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Knowing:

  • Star's radius $r_s$ , luminosity $L_s$, and absolute magnitude $V$

  • Planet's radius $r_p$ , albedo $a$ , and distance $d_s$

How can you calculate the absolute magnitude of a planet in a hypothetical star system?

When plugging Earth value into the answer given by Micheal B. here, it gives me $V_p$ = 19.83, but it should be around -3.99, although this could be an error on my part.

E.UCIT
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    Welcome to Astronomy Stack Exchange! Can you click "Edit" and add to you post what you've tried so far, and which equations you think might apply? Pure "please work this problem for me" questions are generally not well-received, and usually require some evidence of effort. Thanks! – uhoh Aug 07 '22 at 06:32
  • I don’t think absolute magnitude is well defined for planets(ie distance). For stars it’s 10 parsec, tho. – Max0815 Aug 07 '22 at 07:36
  • https://en.wikipedia.org/wiki/Absolute_magnitude – planetmaker Aug 07 '22 at 08:39
  • So this would be the magnitude of the planet at 10 parsecs as viewed from the star (so the phase would be full). You can use HORIZONS https://ssd.jpl.nasa.gov/horizons/app.html or similar to get this for our solar system's planets and curve-fit, assuming your hypothetical planets are similar enough to ours. In particular, that they're Lambertian reflectors: https://en.wikipedia.org/wiki/Lambertian_reflectance – Barry Carter Aug 07 '22 at 12:52
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    This question seems like it might be similar and has some good answers. https://astronomy.stackexchange.com/questions/5957/compute-planets-apparent-visual-magnitude – Greg Miller Aug 07 '22 at 13:45
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    Does this answer your question? Compute Planet's Apparent Visual Magnitude – WarpPrime Aug 07 '22 at 20:48
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    Whenever I plug Earth values into Micheal's answer to test it out, I get $V_p$ = -8.16 when it should be around -3.99; although, I could be doing something wrong.. – E.UCIT Aug 07 '22 at 21:38
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    Are you using the right units? (Parsecs for distance in Michael’s answer, even though it says light-years in the OP.) – Pierre Paquette Aug 07 '22 at 23:36
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    @PierrePaquette I think you may be correct about using the wrong units, but when using parsecs $V_p$ = 19.76 ($r_p$ and $d_s$ in parsecs) or $V_p$ = -62.68 ($d_s$ in parsecs). I think the parsecs unit is exclusive to $d_e-p$. – E.UCIT Aug 08 '22 at 05:15

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