<p>I think I get it - you are trying to calculate a luminosity based on some assumption about how much light is reflected/scattered from the Sun towards the Earth - hence the relevance of the flux from the Sun at the comet.</p>
<p>I think it's very hard to adopt this approach. There is uncertainty about what is the appropriate geometry to use for the comet and how the reflection/scattering should be treated. It is doubtful that assuming a comet diameter of 5 km will be appropriate because I would have thought most of the light we see is scattered from a coma around the comet, not from the surface of the comet itself.</p>

CC BY-SA 4.0
2020-10-22T10:58:21.307Z
2020-10-22T10:58:21.307Z
ProfRob
<p>So, I have been trying to figure out the bolometric luminosity of the comet C/2020 F3 (Neowise) on the 22nd of June, 2020. At that time, Neowise was <span class="math-container">$1.4$</span> AU away from Earth and had an apparent magnitude of <span class="math-container">$m=3$</span>. By using the formula <span class="math-container">$$m=-2.5log\Bigl(\frac{F}{F_O}\Bigl)$$</span> and <span class="math-container">$$F=\frac{L}{4\pi D^2}$$</span> where <span class="math-container">$F_0$</span> is the flux of Vega, <span class="math-container">$F$</span> is the flux of the comet, <span class="math-container">$D$</span> is the distance from Earth to the comet. I found that the bolometric luminosity, <span class="math-container">$L$</span> ,of the comet Neowise was around <span class="math-container">$7.45\cdot10^{14}$</span> watts when it was <span class="math-container">$1.4$</span> AU away from Earth. However, on the 22nd of June, Neowise was also <span class="math-container">$0.4$</span> AU away from the Sun. I have tried to use the flux from the Sun at this distance to determine the luminosity of Neowise again. I have found the flux from the Sun at that distance to be <span class="math-container">$8488.26$</span> <span class="math-container">$Wm^{-2}$</span> using the above formula for flux, but now using <span class="math-container">$D=0.4 AU$</span>. After that I thought of calculating the luminosity by using flux from the sun times the area of Neowise. Assuming the nucleus of Neowise has a diameter of <span class="math-container">$5 km$</span>, and assuming that the comet is fully illuminated,</p>
<p>should I assume that the nucleus of Neowise is spherical or circular? And why should the 2 luminosities be different?</p>

CC BY-SA 4.0
2020-10-22T05:34:54.400Z
It's quite unclear what you are trying to do, what your assumptions are or what meaning you are attaching to the symbols in your formulae.
ProfRob
CC BY-SA 4.0
2020-10-22T10:51:01.870Z
I'm still not with you. Why would these calculations agree if you've used a different value (and definition) for $D$? And of course the "luminosity" of a comet does change with distance from the Sun (and due to all sorts of other factors too).
ProfRob
CC BY-SA 4.0
2020-10-22T11:04:48.873Z
No, my calculations don't agree! That's actually the thing I don't understand
space nerd
<observational-astronomy><comets><luminosity>
Luminosity of Neowise
space nerd
# Luminosity of Neowise

(Asked by: username_0 on 2020-10-21)

So, I have been trying to figure out the bolometric luminosity of the comet C/2020 F3 (Neowise) on the 22nd of June, 2020. At that time, Neowise was $1.4$ AU away from Earth and had an apparent magnitude of $m=3$. By using the formula $$m=-2.5log\Bigl(\frac{F}{F\_O}\Bigl)$$ and $$F=\frac{L}{4\pi D^2}$$ where $F\_0$ is the flux of Vega, $F$ is the flux of the comet, $D$ is the distance from Earth to the comet. I found that the bolometric luminosity, $L$ ,of the comet Neowise was around $7.45\cdot10^{14}$ watts when it was $1.4$ AU away from Earth. However, on the 22nd of June, Neowise was also $0.4$ AU away from the Sun. I have tried to use the flux from the Sun at this distance to determine the luminosity of Neowise again. I have found the flux from the Sun at that distance to be $8488.26$ $Wm^{-2}$ using the above formula for flux, but now using $D=0.4 AU$. After that I thought of calculating the luminosity by using flux from the sun times the area of Neowise. Assuming the nucleus of Neowise has a diameter of $5 km$, and assuming that the comet is fully illuminated,

should I assume that the nucleus of Neowise is spherical or circular? And why should the 2 luminosities be different?

**Tags:** observational-astronomy, comets, luminosity
> **Comments:**
> **username_1**: It's quite unclear what you are trying to do, what your assumptions are or what meaning you are attaching to the symbols in your formulae.  
> **username_1**: I'm still not with you. Why would these calculations agree if you've used a different value (and definition) for $D$? And of course the "luminosity" of a comet does change with distance from the Sun (and due to all sorts of other factors too).  
> **username_0**: No, my calculations don't agree! That's actually the thing I don't understand  

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## 1 answers

### Answer by username_1 on 2020-10-22. Score: 4 | Accepted answer

I think I get it - you are trying to calculate a luminosity based on some assumption about how much light is reflected/scattered from the Sun towards the Earth - hence the relevance of the flux from the Sun at the comet.

I think it's very hard to adopt this approach. There is uncertainty about what is the appropriate geometry to use for the comet and how the reflection/scattering should be treated. It is doubtful that assuming a comet diameter of 5 km will be appropriate because I would have thought most of the light we see is scattered from a coma around the comet, not from the surface of the comet itself.


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