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Precalculus Distance Sun-Earth-Moon Problem

  1. Apr 26, 2007 #1
    Hi,

    I have this homework problem and need help. if anyone can help me please send me a message or email me at gettrinatak007@yahoo.com

    This is the the Precalculus word problem!


    At a certain point in the Earth's orbit around the Sun, its distance from the sun is 92.69 x 10^6 miles. At this point the angle subtended by the sun when observed from the earth is [Angle = 32' (arc minutes)]



    a) Considering the earth's radius to be negligible and making any other reasonable simplifying assumptions, use the law of cosines to find the diameter of the sun giving your answer to the nearest 1000 miles.

    b) using any assumptions and reasonable method, find the distance to the moon when its subtended angle is 32' arc minutes (0.54 degrees) assuming its diameter is 2160miles


    Let me know if you can help me with this. Any help is appreciaited
     
    Last edited: Apr 26, 2007
  2. jcsd
  3. Apr 26, 2007 #2

    Integral

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    Sure we can help you.. BUT..

    You need to show that you have made some effort to complete the problem.

    Can you draw a picture, labing the given quanities, then apply trig to get a answer?
     
  4. Apr 26, 2007 #3

    Dick

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    If you've been assigned this problem, you must know a little trigonometry. If so, then draw a triangle and start filling in angles and distances and if you have a specific problem, someone can help you. If not, then you've been assigned the wrong problem.
     
  5. Apr 26, 2007 #4
    Hi thanks for your quick reply.

    This is due tomrrow and ive been trying so much for past 2-3hrs. i know its too much. and srry i dont have picture to show but anyways, i thought of using the law of cosine formula to solve part a.

    c2 = a2 + b2 - 2abCos Y

    The distance from the earth to sun is 92.69 x 10^6 and i input this into a2 and b2 as well as (2)(a)(b) and for the CosY the angle is 32' (arc minutues which is 0.54 degrees)

    Then I take the sqaure root and find C = 862794.8527

    Is this correct or am I doing something wrong?
    thanks for taking your time looking at this. Hope you can help me with this today. thanks again.
     
  6. Apr 26, 2007 #5

    Dick

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    You are, in fact, doing it correctly. I would put the distance to the sun at the perpendicular bisector of your triangle and solve it as a right triangle problem. But with such a small angle it wouldn't make much difference in the answer.
     
  7. Apr 26, 2007 #6
    I think the formula i used it correct but the answer I'm getting: I dont think it sounds reasonable and thats why I was confused and wanted help with it.

    Found this image on another site...This is same on my homework paper.
    [​IMG]

    How would I use what you are talking about? "The perpendicular bisector" ?
     
  8. Apr 26, 2007 #7

    Dick

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    I know you are right because I know the diameter of the sun is about 10^6 miles. So it's very reasonable. I would have drawn a vertical diameter through the sun and split your isoceles triangle into two right triangles. But now I realize that's an approximation too. Thanks for posting the picture. But I think either approach will give you an answer within 1000 miles. Either one could be considered a 'reasonable simplifying assumption'.
     
  9. Apr 26, 2007 #8
    I had doubts with the answer, but I guess since more people (you) agree with the answer and process of solution I think it is right. Thanks for your help for part (a)...

    Would I be solving the part (b) the same way as I did for part (a)?
    Am I finding the distance to the moon from the Sun or the Earth? Thats what is confusing me because it really doesn't specify. But assuming the solution to part (a) I think it is from the sun? Please let me know what you think and how I would go about solving it...

    Thanks again for your help.
     
  10. Apr 26, 2007 #9

    Dick

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    Yes, it's pretty much the same problem. And you are finding the distance from the moon to the earth. All that changes is that you used to know a and b and want to find c. Now you know c and want to find a and b (and a=b).
     
  11. Apr 26, 2007 #10
    but how would i find a and b when the only information i have available is that the diameter of the moon is 2160 miles and the angle is 32' arc minutes?

    a and b was 92.69 x 10^6 for the distance from earth to sun, but now i need to find distance from earth to moon? i dont know how to do this? what formula do i use?
     
  12. Apr 26, 2007 #11

    Dick

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    Now the diameter of the moon is c. Set b=a and solve for a. There is still only one unknown.
     
  13. Apr 26, 2007 #12
    Well there are two approaches, at least, and in passing must mention what a curious situation we enjoy currently when the moon and the sun subtend the same angular size in the sky. It allows for a just so often total eclipse where the earth falls into the shadow of the moon.

    So you could solve this by similar triangles, that is the ratio of the suns diameter to the distance is the same as the moon to the earth. Or just use

    tan(angle/2)=radius/distance.
     
  14. Apr 26, 2007 #13
    i guess im missing something here. I still dont understand...because
    in part (b) the diameter of the moon is given: 2160 miles. I can't use the diameter of the sun. (I dont know if you meant the same as me...) but anyways, still I dont have enough information to use any of the laws formulas to solve the problem.

    when you say set b=a, what are these values? I dont know them.

    I tried the tan(angle/2) formula but that only gives me a distance in decimal. is that possible?
     
  15. Apr 26, 2007 #14

    Dick

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    Hey doc, How are ya? BTW if you note the diagram that katrina007 thoughtfully provided you will see that tan(angle/2)=radius/distance is subtly wrong. If the angle subtended is very large, it's very wrong. Funny, I hadn't thought of that. BTW also what's with this curious coincidence of the moon's angular size and the sun's? Anthropic? Sorry to the OP for disgressing on the post.
     
  16. Apr 27, 2007 #15
    yea i thought so that the tangent formula isnt going to work...
    can you please explain to me how to work this problem using your way?
     
  17. Apr 27, 2007 #16

    Dick

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    Ooops. Thought you were gone. What do you mean, an distance in decimal? Sure you can use it. If you check it gives you almost the same result as the law of cosines approach.
     
  18. Apr 27, 2007 #17

    Dick

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    Both approaches work quite accurately. And neither approach works completely. The difference is the difference between drawing a vertical diameter through the sun and a vertical line between your two tangent points. The aren't quite the same, right? But for small angles the difference can be ignored.
     
  19. Apr 27, 2007 #18
    I was using what the Doc suggested, using the tan(angle/2) formula. I tried this and got a decimal value which I believe is not correct or maybe I've missed something... Idk.

    But you were talking about letting b=a and solving? The value for C (moon's diameter) is given and its 2160miles. And I know that the angle is 32' arc minutes (0.54 degrees). I'm trying to solve for the distance between earth and moon.... What formula do I use to solve this?

    I dont think I can use law of cosine or sine
     
  20. Apr 27, 2007 #19
    Well that mastermind Dick, has us us both in the dark. I can see where it fails to account for the distance to the center of the earth, but the problem tells us to ignore that.
     
  21. Apr 27, 2007 #20
    I have to go now. past midnight here. need my rest and need a big break from all this calculus stuff.

    Thanks to all that helped me with this hw request. I will check it again tomorrow morning for replies, other hints/tips... will appreciate it if someone can solve the 2nd part of the question.

    Thanks again. Goodnight (if it is night over there...) :D
    Katrina
     
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