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Geometry and Trig: finding an expression for an angle

  1. Sep 17, 2016 #1
    1. The problem statement, all variables and given/known data
    Given the diagram below, showing the path of a geocentric satellite S flying over a ground station G, find an expression for the geocentric semi-angle ##\phi## in terms of ##\epsilon##, the radius of the Earth ##R_E##, and the height of the orbit ##h##.

    2. Relevant equations
    This should just involve basic trigonometry.

    3. The attempt at a solution
    I've extended the line from the centre of the Earth, through G, and out through the orbit path of the satellite. Now, I've drawn a vertical down from the satellite S parallel with the existing straight line until it intersects that extended line. I've called ##r## the height above the Earth of this point where my additional lines intersect.
    $$\cos \phi = \frac{R_E + r}{R_E+h}$$
    $$\cos \phi = \frac{R_E}{R_E+h}+\frac{r}{R_E+h}$$
    satelliteorbit.JPG
    I'm not sure if I've made any progress here, as I need to bring the angle ##\epsilon## into it. I imagine I could use some trig identities as well. I actually know the answer, but can't quite get there. The answer is:
    $$\phi = -\epsilon + \cos^{-1} \left(\frac{R_E}{R_E+h}\cos \epsilon \right)$$
    Any help greatly appreciated.
     
  2. jcsd
  3. Sep 17, 2016 #2
    I have gone for a purely geometrical approach to this. Notice that a tangent gas been drawn at point G that is part of the angle ε. So you can find at least one angle of the larger triangle in terms of ε. From then on you can use trigonometric identities to derive an expression for angle ∅ in terms of the of the variables you mentiones.
     
  4. Sep 17, 2016 #3

    SammyS

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    I take it from the problem statement that h is the height of the orbit above Earth's surface.

    Express ∠EGS in terms of ε and ∠ESG in terms of ε and ϕ . Then use the Law of Sines. .
     
  5. Sep 18, 2016 #4
    The law of the sines:
    ##\frac{\sin \phi}{s}=\frac{\sin (\pi/2+\epsilon)}{R_E + h}=\frac{\sin EGS}{R_E}##

    We can use ##\sin(\pi/2 + \epsilon)=\cos \epsilon##

    So:

    ##\frac{R_E}{R_E+h}\cos \epsilon = \sin EGS##

    I think I can use:
    ##\sin (\pi/2 - \theta) = \cos \theta##
    To change the last expression into:
    ##\frac{R_E}{R_E+h}\cos \epsilon = \cos (\pi/2 - EGS)##
    Then:
    ##\cos^{-1} \left(\frac{R_E}{R_E+h}\cos \epsilon\right) = \pi/2 - EGS##
    Now I know from the geometry of the triangle that:
    ##\phi + \pi/2 + \epsilon + EGS = \pi##
    So: ##\phi + \epsilon = \pi/2 - EGS##
    Oh...I think I've just done it.
    Thanks for your help.
     
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