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13 September, 18:26

A U. S. quarter (coin) is about 3cmindiameter. If you were to hold it upso that it was covering the full Moon, how far (in cm) must you hold it away from your face so that it would pretty much cover the Moon exactly? (That is, at what distance does a quarter have an angular size equal to the Moon's.) Is your arm long enough todo this? (If you happen to see a full Moon sometime and have a quarter in your hand, try this out.)

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  1. 13 September, 18:28
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    R₂ = 3.31 m

    Explanation:

    For this exercise let's use trigonometry. Let's see the angle of the full moon

    θ = x / R

    Where x is the diameter of the moon and R the distance from the Earth to the Moon, the angle is measured in radians

    x = 2 R

    x = 2 1.74 10⁶

    x = 3.48 10⁶ m

    The distance is

    R = 3.84 10⁸ m

    Let's look for the supported angle

    θ = 3.48 10⁶ / 3.84 10⁸

    θ = 9.06 10⁻³ rad

    For the coin to cover the moon it must have the same angle, let's look for the different

    θ = x₂ / R₂

    x₂ = 3 cm = 0.03 m

    R₂ = x2 / θ

    R₂ = 0.03 / 9.06 10⁻³

    R₂ = 3.31 m

    This distance in much greater than the arm length
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