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?

  • $\begingroup$ 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. $\endgroup$ – Rob Jeffries Oct 22 at 5:34
  • $\begingroup$ 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). $\endgroup$ – Rob Jeffries Oct 22 at 10:51
  • $\begingroup$ No, my calculations don't agree! That's actually the thing I don't understand $\endgroup$ – space nerd Oct 22 at 11:04

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.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.