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What would be the surface brightness for a sky full of galaxies each with luminosity, Lgalaxy, and diameter, Dgalaxy? Assume face-on/symmetrical galaxies. Enter your answer in watts per square meter per steradian. Use scientific notation, with 2 decimal places.

a) L galaxy ×D^2 galaxy


b) Lgalaxy​/D^2galaxy


c) L galaxy×Dgalaxy


d) Lgalaxy/Dgalaxy


User Sajjad
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Final answer:

The surface brightness for a sky full of galaxies can be found by dividing the luminosity of a galaxy by its area. Since the area of a face-on galaxy is proportional to the square of its diameter, the surface brightness formula is Lgalaxy/D^2galaxy. This represents the flux, or luminosity per unit area, from a galaxy.

Step-by-step explanation:

The surface brightness of a sky full of galaxies can be determined by taking the luminosity of a single galaxy and dividing it by the area over which that luminosity is spread. The luminosity (L) for a celestial object is the energy emitted per second, while the surface over which this luminosity is spread for a disk-shaped galaxy, seen face-on, would be the area of the circle corresponding to its face, which can be calculated using the formula for the area of a circle, A = π(D/2)^2, where D is the diameter. If we look at the equation L = (A x F) from the provided reference, where A is the surface area and F is the flux, and we want to express this relationship in terms of luminosity per unit area, we would rearrange this to F = L/A. For a galaxy with a circular face, A = π(D/2)^2, which simplifies to A = (π/4)D^2. When we divide the luminosity of a galaxy, L, by its area, we therefore get F = L/(π/4)D^2. Since π is a constant, the critical part of the formula is L/D^2, which is the energy per second per unit area (flux), giving us the surface brightness. Thus, the surface brightness of a galaxy is getting the option (b) Lgalaxy/D^2galaxy.

User Pastapockets
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