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Arcs and possible trig

  1. Mar 6, 2010 #1
    I have not done any maths for a very long time now. And have forgotten a lot of what I learned. Right now I am having to get back into it again. And cant get my head around this problem.

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

    Two people 1.8m tall walk directly away from each other until they can no longer see each other. [Due to the curvature of the earth radius of about 6378km]

    The part of this question I don't get which I am trying to find is how to get this part.

    Assuming nothing else blocks their view, how far do they have to walk to not see each other? [Hint look for the length of the arc.]

    2. Relevant equations

    Dont know how to show my working out. I am in the diagram stage. And trying to get a formula. Just cant remember fully what I am doing.

    3. The attempt at a solution

    For a attempt, well I get know were. And my txt books are no help. All talking about it in graph format.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Mar 6, 2010 #2


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    Homework Helper

    Ok, in your diagram you can simplify things by noticing the symmetry of the landscape between each person. You should draw up one man, the curvature of the Earth, and his line of view from the top of his head just brushing passed the Earth (just as he's about to lose sight of the other man).

    You will need to include the centre of the Earth and create a radius to the man and to the point where the line of sight touches the Earth. This line of sight will be tangent to the circle (Earth) and remember that tangents to circles are perpendicular to the radius. That is, you're now dealing with a right-triangle.
    Let [itex]\theta[/itex] be the angle subtended at the centre of the Earth, r (radius of Earth) be the adjacent side, and r+1.8 (since this is the height of the man and the radius of the Earth) be the hypotenuse. Now using trigonometry, you can find [itex]\theta[/itex] in terms of the radius and the height of the man.

    Lastly, there is a rule that says the length of an arc is [itex]r\theta[/itex]. But remember that the length of the arc in this case is half the length of the actual distance required for the men to walk away from each other (since we're dealing with only one symmetrical half of the problem).

    Can you take it from here?
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