Respuesta :
The apparent brightness is defined by how much light we get pretty much. The formula is:
[tex]AB= \frac{L}{4\pi d^2} [/tex]
This takes into account that the light is spreading out to a sphere that has the surface given by 4πd^2.
We have that d=20 ly and 1 ly is 9.46*10^12 km or 9.46*10^15m.
Hence, d=20*9.46*10^15=189.2 *10^15 m
Now, we can substitute the d value and calculate that L=8.45184* [tex]10^{-10}[/tex] W/m^2
[tex]AB= \frac{L}{4\pi d^2} [/tex]
This takes into account that the light is spreading out to a sphere that has the surface given by 4πd^2.
We have that d=20 ly and 1 ly is 9.46*10^12 km or 9.46*10^15m.
Hence, d=20*9.46*10^15=189.2 *10^15 m
Now, we can substitute the d value and calculate that L=8.45184* [tex]10^{-10}[/tex] W/m^2
The apparent brightness is [tex]8.667\times{10^{-33}}{\text{ W/}}{{\text{m}}^{\text{2}}}[/tex]
Further Explanation:
The apparent brightness of a light source is measured by dividing luminosity with surface area of sphere with radius that is equal to the distance from source of light. The formula for calculating apparent brightness is:
[tex]F{\text{}}={\text{}}\dfrac{L}{{4{\text{}}\pi{\text{}}{d^2}}}{\text{ }}[/tex]
Where,
d: the distance from source of light.
L: luminosity.
Calculate the apparent brightness:
[tex]\begin{aligned}F&=\dfrac{L}{{4\pi d^2}}}\\ &=\frac{{3.8\times1026{\text{watts}}}}{{4\pi{{\left( {{\text{189}}{\text{.2}}\times{\text{1}}{{\text{0}}^{15}}{\text{m}}}\right)}^2}}} \\&=\frac{{3898.8}}{{4.498\times{{10}^{35}}}}\\&=8.667\times{10^{-33}}{\text{ W/}}{{\text{m}}^{\text{2}}}\end{aligned}[/tex]
The apparent brightness is [tex]8.667\times{10^{-33}}{\text{ W/}}{{\text{m}}^{\text{2}}}[/tex]
Working Notes 1:
Conversion of distance in light year to meters:
[tex]\begin{gathered}{\text{d}}=20\times{\text{9}}{\text{.46 }}\times {\text{1}}{{\text{0}}^{15}}\\=189{\text{.2}}\times{\text{1}}{{\text{0}}^{15}}{\text{m}}\end{gathered}[/tex]
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Answer details:
Grade: High School
Subject: Astronomy
Chapter: Luminosity
Keywords: Suppose a star has the same luminosity as our sun (3.8×1026watts) but is located at a distance of 20 light−, what is its apparent brightness.
