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Planet-Star Ratio Method

  1. Mar 9, 2013 #1
    I am currently studying the star Qatar-1 and its orbiting planet Qatar-1b on the Agent Exoplanet website: http://portal.lcogt.net/agentexoplanet/ [Broken] where they are studying exoplanets and their transit times across their stars.

    After going through a 130-image process of using calibrators to track the movement of the planet in its transit across its star, a "lightcurve" is produced (showing the dip in brightness when the transit occurred). From this, the website produces information from your lightcurve. My table of results can be viewed in the attachment.

    What I am trying to ultimately do is to find out how they figured out the planet to star radius ratio to be 0.15. In text on their website, they say that they gained this information from my "lightcurve".

    The lightcurve can be found on this webpage specifically: http://portal.lcogt.net/agentexoplanet/qatar1b/lightcurve/ [Broken] , where they also show their formula on how they figure out the transit time. This formula is as follows:

    [tex]t_t = \frac{P R*}{π a}[/tex]

    Where:
    [itex]t_t[/itex] is the transit time
    R* is the star's radius
    P is the orbital period
    a is the orbital radius

    I have tried emailing the website but am yet to receive a reply and am worried that I never will. Googling has given very few results and I my only theory is that it might have something to do with the drop in brightness during transit - but I am skeptical. If anybody has any idea how a star-planet ratio is determined, it would be most helpful. Thank you in advance!
     

    Attached Files:

    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Mar 9, 2013 #2

    Bandersnatch

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  4. Mar 9, 2013 #3

    Bandersnatch

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    Note that to get just the ratio of radii, you don't have to work out the stellar disc's area. Just assume it's got a radius equal to 1, so the equation I provided reduces to basically taking the square root of the percentage drop in brightness. Here, 0.15 ~= √0.0236.
    Simple as that.
     
  5. Mar 9, 2013 #4
    Thank you, Bandersnatch. That works perfectly. But do you mind me asking why we square root it?
     
  6. Mar 9, 2013 #5
    Bandersnatch's answer worked perfectly... So I don't think there's anything wrong there. I'm just confused as to why you square root it...?
     
  7. Mar 9, 2013 #6

    haruspex

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    Yes, I realised my misunderstanding and deleted the post - not quite quickly enough.
    The percentage drop in brightness represents the planet's area (seen as a disc) as a fraction of the star's area. Since area varies as square of radius, the ratio of the areas is the square of the ratio of the radii.
     
  8. Mar 9, 2013 #7
    Thank you. That makes sense! Is Bandersnatch's equation in his first reply relevant?
     
  9. Mar 10, 2013 #8

    Bandersnatch

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    You can get the "why the square root" from there.

    [tex]\frac{L_o}{L}=\frac{πR_s^2-πR_p^2}{πR_s^2}[/tex]

    So, since we are looking for the ratio of radii,

    [tex]α=\frac{R_p}{R_s}[/tex]

    we can substitute α*Rs for Rp and solve for α, which gives

    [tex]α=\sqrt{1-\frac{L_o}{L}}[/tex]
     
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