Purpose: To understand the life cycle of stars and how stars produce various elements.
Procedural overview: Students can use basic algebra, physics and chemistry principles to estimate the conditions involved in stars and element formation.
Approximate class time: 30-50 minutes.
- Activity guide for students: The Pressure to Be a Star
- Calculators with scientific notation
Notes to the teacher:
This activity makes extensive use of these basic concepts, so if necessary, please review them in advance with your students:
- Scientific notation
- Newton’s law of gravity
- Ideal gas law
In order to keep the math and physics simple, this activity involves some crude approximations and plenty of hand waving. For more detailed calculations that arrive at similar final results, please see for example:
M. Schwarzschild. Structure and Evolution of the Stars. Dover, 1958.
R. Kippenhahn and A. Weigert. Stellar Structure and Evolution. Springer-Verlag, 1990.
S. L. Shapiro and S. A. Teukolsky. Black Holes, White Dwarfs, and Neutron Stars. Wiley, 1983.
Questions for students, with suggested answers:
Defined constants needed in the calculations:
- Speed of light = c = 3.00 x 108 m/sec
- Gravitational constant = G = 6.67 x 10-11 m3/(kg•s2)
1. Our sun is a fairly typical star, so we will do most of our calculations using it. The sun’s average radius is r = 6.96 x108 m. Assuming that the sun is perfectly spherical (it is actually slightly oblate or squished), what is its volume?
Vsun = (4/3) π r3 = 1.41 x 1027 m3
2. For comparison, Earth’s average radius is approximately 6.37 x 106 m. What is Earth’s volume, and how many Earths would fit inside the sun?
VEarth = 1.08 x 1021 m3, so 1.3 million Earths would fit inside the sun
3. The sun’s mass is M = 1.99 x 1030 kg. What is the sun’s average mass density, and how does that compare to the density of water (1000 kg/m3)?
ρavg = M/V = 1410 kg/m3, so 1.41 times the density of water
4. Since the sun is neither contracting nor expanding, the gravitational force pulling matter inward toward the center of the sun must be equal to the outward force of gas pressure generated by the heat of fusion reactions, a condition known as hydrostatic balance. Detailed equations for hydrostatic balance may be found using calculus, but a simple estimate can be made with algebra. From Newton’s law of gravity, the gravitational force pulling inward on matter of density ρavg near the outer edge of the sun would be GρavgM/r2. If the pressure at the center of the sun is Pc, the average pressure gradient between the center of the sun and its edge is Pc/r. Setting those two forces equal would give Pc/r = GρavgM/r2, or Pc = GρavgM/r. In reality, the pressure is much more concentrated in the center of the sun than that linear estimate suggests — roughly 100 times as concentrated — so a decent approximation is Pc = 100 Gρavg M/r.
Using this approximation, what is the pressure at the center of the sun, and how does it compare with the pressure of Earth’s atmosphere at sea level (1.01 x 105 Pa)? Solve the equation for a numerical answer, and also solve the equation using the appropriate units (1 Pa = kg/(m•s2)).
Pc = 100 GρavgM/r = 2.69 x 1016 Pa, so 2.66 x 1011 times Earth’s atmospheric pressure
5. Our assumption was that the pressure is concentrated 100-fold in the center compared with a simple linear estimate. That will compress the density to 100 times the average density. Calculate the density in the center of the sun:
ρc = 100 ρavg = 1.41 x 105 kg/m3
6. Using the ideal gas law with a gas constant for monatomic hydrogen gas, R = 8315 J/(kg•K), the central pressure would be related to the central density and central temperature as Pc = R ρc Tc. But hydrogen in the sun is ionized, so its protons and its electrons each contribute that much pressure, and the total pressure is doubled, Pc = 2 R ρc Tc.
What is the temperature at the center of the sun?
Tc = Pc/(2 R ρc) = 1.15 x 107 K
(Note: The actual value is 1.5 x 107 K, so we did well considering how crude our approximations were.)
7. Normal room temperature is around 295 Kelvin (K). How many times hotter than room temperature is the center of the sun?
Roughly 40,000–50,000 times hotter.
8. Based on the temperature thresholds at which these fusion reactions become significant, which of these reactions are most dominant in the sun?
Tc > 8 x 106 K : proton-proton fusion to produce 4He from protons
Tc > 1.8 x 107 K : carbon-catalyzed (CNO) fusion to produce 4He from protons
Tc > 1 x 108 K : helium burning 4He + 4He → 8Be and 8Be + 4He → 12C + γ
Tc > 1 x 108 to 1 x 109 K: 12C + 4He → 16O + γ and 12C + 12C → 23Na + 1H
Proton-proton fusion is dominant. There is a little carbon-catalyzed fusion.
9. Fusing protons together forms helium-4 nuclei, also called a particles. The mass of one proton is mp = 1.6726 x 10-27 kg. The mass of one a-particle is ma = 6.6442 x 10-27 kg. The total mass of four protons is larger than the mass of the helium-4 nucleus that the protons produced through fusion, so the excess mass is converted into energy. How much energy is produced by forming one a-particle?
E = (4mp – ma)c2 = 4.16×10-12 J
10. The sun produces approximately 3.84 x 1026 W (J/sec) of energy. How many a-particles does it produce per second?
9.24 x 1037a-particles per second
11. How much solar mass gets converted from hydrogen to helium per second?
6.18 x 1011 kg/sec (4 mp x number of a-particles produced per second)
12. If a neutron has a mass mn = 1.6749 x 10-27 kg and an average effective radius of 9.41 x 10-16 m, what is its density?
ρneutron = 4.80 x 1017 kg/m3 (Solve for Vneutron, then ρneutron = mn/Vneutron)
13. If all the matter in the sun were compressed to form a neutron star with the same density as for the neutron you just calculated, how much more dense would that be than the sun’s current average density?
3.40 x 1014 times more dense (ρavg /ρneutron)
14. If all the matter in the sun were compressed to form such a neutron star, what would its radius be?
9970 m = about 10 km (Vneutron star = (4/3) π rneutron star3 = Msun/ρneutron, solve for rneutron star)
Typical neutron stars are only somewhat more massive than the sun, so that is actually a good estimate of the radius of real neutron stars.
15. A mass M must be compressed to a Schwarzschild radius of RS to become a black hole. RS is also the point at which light cannot escape. The escape velocity for an object from a given radius is that such that the object’s positive kinetic energy is sufficient enough to cancel out the object’s negative gravitational potential energy, mv2/2 – GMm/R = 0, or R = 2GM/v2. If light cannot escape, v=c, one finds the Schwarzschild radius RS = 2GM/c2. If the sun were compressed to form a black hole, what would its Schwarzschild radius be?
RS = 2GM/c2 = 2950 m = about 3 km
16. Out of all of these facts and numbers, what surprises you the most?
Student answers will vary.
17. What other relevant characteristics or effects can you calculate using your knowledge of math and science?
Student answers will vary.