Question
Please only use the information provided.
3. The Inner Workings of the Sun: Pressure measures force per unit area. The pres- 15 marks. sure at the center of the Sun comes from supporting the weight per unit area of a column of material all the way to the surface. From Newton's law of gravity, we can approximate the average gravitational field felt by the column as GMs/ R3, 2.0 x 1023 gm), Rs is the radius of the Sun where Ms is the mass of the Sun (Ms (Rs 6.96 x 10 cm and G is Netwon's gravitational constant (G 6.67 x 10 s2)). For the mass per unit area, we can divide the mass of the Sun by the square of the only parameter in the problem with length, the radius. So, the mass per unit area becomes Ms/R Therefore the approximate weight per unit area the pressure at the center of the Sun is the average gravitational field times the mass per unit area, P (MS/R2) x (GMs/R GM /R (a) Using the approximation P GM R1, what is the pressure at the center of the Sun? Check to make sure that the units of your answer is gm s2) [This unit is called the barye.J The actual Sun's internal pressure is only 19 times greater than this approximation. Correcting your answer by this factor of 19, how does the pressure at the center of the Sun compare with the Earth's atmospheric pressure at sea level, 1.01 x 100 barye? (b) The central density of the Sun is p 150 gm/cm3. Suppose that the center of the Sun is composed only of free protons with a mass of m 1.67 x 10 g What is the number of protons per cm3 at the Sun's center? (c) The ideal gas law tells us that P nkT where P is the pressure, n, is the particle number density, k is Boltzmann's constant (k 1.38 x 10 16 erg/K), and T is the temperature in Kelvin. Using the corrected value for the Sun's internal pressure that you calculated in part (a) and the proton number density that you calculated in part (b), what is the temperature at the center of the Sun?
Explanation / Answer
a) Mass of sun, Ms = 2*10^33 gm
Radius of sun, Rs = 6.96*10^10 cm
P = GMs^2/Rs^4 = 6.67*10^-8 * (2*10^33)^2/(6.96*10^10)^4 = 1.13696*10^16 gm / (cm s^2)
ACtual pressure = P' = 19*P = 2.16024*10^17 Barye
P'/Patm = 2.16024*10^17/1.01*10^6 = 2.1388537*10^11 times
b) rho = 150 gm/cm^3
mp = 1,67*10^-24 grams
number of protons = n
n*mp = rho*V ( V is volume of sun)
so, n*1.67*10^-24 = 150*4*pi*Rs^3 / 3 = 150*4*pi*(6.96*10^10)^3 / 3
n = 1.2678*10^59 protons
c) P = nkT
P = 2.16024*10^17 Barye
n = 1.2678*10^59 protons / V
V = 4*pi*Rs^3/3
k = 1.38*10^-16 erg/K
T = 17428831.58 K