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Homework Help: Smallest angle of inclination between two stars for eclipse

  1. Feb 16, 2015 #1
    1. The problem statement, all variables and given/known data
    Assume two stars are in circular orbit around a mutual center of mass and are separated by d = a. Assume angle of inclination is i and stellar radii are r1 and r2. Find an expression for the smallest angle of inclination that will just barely produce an eclipse.

    2. Relevant equations
    None that I know of. Question asking for you to invent one.

    3. The attempt at a solution
    I drew a diagram and used trigonometry to find acos(i) = ???
    I can't find either of the legs of the triangle because those legs will change depending on the position of the star in back, or the size of the star in back (the star having its light blocked for the eclipse), but what variables could I set that equation equal to so that the size of the star in back is accounted for? Also, the size of the star in front matters too, so we must assume that the star in back is smaller or of equal size to the star in front, correct? Or can an eclipse occur with something bigger in back? The solution manual sets the equation equal to r1+r2 but I don't understand the reasoning behind that. That would imply the adjacent leg of the triangle is equal to r1+r2, which I don't understand at all.
  2. jcsd
  3. Feb 16, 2015 #2
    You ask a number of questions that makes me think you have missed something or I have missed something. I sketched it out and it seems pretty straight forward. Maybe you could provide your sketch for us?
  4. Feb 17, 2015 #3


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    Staff: Mentor

    The question is vague about what constitutes an eclipse. There are total and partial eclipses. Does a partial eclipse count? The problem also doesn't mention the relative masses of the stars, so how are we to locate the center of gravity? The axis of rotation of the plane of the orbits for the inclination must pass though the center of gravity.
  5. Feb 17, 2015 #4
    Couldn't you assume that the angle of inclination can be very closely approximated by taking the angle with the star being eclipsed?
  6. Feb 17, 2015 #5
    I uploaded a picture of the sketch. I have 'a' as the hypotenuse of the right triangle, but I don't know how to get the adjacent side of that triangle, which is necessary for me to have an expression.

    I assume it meant full eclipse where the entire star that's in back is obscured by the front star. Which begs the question, what happens when the small star is in front? Then it won't be an eclipse anymore. I guess at that point, all that matters is all of the smaller star is either in front or in back of the larger star.

    I don't think locating the center of gravity is relevant. We just know that they orbit a mutual center of gravity, and I assume that's enough to get the expression I'm looking for.
    What do you mean by taking the angle with the star being eclipsed?

    Thanks for the replies, guys.

    Attached Files:

  7. Feb 17, 2015 #6


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    Staff: Mentor

    Astronomers would take exception to this :smile: To them any overlap of two bodies along the line of sight counts as an eclipse of some form.

    Instead of trying to perform rotations of the plane of the orbit to try to line things up the way you want, try leaving the plane horizontal and moving the "viewer" into a position that lines up the edges you're interested in.
  8. Feb 17, 2015 #7
    I think you have to assume a partial eclipse counts as an eclipse too, as gneil suggests. In that case it doesn't matter which star is in front.

    I agree with you here.

    The vertical line you drew in your sketch represents the line of sight to the person observing the eclipse. You drew the line presumably through the center of mass. Would the angle "i" change significantly if instead you drew the vertical line passing through the center of one of the stars? Try re-drawing the sketch with this and show just the moment when the smallest partial eclipse occurs.
  9. Feb 18, 2015 #8
    I got it. Thanks for the responses guys. My problem was assuming that an eclipse occurs only when the back star is completely obscured from vision. When in fact, an eclipse begins to occur right when any part of the back star is obscured from vision. In which case, that begins to happen when the bottom edge of the back star meets the top edge of the front star, which makes my adjacent leg equal to the sum of both radii.
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