I am wondering how I would be able to predict the Magnitude of fictional stars, given we know enough about them. As the formulae I can find online are mostly getting overly high values that don't seem accurate, and I am also left wondering about how to estimate temperature.

The star I am thinking of is a hypothetical G-type main-sequence star of 1.0908 Solar Masses (falls within the 0.9 to 1.1 M range for G-types), around the same density of the sun if that doesn't cause problems. What would the magnitude of such a star come out as? And can you explain while showing the process of finding that? (And how to estimate what temperature a star would be given mass and radius, if you feel like it, please).

I apologize for the bother, and thank you if you decide to help.

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    $\begingroup$ What did you find and where? What did Wikipedia say? $\endgroup$ Commented Mar 3, 2022 at 23:32
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    $\begingroup$ Wikipedia says m 1 - m ref = -2.5 log10(I 1/I ref), but can't find the other formula I tried as I am having difficulty finding the intensities as I don't know how to find the "net power radiated" P for P/(4πr^2) $\endgroup$
    – Zoey
    Commented Mar 4, 2022 at 0:06

1 Answer 1


I'll assume you mean absolute magnitude, since apparent magnitude depends on the distance between the observer and the star.

Let us assume the Sun has absolute magnitude +4.83. I'll use the letter "x" for magnitude since M will be used later for mass.

Absolute Magnitude formula: $x = x_{\odot} + 2.5 \log{L_{\odot}/L}$.

In your interested mass range, $L/L_{\odot} = (M/M_{\odot})^{4} = 1.0908^4$ so $x = x_{\odot} - 0.377 = 4.45$.

To learn about the physics, the main thing you want to read about is the main sequence, which relates absolute magnitude with temperature. This relation comes from the way temperature, luminosity, and radius all depend on the mass of a star. I'll elaborate on these next, mostly in hand-wavy words, as a fully accurate description would be a stellar astrophysics course.

  1. $M \propto R$

Stars exist in a balance between pressure and gravity called hydrostatic equilibrium. Stars with stronger gravity also need more pressure, and more pressure means higher central temperature. But stars mostly have similar central temperature, since if you crank up the temperature too much, that would imply a fusion rate which would blow the star apart, and stars fuse hydrogen just enough to maintain equilibrium. $\frac{dP}{dr} = -\rho g $ becomes $\sim T/R = \frac{GM}{R^{2}}$ so $T \propto M/R$. So if temperature is roughly constant between stars, then larger stars are more massive stars: $M \propto R$.

  1. Mass-Luminosity relation

Given a certain mass and radius, stars require a certain amount of pressure gradient against gravity. Well, it turns out the pressure gradient also means that energy is going to flow out of the star from hot to cold, from center to surface. So, after the algebra settles, you find some complicated relationships predicting stellar luminosity as a function of mass.

  1. Stefan-Boltzmann law

I can't believe I'm going to say "Stefan-Boltzmann" and "simple" in the same sentence, but the Stefan-Boltzmann law is actually the simplest of the 3 (depending on who you ask) as the other topics belong squarely in a stellar astrophysics course and this is an intro physics topic. Blackbodies radiate more power if they are larger and if they are hotter. $L = 4\pi R^{2} \sigma T^{4}$.

With these three ingredients, given a certain mass, you can find the radius and the luminosity, and hence the temperature.

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    $\begingroup$ Yes, I was referring to Absolute as that is the most important and constant one. Apparent is just based on distance. $\endgroup$
    – Zoey
    Commented Mar 4, 2022 at 23:39
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    $\begingroup$ Also thanks for the explanation. $\endgroup$
    – Zoey
    Commented Mar 4, 2022 at 23:41
  • $\begingroup$ Cheers, best of luck $\endgroup$
    – Alwin
    Commented Mar 5, 2022 at 0:28
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    $\begingroup$ Thanks, and also to anyone pedantic, yes, I meant distance and Absolute Magnitude, not just distance. I known some people on SE get extremely pedantic. $\endgroup$
    – Zoey
    Commented Mar 5, 2022 at 2:30

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