# Luminosity of Neowise

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?

• 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 Oct 22 '20 at 5:34
• 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 Oct 22 '20 at 10:51
• No, my calculations don't agree! That's actually the thing I don't understand – space nerd Oct 22 '20 at 11:04