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Find gravitational acceleration through luminosity

  1. Feb 15, 2013 #1
    1. The problem statement, all variables and given/known data
    Use equation 1.3 to estimate the solar radius Ro from its luminosity and effective temperature. Show that the gravitational acceleration g at the surface is about 30 times larger than that on Earth.


    2. Relevant equations
    (1.3) L=4∏R2σSBT4


    3. The attempt at a solution
    Well we know the gravity of something can be determined through g=[itex]\frac{-GM}{r*r}[/itex]. Solving for r in the gravitational equation will allow us to substitute that into the luminosity equation. With all of those known, we can solve for g and see how it is 30 times larger. This however did not work. My answer was -27.79 m*s-2.

    Now since this number is pretty much off by a factor of ten, I feel that I'm on the right track. Maybe I messed something up in my calculation that you guys will see. My substitution will be below if it helps.

    1. L=4∏R2σSBT4

    2. g=[itex]\frac{-GM}{r*r}[/itex]

    3. r2=[itex]\frac{-GM}{g}[/itex]

    4. L=4∏[itex]\frac{-GM}{g}[/itex]σSBT4

    5. g=[itex]\frac{-GMT^{4}4∏σ}{L}[/itex]

    6. Then solve for g!
     
    Last edited: Feb 15, 2013
  2. jcsd
  3. Feb 15, 2013 #2

    Simon Bridge

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    I'm not clear on your reasoning here - doesn;t the question expect you to do this the exact opposite way around?
     
  4. Feb 15, 2013 #3
    Hey sorry about that, meant to delete that sentence about the effective temperature. It was a method I was using before. Does it make more sense now?
     
  5. Feb 15, 2013 #4

    Dick

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    It looks to me like you solved correctly for g in terms of L, T and M. Now they must have given you some numbers to plug in for those variables. What were they?
     
    Last edited: Feb 15, 2013
  6. Feb 15, 2013 #5

    Simon Bridge

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    It's the reasoning that puzzles me - wouldn't it be more direct to solve for R2 in eq. 1.3 and sub into the equation for g?

    $$g_{Sol} = \frac{M_{Sol}}{m_E} \frac{r^2_{E}}{R^2_{Sol}} g_{E}\; : \; R^2=\frac{L}{4\pi\sigma T^4}$$

    Anyway - you got the same equation so ... provided you have not made some arithmetic error, or used bad numbers, what does the mismatch tell you about the estimate? What assumptions are made to get equation 1.3?
     
  7. Feb 15, 2013 #6
    In terms of some numbers that were used I had:
    M = 2x1030kg
    T = 5780K
    σSB = 5.67x10-8Wm-2K-4
    L = 3.86x1026W
    G = 6.67x10-11m3kg-1s-2
     
  8. Feb 15, 2013 #7
    Simpler problem than I thought. Got the answer and it's right.

    I solved for R like the problem says and like Simon said. Then plugged that into the accel due to gravity equation. This gave me a number around 280 which is close to 30 times that of Earth's gravity.

    Thanks guys!
     
  9. Feb 16, 2013 #8

    tms

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    The primary assumption is that the sun is a black body.
     
  10. Feb 16, 2013 #9

    Simon Bridge

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    I had wondered if the teacher wanted to provide an example of a model that did not work well :) If the numbers were correct, and there was no finger-slippage on the calculator, then the result disagreeing with reality would mean that the model contained a flaw.

    Occasionally someone will do that to you just to see who's alert.
    You get marks by identifying the flaw - hence the suggestion to look at the assumptions.

    But that was not the case this time.
     
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