Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Surface gravity as a function of luminosity?

  1. Jan 25, 2012 #1
    I've been using these forums for awhile but this is my first post. So thanks in advance for taking the time to consider this problem. It is much appreciated.

    The question is: How does surface gravity vary as a function of luminosity along the main sequence?

    By surface gravity, I'm sure the question is asking for [itex]g=\frac{G M}{R^2}[/itex]. Knowing that luminosity is [itex]L=4 \pi R^2\sigma _{\text{SB}} T^4[/itex]. I've solved for R in the luminosity equation and obtained [itex]g=\frac{4 \pi G M T^4 \sigma _{\text{SB}}}{L}[/itex], but I'm pretty sure the problem is supposed to be harder than this. It's obvious (according to this relationship that's probably wrong) that if luminosity increases, then the surface gravity becomes weaker. So am I looking at this problem too simplistically?
     
  2. jcsd
  3. Jan 25, 2012 #2

    mathman

    User Avatar
    Science Advisor
    Gold Member

    You are assuming all the other terms in the expression for g as a function of L remain constant. I am not a student of this subject, but that appears to be a very strong assumption.
     
  4. Jan 25, 2012 #3
    All of them are constants except for L and possibly T. I'm pretty sure M is a constant because surface gravity is the gravity of the total mass.
     
  5. Jan 26, 2012 #4

    Chronos

    User Avatar
    Science Advisor
    Gold Member

    There is no such thing as surface gravity. Gravity is always calculated based on the center of mass.
     
  6. Jan 26, 2012 #5

    mathman

    User Avatar
    Science Advisor
    Gold Member

    I suspect that the dependence on T is quite strong, weakening your original assertion.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook