<p>I believe, givn the paucity of information in the question, that what is meant is that if the comet is only visible by virtue of light it reflects, then what is its luminosity.</p>
<p>This can be calculated from how much light it receives, which is given by the flux of sunlight at its position (which you have calculated), multiplied by its cross-sectional area, which is <span class="math-container">$\pi r^2$</span>, where <span class="math-container">$r$</span> is the comet radius.
A fraction <span class="math-container">$A$</span>, where <span class="math-container">$A$</span> is the albedo, is then reflected. i.e.
<span class="math-container">$$L \simeq A \pi r^2 \left( \frac{L_\odot}{4\pi d^2}\right)\, , $$</span>
where <span class="math-container">$d$</span> is the distance from the Sun.</p>
<p>If instead the question is asking what the <em>intrinsic</em> luminosity of the comet is by virtue of it being heated by the sunlight then you would have to make all sorts of assumptions about whether it is in equilibrium or not (doubtful), whether it is heating up or cooling down, whether there is sublimination from the surface, and then possibly what its temperature is, and have a model for its emissivity.</p>
<p>If it were in equilibrium and you could assume that everything it absorbs is re-emitted (rather than say going into powering sublimation) then
<span class="math-container">$$ L \simeq (1-A)\pi r^2 \left( \frac{L_\odot}{4\pi d^2}\right)\, , $$</span>
which for <span class="math-container">$A=0.5$</span> is exactly the same answer (<span class="math-container">$A=0.5$</span> is a huge overestimate of the albedo of a cometary nucleus by the way).</p>

CC BY-SA 4.0
2022-01-28T16:30:11.577Z
2022-01-28T16:30:11.577Z
ProfRob
<p>Geez… only antlersoft gets it. AmeliaGrace wants the effective cross-sec of the <em>comet</em> , not a <em>nucleus</em> , which would be straightforward. A comet is a phenomenon of nucleus plus coma plus tail plus whatever adnexa. (Ulysses showed that tails, at multiple comets, are ridiculously long. A human notion of “tail” is limited by our eyesight, not what’s actually there or not.)</p>
<p>The extent of a coma, let alone tail, is inherently arbitrary- Rosetta flying through 67P/C-G’s coma was “filthy” by spacecraft-engineering standards but still cleaner than a cleanroom on Earth. The definition/calculation of coma radius/tail length then arbitrates with instrument aperture, exposure time, etc. And that was 67P/C-G. ‘Oumuamua was dust-poor, gas-rich to the point that its very comethood could only be showed mathematically, not directly observed (imaging) at all. And there is every shade of grey between these two extremes- 2/Encke, for one, is considered “dust-poor” because its solid emission is mainly macroscopic grains, which violate the standard for “dust” and show up poorly in most telescopes.</p>
<p>Faced with a continuum debate, the simplest path forward is to define some line in the sand, then proceed with that (arbitrary) standard, acknowledging its limited domain. AmeliaGrace even started down this trail in the woods:</p>
<blockquote>
<p>I was thinking that I could calculate the area of the comet seen from the Earth</p>
</blockquote>
<p>Assuming visible-light telescopes/detectors, this is somewhat straightforward. If there’s an issue, it’s an arbitrary line separating faint coma from noise/hot pixels, but we do that already, no choice but to. <em>Even then</em>, that line is drawn at different radii, for active comets versus comets still in the outer Solar System versus comets just not releasing much.</p>
<blockquote>
<p>However, I have no idea how to calculate the effective area of the comet.</p>
</blockquote>
<p>Given <em>a</em> comet, go ahead and take <em>the</em> image and draw <em>your</em> radius. But for comet <em>s</em> , you cannot give <em>the radii</em> of <em>their whole class</em> .</p>

CC BY-SA 4.0
2022-08-12T15:10:36.720Z
2022-08-12T15:10:36.720Z
caInstrument
<p>We can calculate the luminosity &quot;H&quot; (absolute magnitude) by means of:</p>
<p><span class="math-container">$$H=5 \ \log \frac{1329}{d \cdot \sqrt p}$$</span></p>
<p><strong>log</strong> is the decimal logarithm</p>
<p><strong>d</strong> is the diameter in km</p>
<p><strong>p</strong> is the albedo</p>
<p>For d=9 km and p=0.5 we obtain <strong>H=11.6</strong></p>
<p>We cannot calculate the apparent luminosity, <strong>since the distance to Earth is not given</strong>. If we knew the distance to the Earth, then we could calculate the apparent luminosity (apparent magnitude) by means of:</p>
<p><span class="math-container">$$m=H+5 \ \log (D_s \cdot D_e)-2.5 \ \log f(F)$$</span></p>
<p><strong>m</strong> is the apparent magnitude</p>
<p><strong>Ds</strong> is the distance to the Sun in Astronomical Units (AU)</p>
<p><strong>De</strong> is the distance to Earth in Astronomical Units (AU)</p>
<p><strong>f(F)</strong> is the phase function</p>
<p>For more details, consult:</p>
<p><a href="https://astronomy.stackexchange.com/a/48213/31201">Will the James Webb Space Telescope be visible from earth?</a></p>
<p>and:</p>
<p><a href="https://astronomy.stackexchange.com/a/48218/31201">Good source for the relationship between absolute magnitude, diameter, and albedo?</a></p>

CC BY-SA 4.0
2022-01-27T17:39:59.313Z
Is that a bolometric magnitude? If not can you say how we convert that into Watts (the units of luminosity). If I understand correctly, the "absolute magnitude" is actually a V-band magnitude assuming the object is 1 au from the Sun and viewed at 1 au from the Earth. So not really what's asked for. It would have to be scaled in some way.
ProfRob
CC BY-SA 4.0
2022-01-28T15:20:37.057Z
Note that it is not possible to know how many "watts" the observer receives from the comet, if the distance from the comet to the observer is not known.
I think the problem is missing data.
Albert
CC BY-SA 4.0
2022-01-28T16:13:20.447Z
How many Watts (per square metre) the observer receives is not the luminosity.
ProfRob
CC BY-SA 4.0
2022-01-27T15:29:22.057Z
2022-01-27T15:29:22.057Z
Albert
<p>Assuming a 9-km comet is heading towards the Earth and it is at distance 5.2 AU away from the Sun, what is its luminosity? Assume that it has an albedo of 0.5?</p>
<p>For now, I have calculated the Sun's flux at the comet's location.</p>
<p><span class="math-container">$$F=\frac{3.8\times10^{26}W}{4\pi\left(7.78\times10^{11}m\right)^2}=50.74\frac{W}{m^2}$$</span></p>
<p>I have used the flux of the Sun and the distance of the comet from the Sun to the in meters. I was thinking that I could calculate the area of the comet seen  from the Earth and then calculate the luminosity of the comet. However, I have no idea how to calculate the effective area of the comet.</p>

CC BY-SA 4.0
2022-01-26T03:57:09.147Z
The luminosity of comets is highly variable depending on the precise mix of volatiles and dust.  The visible coma and tail of a comet can be huge (if highly diffuse); 10^5 times or more the size of the nucleus.  On the other hand, at 5 AU distance the comet may not have much of a coma yet, and so this may just be a simple function of diameter and albedo as the exercise implies.  Determining the angular diameter of the comet as seen from earth is a simple exercise in trigonometry.
antlersoft
CC BY-SA 4.0
2022-01-26T09:36:44.583Z
For a better estimate on an object with an extended coma you also want to consider the phase angle as it matters for (back)scatter on small particles.
planetmaker
CC BY-SA 4.0
2022-01-26T17:46:58.243Z
@antlersoft the luminosity is not a function of how far awy it is from the Earth, only of how far awy it is from the Sun,. It isn't clear to me actually what is meant by the "luminosity". The intrinsic luminosity requires some sort of temperature and an emissivity model and/or an assumption of equilibrium. This probably matters whether it is moving towards the Sun or away, since I doubt it is in equilibrium.
ProfRob
<asteroids><luminosity><albedo><flux>
Calculating the luminosity of a comet
AmeliaGrace
# Calculating the luminosity of a comet

(Asked by: username_0 on 2022-01-25)

Assuming a 9-km comet is heading towards the Earth and it is at distance 5.2 AU away from the Sun, what is its luminosity? Assume that it has an albedo of 0.5?

For now, I have calculated the Sun's flux at the comet's location.

$$F=\frac{3.8\times10^{26}W}{4\pi\left(7.78\times10^{11}m\right)^2}=50.74\frac{W}{m^2}$$

I have used the flux of the Sun and the distance of the comet from the Sun to the in meters. I was thinking that I could calculate the area of the comet seen from the Earth and then calculate the luminosity of the comet. However, I have no idea how to calculate the effective area of the comet.

**Tags:** asteroids, luminosity, albedo, flux

> **Comments:**
> **username_1**: The luminosity of comets is highly variable depending on the precise mix of volatiles and dust.  The visible coma and tail of a comet can be huge (if highly diffuse); 10^5 times or more the size of the nucleus.  On the other hand, at 5 AU distance the comet may not have much of a coma yet, and so this may just be a simple function of diameter and albedo as the exercise implies.  Determining the angular diameter of the comet as seen from earth is a simple exercise in trigonometry.  
> **username_2**: For a better estimate on an object with an extended coma you also want to consider the phase angle as it matters for (back)scatter on small particles.  
> **username_3**: @username_1 the luminosity is not a function of how far awy it is from the Earth, only of how far awy it is from the Sun,. It isn't clear to me actually what is meant by the "luminosity". The intrinsic luminosity requires some sort of temperature and an emissivity model and/or an assumption of equilibrium. This probably matters whether it is moving towards the Sun or away, since I doubt it is in equilibrium.  

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## 3 answers

### Answer by username_3 on 2022-01-28. Score: 1 | Accepted answer

I believe, givn the paucity of information in the question, that what is meant is that if the comet is only visible by virtue of light it reflects, then what is its luminosity.

This can be calculated from how much light it receives, which is given by the flux of sunlight at its position (which you have calculated), multiplied by its cross-sectional area, which is $\pi r^2$, where $r$ is the comet radius.
A fraction $A$, where $A$ is the albedo, is then reflected. i.e.
$$L \simeq A \pi r^2 \left( \frac{L\_\odot}{4\pi d^2}\right)\, , $$
where $d$ is the distance from the Sun.

If instead the question is asking what the *intrinsic* luminosity of the comet is by virtue of it being heated by the sunlight then you would have to make all sorts of assumptions about whether it is in equilibrium or not (doubtful), whether it is heating up or cooling down, whether there is sublimination from the surface, and then possibly what its temperature is, and have a model for its emissivity.

If it were in equilibrium and you could assume that everything it absorbs is re-emitted (rather than say going into powering sublimation) then
$$ L \simeq (1-A)\pi r^2 \left( \frac{L\_\odot}{4\pi d^2}\right)\, , $$
which for $A=0.5$ is exactly the same answer ($A=0.5$ is a huge overestimate of the albedo of a cometary nucleus by the way).


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### Answer by username_4 on 2022-08-12. Score: 0

Geez… only username_1 gets it. username_0 wants the effective cross-sec of the *comet* , not a *nucleus* , which would be straightforward. A comet is a phenomenon of nucleus plus coma plus tail plus whatever adnexa. (Ulysses showed that tails, at multiple comets, are ridiculously long. A human notion of “tail” is limited by our eyesight, not what’s actually there or not.)

The extent of a coma, let alone tail, is inherently arbitrary- Rosetta flying through 67P/C-G’s coma was “filthy” by spacecraft-engineering standards but still cleaner than a cleanroom on Earth. The definition/calculation of coma radius/tail length then arbitrates with instrument aperture, exposure time, etc. And that was 67P/C-G. ‘Oumuamua was dust-poor, gas-rich to the point that its very comethood could only be showed mathematically, not directly observed (imaging) at all. And there is every shade of grey between these two extremes- 2/Encke, for one, is considered “dust-poor” because its solid emission is mainly macroscopic grains, which violate the standard for “dust” and show up poorly in most telescopes.

Faced with a continuum debate, the simplest path forward is to define some line in the sand, then proceed with that (arbitrary) standard, acknowledging its limited domain. username_0 even started down this trail in the woods:

> I was thinking that I could calculate the area of the comet seen from the Earth

Assuming visible-light telescopes/detectors, this is somewhat straightforward. If there’s an issue, it’s an arbitrary line separating faint coma from noise/hot pixels, but we do that already, no choice but to. *Even then*, that line is drawn at different radii, for active comets versus comets still in the outer Solar System versus comets just not releasing much.

> However, I have no idea how to calculate the effective area of the comet.

Given *a* comet, go ahead and take *the* image and draw *your* radius. But for comet *s* , you cannot give *the radii* of *their whole class* .


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### Answer by username_5 on 2022-01-27. Score: 0

We can calculate the luminosity "H" (absolute magnitude) by means of:

$$H=5 \ \log \frac{1329}{d \cdot \sqrt p}$$

**log** is the decimal logarithm

**d** is the diameter in km

**p** is the albedo

For d=9 km and p=0.5 we obtain **H=11.6**

We cannot calculate the apparent luminosity, **since the distance to Earth is not given**. If we knew the distance to the Earth, then we could calculate the apparent luminosity (apparent magnitude) by means of:

$$m=H+5 \ \log (D\_s \cdot D\_e)-2.5 \ \log f(F)$$

**m** is the apparent magnitude

**Ds** is the distance to the Sun in Astronomical Units (AU)

**De** is the distance to Earth in Astronomical Units (AU)

**f(F)** is the phase function

For more details, consult:

[Will the James Webb Space Telescope be visible from earth?](https://astronomy.stackexchange.com/a/48213/31201)

and:

[Good source for the relationship between absolute magnitude, diameter, and albedo?](https://astronomy.stackexchange.com/a/48218/31201)

> **Comments:**
> **username_3**: Is that a bolometric magnitude? If not can you say how we convert that into Watts (the units of luminosity). If I understand correctly, the "absolute magnitude" is actually a V-band magnitude assuming the object is 1 au from the Sun and viewed at 1 au from the Earth. So not really what's asked for. It would have to be scaled in some way.  
> **username_5**: Note that it is not possible to know how many "watts" the observer receives from the comet, if the distance from the comet to the observer is not known. I think the problem is missing data.  
> **username_3**: How many Watts (per square metre) the observer receives is not the luminosity.  

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