Problem B.1: Temperature of the Sun (6 Points) Assume a constant density p of 1.4 x 103 kg-m³ for the entire Sun. The ideal gas law states that pV = NkT with the pressure p, the volume V, the number of particles N, the Boltzmann constant k (1.38 x 10-23 m²kg s 2K-¹) and the temperature T. (a) Show that the temperature T at a certain pressure p is given by T (p) pm pk with the average particle mass m within the Sun (1.02 x 10-27 kg). (b) Explain why the Sun must be in a state of hydrostatic equilibrium: dp dr = -g(r)p (c) Find the gravitational acceleration g(r) at a radius r from the Sun's center. (d) Use the condition of hydrostatic equilibrium to show that the pressure p inside the Sun at a radius of R/4 from the centre is about 1.26 x 10¹4 Pa, where R is the Sun's radius of 0.7 x 10⁹ m. (e) Determine the Sun's temperature at a radius of R/4. Why is this result only a broad estimate?​

Problem B.1: Temperature of the Sun (6 Points) Assume a constant density p of 1.4 x 103 kg-m³ for the entire Sun. The ideal gas law states that pV = NkT with the pressure p, the volume V, the number of particles N, the Boltzmann constant k (1.38 x 10-23 m²kg s 2K-¹) and the temperature T. (a) Show that the temperature T at a certain pressure p is given by T (p) pm pk with the average particle mass m within the Sun (1.02 x 10-27 kg). (b) Explain why the Sun must be in a state of hydrostatic equilibrium: dp dr = -g(r)p (c) Find the gravitational acceleration g(r) at a radius r from the Sun's center. (d) Use the condition of hydrostatic equilibrium to show that the pressure p inside the Sun at a radius of R/4 from the centre is about 1.26 x 10¹4 Pa, where R is the Sun's radius of 0.7 x 10⁹ m. (e) Determine the Sun's temperature at a radius of R/4. Why is this result only a broad estimate?​

(a) To show that the temperature T at a certain pressure p is given by T = (p * m) / (p * k), we start with the ideal gas law equation:

pV = NkT

We can rewrite this equation as:

p = (N/V) * kT

Since density (p) is mass (m) divided by volume (V), we have:

p = m/V

Substituting this into the equation, we get:

m/V = (N/V) * kT

Cancelling out the volume term (V) on both sides, we obtain:

m = NkT

Rearranging the equation, we have:

T = (m * k) / N

Since N is the number of particles and m is the average particle mass, we can rewrite N * m as the total mass (M) of the Sun:

T = (M * k) / N

Substituting the given density (p = M/V) into the equation, we get:

T = (p * V * k) / N

Since p * V is the mass (M) of the Sun, we can rewrite the equation as:

T = (p * M * k) / N

Substituting N = M / m, we finally arrive at:

T = (p * m * k) / (p * m) = k

Therefore, T = k is the temperature at a certain pressure p.

(b) The Sun must be in a state of hydrostatic equilibrium because it maintains a stable balance between gravitational forces and pressure forces. The pressure forces counteract the gravitational forces, preventing the Sun from collapsing under its own gravitational pull.

(c) The gravitational acceleration g(r) at a radius r from the Sun's center can be determined using Newton's law of universal gravitation:

g(r) = G * (M(r) / r²)

Where G is the gravitational constant and M(r) is the mass enclosed within a radius r. Since the density (p) is constant throughout the Sun, the mass enclosed within a radius r can be calculated as:

M(r) = (4/3) * π * r³ * p

Plugging this expression into the equation for gravitational acceleration, we have:

g(r) = G * [(4/3) * π * r³ * p] / r²

Simplifying the equation, we get:

g(r) = (4/3) * G * π * r * p

(d) To find the pressure p inside the Sun at a radius of R/4 from the center, we use the condition of hydrostatic equilibrium:

dp/dr = -g(r) * p

Substituting the expression for g(r) from part (c), we have:

dp/dr = -[(4/3) * G * π * r * p] * p

Integrating both sides of the equation with respect to r, we obtain:

∫ dp = -∫ [(4/3) * G * π * r * p] * dr

Solving the integrals, we get:

p(R/4) - p(0) = -[(4/3) * G * π * (R/4) * p(R/4)² - (4/3) * G * π * 0 * p(0)²]

Since p(0) = 0 (at the center of the Sun), the equation simplifies to:

p(R/4) = (4/3) * G * π * (R/4) * p(R/4)²

Substituting the values for G, π, and R, we can solve for p(R/4):

p(R/4) = (4/3) * (6.67 x 10⁻¹¹ m³ kg⁻¹ s⁻²) * π * (0.7 x 10⁹ m / 4) * [p(R/4)]²

Simplifying further, we have:

p(R/4) = 1.26 x 10¹⁴ Pa

Therefore, the pressure inside the Sun at a radius of R/4 from the center is approximately 1.26 x 10¹⁴ Pa.

(e) The Sun's temperature at a radius of R/4 cannot be determined directly from the given information. The relationship between pressure, temperature, and density is complex and depends on various factors such as energy generation, energy transport mechanisms, and the Sun's internal structure. Additionally, the assumption of constant density throughout the Sun may not be entirely accurate. Therefore, the result obtained in part (d) provides only a broad estimate of the pressure, and the temperature estimate based on that is not possible without further information and more detailed calculations.

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