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Determining the Radius of the Moon (Ratio and shadow measurements)

  1. May 16, 2012 #1
    Hello.

    I have been going through some old OU books on the sciences and started working through an excercise when I came to a stumbling block. I have come here in the hope that someone may be able to shed some light on my problem. This is all for personal study as I am not enrolled on any courses in the subject and just wish to expand my mind.

    1. The problem statement, all variables and given/known data

    The exercise I am going through is part of an old OU science module that belonged to my mother. Unfortunantely she no longer has the videos which I need to explain to me what I need to do next to correct my results.

    Basically the exercise makes the assumption that the shadow cast by the earth on to the surface of the moon during a solar eclipse is the same size as the earth itself. Now whilst I have an idea why this is not the case the assumption stands through most of the exercise in order to show a method of calculation to the student.

    Through the use of the photos of the moon showing different phases of a partial eclipse I am asked to make measurements to estimate the radius of the earth in the photograph.

    Before I explain any more I would like to show everything I have calculated thus far in the exercise.

    (Exact correct figures may not be used during the exercise as the learning experience is more to do with the approach than the results)

    2. Relevant equations


    3. The attempt at a solution

    The radius of the Earth's shadow in the photograph is:

    less than 7.2 cm (upper limit)
    more than 6.4 cm (lower limit)

    Radius of Earth's shadow in the photograph (RE) = 6.80cm +/- 0.40cm

    Radius of the Moon in the photograph = (RM) 2.75cm +/- 0.1cm

    Therefore the Radius of the Earth is 2.47 times bigger than the radius of the moon.

    RE = 2.47 RM

    Limits of Uncertainty


    Max Ratio = 7.2/2.65 = 2.72
    Min Ratio = 6.4/2.85 = 2.25

    Therefore RE = (2.47 +/- 0.22) RM

    Now RE = 2.47 RM (Text book uses a figure of 6200km for earths radius)

    So 6200/2.47 = RM

    RM = 2510

    I then calculate the limits of uncertainty for distance using the earlier numbers and get:

    RM = 2510 +/- 231 km

    This is the extent of my working out

    Having gotten this far through the exercise I read the following:

    "One word of caution about this calculation of RM: throughout Section 3 [the above] you make the assumption that the shadow of the Earth at the Moon is the same size as the Earth itself. As you will see in the TV programme, this is an unjustified assumption that leads to a considerable error in the value obtained for RM. However, the programme also shows how to correct this error, so be prepared to adjust your value of RM after viewing."

    Unfortunately I don't have this TV program and have been unable to obtain a copy or find anything similar avaliable. So I have come here in the hope that one of you brilliant minds can assist me in completeing this exercise by explaining to me what I am missing and how to correct my inital results.
     
    Last edited: May 16, 2012
  2. jcsd
  3. May 16, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    It's possible that they are referring to the fact that the Sun is not at an infinite distance, so it's rays are not strictly parallel. During a lunar eclipse the Moon is an additional distance d ~ 384400 km from the Earth's distance to the Sun of about D ~ 1.496 x 1011 m. There will be a "projection effect". Draw a diagram and work out the magnification.
     
  4. May 17, 2012 #3

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The sun is not a "point source". We see only the full shadow (umbra) really dark on the Moon, which is smaller than the size of Earth.

    LEDiagram1c.JPG

    ehild
     
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