The temperature of the Sun’s surface (the so-called photosphere) is about 5500 C. Assume that the Sun’s surface is pretty close to a perfect black-body emitter.

(a) At what wavelength does the peak intensity of Sun’s radiation lie?
(b) Estimate the total power radiated by Sun into space. The Sun’s radius is 7.0E10^8 m.
(c) This radiated energy comes from a decrease in the Sun’s internal energy, which as
we know is equivalent to a decrease in mass. From the relativistic identity, how many
kilograms of mass are lost per second?
(d) From the amount of radiated energy, determine the power per unit area arriving at theEarth, 1.5E10^11 m away.
(e) The average power consumption in the US is about 2.9 TW. The best solar panels
todays convert about 20% of incident radiation into electricity. Estimate the area that
would need to be covered by solar panels if all this consumption were to be satisfied by
solar energy. Express your answer (i) in square kilometers and (ii) in fraction of the land
area of California.

(a) To find the wavelength at which the peak intensity of the Sun's radiation lies, we can use Wien's displacement law. According to this law, the peak wavelength (λmax) is inversely proportional to the temperature (T) of the black body emitter. The formula is given by:

λmax = b / T

where b is Wien's displacement constant, which is approximately equal to 2.898 × 10^-3 m·K.

Substituting the given temperature of the Sun's surface (5500°C = 5773 K) into the formula:

λmax = (2.898 × 10^-3 m·K) / 5773 K

Calculating the result, we find:

λmax ≈ 5.022 × 10^-7 m

So, the peak intensity of the Sun's radiation lies at a wavelength of approximately 502.2 nm.

(b) To estimate the total power radiated by the Sun into space, we can use the Stefan-Boltzmann law, which states that the total power radiated per unit area (P) is proportional to the fourth power of the absolute temperature (T) of the black body emitter. The formula is given by:

P = σ * A * T^4

where σ is the Stefan-Boltzmann constant, which is approximately equal to 5.67 × 10^-8 W/(m^2·K^4). A is the surface area of the Sun, given by:

A = 4 * π * r^2

where r is the radius of the Sun (7.0 × 10^8 m).

Substituting the values into the formula:

P = (5.67 × 10^-8 W/(m^2·K^4)) * (4 * π * (7.0 × 10^8 m)^2) * (5773 K)^4

Calculating the result, we find:

P ≈ 3.85 × 10^26 W

Therefore, the total power radiated by the Sun into space is approximately 3.85 × 10^26 watts.

(c) From the relativistic identity (E = mc^2), we know that a decrease in the Sun's internal energy is equivalent to a decrease in its mass. We can calculate the mass lost per second by dividing the radiated energy per second by the square of the speed of light (c^2):

Δm = E / c^2

Substituting the radiated energy (P) into the formula, we get:

Δm = (3.85 × 10^26 W) / (3.00 × 10^8 m/s)^2

Calculating the result, we find:

Δm ≈ 4.27 × 10^9 kg/s

Therefore, approximately 4.27 × 10^9 kilograms of mass are lost by the Sun per second.

(d) To determine the power per unit area arriving at the Earth, we need to consider the inverse square law, which states that the power per unit area decreases with the square of the distance. The formula is given by:

P/A = P0 / (4πr^2)

where P0 is the total power radiated by the Sun (3.85 × 10^26 W) and r is the distance from the Earth to the Sun (1.5 × 10^11 m).

Substituting the values into the formula:

P/A = (3.85 × 10^26 W) / (4π * (1.5 × 10^11 m)^2)

Calculating the result, we find:

P/A ≈ 1361 W/m^2

Therefore, the power per unit area arriving at the Earth is approximately 1361 watts per square meter.

(e) To estimate the area needed to satisfy the average power consumption in the US (2.9 TW) using solar energy, taking into account the conversion efficiency of solar panels (20%), we can use the formula:

Area = (Power needed) / (Conversion efficiency * Power per unit area)

The power needed is 2.9 TW, which is equivalent to 2.9 × 10^12 W.

Substituting the values into the formula:

Area = (2.9 × 10^12 W) / (0.2 * (1361 W/m^2))

Calculating the result, we find:

Area ≈ 1.07 × 10^9 m^2

To convert this area into square kilometers, we divide by 1 million:

Area ≈ 1.07 × 10^3 km^2

Therefore, the area needed to satisfy the average power consumption in the US using solar energy is approximately 1,070 square kilometers.

To express this in terms of the fraction of the land area of California, we need to know the total land area of California. The land area of California is approximately 423,970 square kilometers. Therefore, the fraction of the land area of California needed to be covered by solar panels would be:

Fraction = (Area needed) / (Land area of California)

Fraction = (1.07 × 10^3 km^2) / (423,970 km^2)

Calculating the result, we find:

Fraction ≈ 0.002527

So, approximately 0.2527% of the land area of California would need to be covered by solar panels to satisfy the average power consumption in the US.