use your answer to (a) to find the Schwarzschild radius of a 42 solar mass black hole. (Mass of Sun = 1.99 × 1030 kg, G = 6.67 × 10−11 N m² kg−2, c = 3.00 × 108 m s−1)
Show solution
Convert first: M = 21 × 1.99 × 1030 = 4.179 × 1031 kg
F: Rs = 2GM ÷ c²
I: Rs = 2 × 6.67 × 10−11 × 4.179 × 1031 ÷ (3.00 × 108)²
F: Rs is already the subject
A: (a) 61 942.0… m = 62 km (2 s.f.). (b) Rs ∝ M and the mass is doubled, so Rs = 2 × 61 942.0… m = 123 884.1… m = 120 km (2 s.f.)
Watch out: Rs is directly proportional to M, not to M² or √M. Doubling the mass doubles the radius.
Q7 (Exam) The first black hole to be imaged, at the centre of the galaxy M87, has a mass of 6.5 × 109 solar masses. (a) Calculate its Schwarzschild radius in AU, to 2 significant figures, and compare it with Neptune’s orbital radius of 30 AU. (b) Calculate the time light would take to travel across the diameter of its event horizon, in hours, to 2 significant figures. (Mass of Sun = 1.99 × 1030 kg, 1 AU = 1.50 × 1011 m, G = 6.67 × 10−11 N m² kg−2, c = 3.00 × 108 m s−1)
Show solution
Convert first: M = 6.5 × 109 × 1.99 × 1030 = 1.2935 × 1040 kg
Step 1 – (a) Schwarzschild radius
F: Rs = 2GM ÷ c²
I: Rs = 2 × 6.67 × 10−11 × 1.2935 × 1040 ÷ (3.00 × 108)² = 1.917… × 1013 m
A: 1.917… × 1013 ÷ 1.50 × 1011 = 127.8… AU = 130 AU (2 s.f.), about 4 times Neptune’s orbital radius
Step 2 – (b) light-crossing time
t = 2Rs ÷ c = 2 × 1.917… × 1013 ÷ 3.00 × 108 = 127 816.9… s
A: 127 816.9… ÷ 3600 = 35.50… h = 36 hours (2 s.f.)
Q8 (Exam) A neutron star formed in a supernova has a mass of 1.4 solar masses and a radius of 12 km. (a) Calculate the Schwarzschild radius for this mass in km, to 2 significant figures, and explain whether the neutron star is a black hole. (b) Calculate the escape velocity from its surface as a fraction of c, to 2 significant figures, using v = √(2GM ÷ r). (Mass of Sun = 1.99 × 1030 kg, G = 6.67 × 10−11 N m² kg−2, c = 3.00 × 108 m s−1)
Show solution
Convert first: M = 1.4 × 1.99 × 1030 = 2.786 × 1030 kg; 12 km = 1.2 × 104 m
Step 1 – (a) Schwarzschild radius
F: Rs = 2GM ÷ c²
I: Rs = 2 × 6.67 × 10−11 × 2.786 × 1030 ÷ (3.00 × 108)²
A: 4129.4… m = 4.1 km (2 s.f.). The star’s radius (12 km) is larger than Rs, so it is not a black hole: light can escape from its surface.
Step 2 – (b) escape velocity
v = √(2 × 6.67 × 10−11 × 2.786 × 1030 ÷ 1.2 × 104) = 1.759… × 108 m s−1
A: 1.759… × 108 ÷ 3.00 × 108 = 0.59c (2 s.f.)
Watch out: Rs ≈ 2GM ÷ c² comes from setting the escape velocity equal to c in v = √(2GM ÷ r). It is the radius the mass would need to be squeezed inside, not the radius of the object.
Q9 (Stretch) The mean density of a black hole can be defined as its mass divided by the volume inside its event horizon, (4/3)πRs³. (a) Calculate the mean density of a 1 solar mass black hole, to 2 significant figures, and compare it with the density of a neutron star (about 1017 kg m−3). (b) Show that the mean density is proportional to 1 ÷ M², and use this to find the mean density of the 6.5 × 109 solar mass black hole in M87, to 2 significant figures. (Mass of Sun = 1.99 × 1030 kg, G = 6.67 × 10−11 N m² kg−2, c = 3.00 × 108 m s−1)
Show solution
Step 1 – (a) Schwarzschild radius
Rs = 2 × 6.67 × 10−11 × 1.99 × 1030 ÷ (3.00 × 108)² = 2949.6… m
Step 2 – (a) volume and density
V = (4/3)π × 2949.6…³ = 1.074… × 1011 m³
F: ρ = M ÷ V
I: ρ = 1.99 × 1030 ÷ 1.074… × 1011
A: 1.851… × 1019 kg m−3 = 1.9 × 1019 kg m−3 (2 s.f.), about 200 times a neutron star’s density
Step 3 – (b) proportionality
ρ = M ÷ (4/3)πRs³ and Rs ∝ M, so ρ ∝ M ÷ M³ = 1 ÷ M²
Step 4 – (b) M87 black hole
ρ = 1.851… × 1019 ÷ (6.5 × 109)² = 0.4381… kg m−3
A: 0.44 kg m−3 (2 s.f.), less than the density of air at sea level (about 1.2 kg m−3)
Watch out: a supermassive black hole is not dense in this sense. Its mean density inside the event horizon falls as its mass increases.
Q10 (Stretch) The star S2 orbits Sagittarius A* with a period of 16.0 years. Treat the orbit as a circle of radius 1000 AU. (a) Use T² = (4π² ÷ GM)r³ to calculate the mass of Sagittarius A* in solar masses, to 2 significant figures. (b) Calculate its Schwarzschild radius, to 2 significant figures. (c) Calculate how many times larger the radius of S2’s orbit is than the Schwarzschild radius, to 2 significant figures. (1 year = 3.16 × 107 s, 1 AU = 1.50 × 1011 m, mass of Sun = 1.99 × 1030 kg, G = 6.67 × 10−11 N m² kg−2, c = 3.00 × 108 m s−1)
Show solution
Convert first: T = 16.0 × 3.16 × 107 = 5.056 × 108 s; r = 1000 × 1.50 × 1011 = 1.50 × 1014 m
Step 1 – (a) mass from the orbit
F: T² = (4π² ÷ GM)r³, so M = 4π²r³ ÷ GT²
I: M = 4π² × (1.50 × 1014)³ ÷ (6.67 × 10−11 × (5.056 × 108)²) = 1.332… × 1044 ÷ 1.705… × 107
A: 7.814… × 1036 kg, and 7.814… × 1036 ÷ 1.99 × 1030 = 3.9 × 106 solar masses (2 s.f.)
Step 2 – (b) Schwarzschild radius
F: Rs = 2GM ÷ c²
I: Rs = 2 × 6.67 × 10−11 × 7.814… × 1036 ÷ (3.00 × 108)²
A: 1.158… × 1010 m = 1.2 × 1010 m (2 s.f.)
Step 3 – (c) comparison
1.50 × 1014 ÷ 1.158… × 1010 = 12 950.4…
A: 13 000 times (2 s.f.). An object this massive packed into a region this small and producing almost no light is strong evidence for a supermassive black hole.
Watch out: forgetting to convert years to seconds and AU to metres. Every quantity in M = 4π²r³ ÷ GT² must be in SI units.
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