B1-4. Astrophysics

1. Luminosity & Apparent Brightness

When observing stars, we must distinguish between the total energy they emit and the energy we physically receive on Earth.

Apparent Brightness ($b$): The power of the star's radiation received per unit area on Earth.

$$b = \dfrac{L}{4 \pi d^2}$$

$b$: Apparent brightness (W m⁻²)
$L$: Luminosity, the total power radiated by the star (W). Note that for stars, $L$ is equivalent to $P$ in the Stefan-Boltzmann law ($L = \sigma A T^4$).
$d$: Distance from the star to Earth (m)

Example 1: Brightness of the Sun

Problem: The Sun has a total luminosity of approximately $3.83 \times 10^{26}$ W. The average distance from the Earth to the Sun is $1.50 \times 10^{11}$ m. Calculate the apparent brightness of the Sun as observed from Earth (this value is also known as the Solar Constant).

Solution:

Step 1: Apply the Apparent Brightness formula.
$$b = \dfrac{L}{4 \pi d^2}$$
Step 2: Substitute the known values.
$$b = \dfrac{3.83 \times 10^{26}}{4 \times \pi \times (1.50 \times 10^{11})^2} = \dfrac{3.83 \times 10^{26}}{4 \times \pi \times 2.25 \times 10^{22}} = \dfrac{3.83 \times 10^{26}}{2.83 \times 10^{23}}$$
Step 3: Solve.
$$b \approx \mathbf{1350 \text{ W m⁻²}}$$

2. Wien's Displacement Law

The wavelength where a black body's emission is most intense ($\lambda_{\text{max}}$) is determined entirely by its temperature.

Wien's Law Equation:

$$\lambda_{\text{max}} T = 2.9 \times 10^{-3}\text{ m K}$$

$\lambda_{\text{max}}$: The peak wavelength of emitted radiation (m)
$T$: Absolute surface temperature (K)

Interpretation: Hotter stars peak at shorter wavelengths (blue light), while cooler stars peak at longer wavelengths (red light).

Example 2: Star Temperature Comparison

Problem: The spectrum of the star Rigel peaks at a wavelength of $263$ nm, while the spectrum of Betelgeuse peaks at $828$ nm. Determine which of these two stars is cooler and calculate its surface temperature.