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Estimate diameter of the moon

  1. Nov 6, 2012 #1
    Assume that the average distance of the sun from the earth is 400 times the average distance of the moon from the earth. Now consider a total eclipse of the sun and state conclusions that can be drawn about:
    A. The relation between the sun's diameter and the moon's diameter.
    B. The relative volumes of the sun and the moon.
    C. Find the angle intercepted at the eye by a dime that just eclipses the full moon and from this experimental result and the given distance between the moon and earth (=3.80×105 km) estimate the diameter of the moon.


    Here is my attempted answer to the first parts:

    Part A.
    Since the sun is 400 times a moon's distance from earth, that must mean it looks 400 times smaller than it would if it were a moon's distance away. If the moon eclipses the sun perfectly, then that means that the sun is 400 times larger than the moon. Therefore the diameter of the sun is also 400 times larger than the diameter of the moon.

    Part B.
    If the sun is 400 times bigger than the moon, then the radius of the sun is also 400 times bigger than the radius of the moon. The volume of a sphere is given by [itex]\frac{4}{3}\pi r^3[/itex]. If the moon's radius is r, then the sun's radius is 400r. The volume of the sun will then be 6.4×107 times larger than the volume of the moon.

    Part C.
    I am unsure how to proceed with this part. I'm not totally clear on what is meant by "the angle intercepted by the eye", for one. It says "experimental result"; does it want me to actually go out and measure this angle?
     
  2. jcsd
  3. Nov 6, 2012 #2
    Yes, you should measure it.
    You just put the dime between one eye (close the other one) and move it along the eye-moon axis until it covers the moon (barely). You have to measure the distance between the dime and the eye in this position.
    Then you can go inside and calculate the angular size of the dime at this distance. You will need to measure diameter of the dime too.

    Now, it's interesting to know the angular diameter of the moon (about half degree) but you don't need the actual value to calculate the diameter of the moon. You can use similar triangles, with the values that you measured and the given distance (earth-moon).
     
  4. Nov 7, 2012 #3
    Thank you! I guess I will have to wait for the full moon for this problem...or at least a non-cloudy night. I will run through the method anyway since I am unsure of my trigonometry...

    So say I measured the distance where the dime just eclipses the moon and I got 10 centimeters. A dime is about 1.8 centimeters in diameter. Would the angular size of the dime at 10 centimeters then be the tangent of [itex]\frac{1.8}{10}[/itex]?
     
  5. Nov 7, 2012 #4
    Approximately right, due to the small value of the angle.
    To get the actual formula, draw a triangle with base the diameter of the dime and the opposite point in your eye. This is an isosceles triangle. The height of this triangle is 10 cm. The base is 1.8 cm. The angular size is the angle between the sides going from you eye to the dime.
     
    Last edited: Nov 7, 2012
  6. Nov 7, 2012 #5
    Ah, okay. So if you divide the isosceles triangle into two right triangles with a base of 0.9 and a height of 10, then the angle α opposite the 0.9 side will be [itex]\arctan\frac{0.9}{10}[/itex]. The angular size will be twice this value. Is that right?
     
  7. Nov 7, 2012 #6
    Yes, right.

    The next full moon will be in about a week, I believe. But you can try now for a first attempt. The moon is now half full, isn't it? You can see its diameter.

    Edit.
    Sorry, the moon is at last quarter now (not first) so it will be 3 weeks until the next full moon.
    One more reason to try it now.
     
    Last edited: Nov 7, 2012
  8. Nov 7, 2012 #7
    Part C is asking for an angle. Do you know a trigonomic function that will give you an angle if you input the diameter and distance? With tangents you input the angle not the ratio of diameter to distance.
     
  9. Nov 7, 2012 #8
    Thank you both. I will try it tonight if there are no clouds!

    Continuing my practice example though, the angular size is [itex]2\arctan 0.09\approx10.286[/itex]. Using similar triangles, where the height of the new triangle is the given distance to the moon from Earth, then I would calculate the diameter of the moon as [itex]2\cdot(3.8\times10^{5})\tan5.14[/itex] kilometers [itex]\approx6.84\times10^4[/itex] kilometers. Right?
     
  10. Nov 7, 2012 #9

    PeterO

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    Homework Helper

    The moon is often visible during the daytime - you may be able to go outside now rather than wait until tonight.
     
  11. Nov 7, 2012 #10
    How do you know is still daytime where he lives? (the mention of a dime?).
    In this phase, the moon is not visible in the afternoon.
    It raises around midnight. You may be able to see it in the morning, until about noon.
    The exact times varies according to location. And will change every day. It will raise even later.


    @skeptic
    He said he is using the Arctan, a function which "provides" the angle.

    @Ragnarok7
    The sample values look OK. In actual practice you will need to hold it farther away to have an angular size comparable with that of the moon. If you are talking about US (or Canada) dime, at arm length it covers about 2 degrees, about the same as your thumb (maybe a little less). This is still too much. You will need some holder. Or some other, smaller object.
     
  12. Nov 7, 2012 #11
    I live in the US on the east coast and tonight the moon doesn't rise til after one in the morning, apparently. I'll see if it's visible tomorrow morning. Maybe I can get someone to help me, we'll see.
     
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