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Transit depth and the size of a planet...

  1. Dec 8, 2015 #1
    1. The problem statement, all variables and given/known data

    Astronomers discover a new transiting planet named Jakku. The parent star, Abrams, is known to have a radius equal to ½ that of the Sun’s. a) The transit depth of Jakku is 757 parts per million (757 x 10-6). How large is the planet? Convert your answer into units of Earth radii. b) The blackbody temperature of Jakku is 290K. The scientists who discovered it make a press release claiming this planet is definitely habitable. What criticism might you have of this statement?



    2. Relevant equations

    transit depth = (Rp/R*)2sq

    3. The attempt at a solution

    since I already have the transit depth 757 x 10-6 =(Rp/216237)2sq
    I know I have to solve for Rp...but this is where I need help
     
  2. jcsd
  3. Dec 8, 2015 #2

    gneill

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    How did you determine the radius of the star? What value did you use for the radius of the Sun?

    After you've dealt with that it looks like it's just a matter of a bit of algebra to isolate Rp.
     
  4. Dec 8, 2015 #3
    i halved the radius of the sun and that gave me the radius of adam
     
  5. Dec 8, 2015 #4

    gneill

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    Yes, but what value did you use for the radius of the Sun?

    Edit: Okay, I think what you've done is used miles for the units of length. You should have stated what unit system you were using.
     
  6. Dec 8, 2015 #5
    not quite sure what you mean,, but thanks for your patience...would that be 1/2*?
     
  7. Dec 8, 2015 #6

    gneill

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    I did a bit of checking of your value for the star radius and determined that you used a Sun radius in units of miles. What was confusing me is that i didn't recognize your star radius value compared to what I calculated myself (I used kilometers for the units).

    So you're left with isolating Rp from your expression:

    $$TD= \left( \frac{R_p}{R_s} \right)^2$$

    correct?
     
  8. Dec 8, 2015 #7
    Hmmm but I thought we already have transit depth?
     
  9. Dec 8, 2015 #8

    gneill

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    Yes. You have both transit depth (TD) and the radius of the star (Rs).
     
  10. Dec 8, 2015 #9
    Oh ok. This is where I get lost. How to find Rp
     
  11. Dec 8, 2015 #10

    gneill

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    So take a stab at the algebra. What are your first thoughts? What is worrying you?
     
  12. Dec 8, 2015 #11
    What's worrying me is that I don't know algebra... I've hired a tutor but I'm trying to this without him.
    Thanks for your time and patience
     
  13. Dec 8, 2015 #12

    gneill

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    You haven't done any manipulation of mathematical expressions? If I may enquire, from what class or course does this question come?
     
  14. Dec 8, 2015 #13
    An introductory astronomy course which is more like a physics course
     
  15. Dec 8, 2015 #14

    gneill

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    Okay, well I think you'll want to find some time to brush up on your basic algebra. You said that you've hired a tutor, so no doubt he can help you there.

    In the meantime, for your problem at hand, what happens if you take the square root of both sized of the equation? How will it look then?
     
  16. Dec 8, 2015 #15
    I don't know what the square root of TD would look like


    Oh, would it just be what ever the square root of the transit depth is? 757x10^-6??!
     
  17. Dec 8, 2015 #16

    gneill

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    Yes. But you can leave it in symbolic form while you are carrying out the algebra. Thus: ##\sqrt{TD}##.

    When you start plugging in numbers early on it's a lot more work to carry them through the manipulations without introducing transcription errors along the way. And it's easier to "debug" symbols rather than digits, because unlike digits, symbols tell you what they're supposed to represent at any given time :smile:
     
  18. Dec 8, 2015 #17
    Why do we square the transit depth?
     
  19. Dec 8, 2015 #18

    gneill

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    Square root. You want to get rid of the square on the right hand side, so you take the square root of both sides. Whatever you do to one side of an equation you must do to the other. Otherwise it would no longer be an equation (implying equality).
     
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