he sun radiates like a perfect blackbody with an emissivity of exactly 1.

(a) Calculate the surface temperature of the sun, given it is a sphere with a 7.00 ✕ 108 m radius that radiates 3.80 ✕ 1026 W into 3 K space.
K
(b) How much power does the sun radiate per square meter of its surface?
W/m2
(c) How much power in watts per square meter is this at the distance of the earth, 1.50 ✕ 1011 m away? (This number is called the solar constant.)
W/m2

To calculate the values, we can use the Stefan-Boltzmann Law, which states that the power radiated by a blackbody is directly proportional to the fourth power of its temperature. The formula is given as:

P = σ * ε * A * T^4

Where:
P is the power radiated (in watts),
σ is the Stefan-Boltzmann constant (approximately 5.67 × 10^(-8) W/m^2K^4),
ε is the emissivity (which is 1 for a perfect blackbody),
A is the surface area of the object (in m²),
and T is the temperature (in Kelvin).

(a) To find the surface temperature of the Sun, we are given that it has a radius of 7.00 × 10^8 m and radiates 3.80 × 10^26 W. We need to calculate T.

Using the formula:
3.80 × 10^26 W = 5.67 × 10^(-8) W/m^2K^4 * 1 * 4π * (7.00 × 10^8 m)^2 * T^4

Simplifying the equation and solving for T:
T^4 = (3.80 × 10^26 W) / (5.67 × 10^(-8) W/m^2K^4 * 1 * 4π * (7.00 × 10^8 m)^2)

T^4 ≈ 1.267 × 10^7 K^4

Taking the fourth root of both sides:
T ≈ (1.267 × 10^7 K^4)^(1/4)

Using a calculator:
T ≈ 5778 K

Therefore, the surface temperature of the Sun is approximately 5778 Kelvin.

(b) To calculate the power radiated per square meter of the Sun's surface, we use the same formula:

P = σ * ε * A * T^4

The value of P here represents the total power radiated by the Sun, which we know from the previous calculation is 3.80 × 10^26 W. The value of A represents the surface area of the Sun, which can be calculated using the formula for the surface area of a sphere:

A = 4πr²

A ≈ 4π * (7.00 × 10^8 m)^2

Simplifying the equation, we have:
P = 5.67 × 10^(-8) W/m^2K^4 * 1 * 4π * (7.00 × 10^8 m)^2 * T^4

Substituting the values we know:
P / A = 3.80 × 10^26 W / [4π * (7.00 × 10^8 m)^2]

Simplifying the equation and solving for P/A:
P / A ≈ 6.16 × 10^7 W/m^2

Therefore, the Sun radiates approximately 6.16 × 10^7 watts per square meter of its surface.

(c) To calculate the power in watts per square meter at the distance of the Earth (1.50 × 10^11 m away), we can use the inverse square law for radiation:

P1 / P2 = (r2 / r1)^2

Where:
P1 is the power at the Sun's surface (which we calculated in part (b)),
P2 is the power at the distance of the Earth (which we need to find),
r1 is the radius of the Sun,
and r2 is the distance of the Earth from the Sun.

Substituting the known values:
P1 / P2 = (7.00 × 10^8 m / 1.50 × 10^11 m)^2

Simplifying and solving for P2:
P2 = P1 * (r1 / r2)^2

Substituting the values of P1 and r1 from the previous calculations:
P2 = 6.16 × 10^7 W/m^2 * (7.00 × 10^8 m / 1.50 × 10^11 m)^2

Simplifying the equation:
P2 ≈ 1.37 × 10^3 W/m^2

Therefore, at the distance of the Earth, the Sun radiates approximately 1.37 × 10^3 watts per square meter, which is known as the solar constant.

(a) To calculate the surface temperature of the sun, we can use the Stefan-Boltzmann Law, which relates the power radiated by a blackbody to its temperature.

The formula for the power radiated by a blackbody is:
P = σ*A*T^4

where P is the power radiated, σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4), A is the surface area of the blackbody, and T is the temperature in Kelvin.

In this case, the power radiated by the sun is given as 3.80 x 10^26 W, and the radius of the sphere is 7.00 x 10^8 m. The surface area of a sphere is given by the formula:
A = 4πr^2

Substituting these values into the equation, we have:
3.80 x 10^26 = 5.67 x 10^-8 * 4π(7.00 x 10^8)^2 * T^4

Simplifying this equation and solving for T, we get:
T^4 = (3.80 x 10^26) / (5.67 x 10^-8 * 4π(7.00 x 10^8)^2)
T^4 ≈ 3.48 x 10^15
T ≈ ∛(3.48 x 10^15)
T ≈ 6179 K

Therefore, the surface temperature of the sun is approximately 6179 K.

(b) To calculate the power radiated per square meter of the sun's surface, we can divide the total power radiated by the surface area of the sun.

The surface area of a sphere is given by:
A = 4πr^2

Substituting the given radius of the sun (7.00 x 10^8 m) into the formula, we get:
A = 4π(7.00 x 10^8)^2
A ≈ 6.16 x 10^18 m^2

Dividing the total power radiated by the surface area, we have:
Power per square meter = (3.80 x 10^26) / (6.16 x 10^18)
Power per square meter ≈ 6.17 x 10^7 W/m^2

Therefore, the power radiated per square meter of the sun's surface is approximately 6.17 x 10^7 W/m^2.

(c) To calculate the power per square meter at the distance of the earth, we can use the inverse square law, which states that the intensity of radiation decreases with the square of the distance.

The power per square meter at the distance of the earth is given by:
Power per square meter = (Power per square meter at the sun's surface) / (Distance from the sun to the earth)^2

Substituting the values, we have:
Power per square meter = (6.17 x 10^7) / (1.50 x 10^11)^2

Calculating the result, we get:
Power per square meter ≈ 1.37 x 10^3 W/m^2

Therefore, the power per square meter at the distance of the earth is approximately 1.37 x 10^3 W/m^2. This value is known as the solar constant.