- #1
rohanlol7
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Homework Statement
http://www.physics.ox.ac.uk/olympiad/Downloads/PastPapers/BPhO_Paper3_2004_QP.pdf
question 3. Should i post a picture of it if its more appropriate ?
Note: I'm expected to do this in about an hour
[Mentor note: Image of the relevant portion of Q3 inserted]
Homework Equations
T=1/f
tan=cos/sin
The Attempt at a Solution
the first two parts are fine. Part (c) starts to screw with me
assume that at t=0 a wave has just hit the observer.
So at t=T a wave is emitted and a second wave hits the observer
at t=L/c*cos(theta) +T, a wave hits the observer again ( L is the horizontal distance from the observer to the satellite and T is the period and c is the speed of light )
at t=(L+/-vT)/(c*cos(theta+d(theta)) +2T another wave hit the observer.
So the period is T+/- vT/c*cos(theta)
this unfortunately does not lead to the required expression
I tried considering the problem as emission of photons but i just get the same result.
My guess is that relativity is what's missing in my approach. If that's the case what exactly should i know before i am able to do this problem? ( i have not studied the mathematical treatment of relativity and this i for preparation for a contest )
For the next part i have 2 approaches, first i consider the diagram given and derive and equation for cos(theta) in terms of t. this gives me cos(theta)= d-vt+h^2/(d-vt)
d is the distance at the moment where the satellite is in range.
This equation however only seems reliable when the satellite is passing over the head of the observer as this is the only point where the curvature of the Earth and path of the satellite can be ignored.
So I went on to derive a general equation for cos(theta), but this time is used the geometry of the circular path and the curvature of the Earth etc. this gave me a weird long equation of the form kcos^2(theta)= (a-(vt)^2+bvt)/(a+bvt) . Now this is a weird one. I'm not sure if I'm expected to really sketch this or is there another simpler approach that they are looking for here ?
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