Questions: Flux, Parallax, Luminosity Star 1, 4.19 × 10^-9 W / m^2, 0.094'2, Ls

Solution Steps

We are given the following data for Star 1:

• Flux (\(F\)): \(4.19 \times 10^{-9} \, \mathrm{W/m^2}\)
• Parallax (\(p\)): \(0.094 \, \text{arcseconds}\)
• Luminosity (\(L\)): \(L_s\) (solar luminosities)

The distance \(d\) in parsecs (pc) can be calculated from the parallax \(p\) in arcseconds using the formula: \[ d = \frac{1}{p} \] Given \(p = 0.094 \, \text{arcseconds}\): \[ d = \frac{1}{0.094} \approx 10.6383 \, \text{pc} \]

The luminosity \(L\) of a star can be calculated using the inverse square law for flux: \[ F = \frac{L}{4 \pi d^2} \] Rearranging to solve for \(L\): \[ L = F \cdot 4 \pi d^2 \] Substituting the given values: \[ L = (4.19 \times 10^{-9} \, \mathrm{W/m^2}) \cdot 4 \pi (10.6383 \, \text{pc})^2 \]

1 parsec (pc) is approximately \(3.086 \times 10^{16} \, \text{m}\): \[ d = 10.6383 \, \text{pc} \times 3.086 \times 10^{16} \, \text{m/pc} \approx 3.282 \times 10^{17} \, \text{m} \]

Substituting the distance in meters: \[ L = (4.19 \times 10^{-9} \, \mathrm{W/m^2}) \cdot 4 \pi (3.282 \times 10^{17} \, \text{m})^2 \] \[ L \approx (4.19 \times 10^{-9}) \cdot 4 \pi (1.076 \times 10^{35}) \] \[ L \approx (4.19 \times 10^{-9}) \cdot (1.351 \times 10^{36}) \] \[ L \approx 5.658 \times 10^{27} \, \text{W} \]

Final Answer