Doppler effect of a satellite (extended)

[rohanlol7]

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 ?

 

[gneill]

question 3. Should i post a picture of it if its more appropriate ?

It's always better to have the question statement visible in the post. Some helpers are not keen to follow off-site links to see if the question is of interest, or to download potentially large PDF's wherein the question is buried.

Look up the Doppler effect. Your course text should show a derivation (if not there are plenty of online sources).

 

[rohanlol7]

It's always better to have the question statement visible in the post. Some helpers are not keen to follow off-site links to see if the question is of interest, or to download potentially large PDF's wherein the question is buried.

Look up the Doppler effect. Your course text should show a derivation (if not there are plenty of online sources).

Ok I'll try to make it visible.
When I couldn't do it, I went to try and derive the equation for frequency when theta is zero. My result was correct , I still can't get it

 

[gneill]

When I couldn't do it, I went to try and derive the equation for frequency when theta is zero. My result was correct , I still can't get it

Can you show your work? What distinguishes the zero-angle case form the general case?

 

[rohanlol7]

Can you show your work? What distinguishes the zero-angle case form the general case?

Okay so the zero angle case is just when the observer and source are on the same line.
suppose at a time t=0 a wavefront just hits the observer. at time t=T (period) a second wave is emitted and the observer has moved a distance vT and the distance from the observer to source is d-vT. this second wave takes (d-vT)/c to reach the observer and so at t=T+(d-vT)/c the second wave is recieved. the thrid wave is received at t=T+(d-2vT)/c etc. so the time between waves = T-vT/c and frequency is the reciprocal of that and this result is fine i think

 

[gneill]

Okay so the zero angle case is just when the observer and source are on the same line.
suppose at a time t=0 a wavefront just hits the observer. at time t=T (period) a second wave is emitted and the observer has moved a distance vT and the distance from the observer to source is d-vT. this second wave takes (d-vT)/c to reach the observer and so at t=T+(d-vT)/c the second wave is recieved. the thrid wave is received at t=T+(d-2vT)/c etc. so the time between waves = T-vT/c and frequency is the reciprocal of that and this result is fine i think

It's a little more complicated than that. Letting $T_o$ be the original period of the source and $T$ the received period, then you have found:

$T = T_o - \frac{v T_o}{c} = (1 - \frac{v}{c}) T_o = \left( \frac{c - v}{c} \right) T_o$

If you take the reciprocal to find the new frequency then you get:

$f = \left( \frac{c}{c - v} \right) f_o$

This doesn't look like your desired form of $f = f_o - f_o \frac{v}{c}$ . To get there you'll have to consider the given assumption: Assume v << c where c is the speed of light. Usually this sort of hint implies that some term or terms will become negligible and can be ignored, especially if some form of series approximation is involved.

 

[rohanlol7]

Can you show your work? What distinguishes the zero-angle case form the general case?

It's a little more complicated than that. Letting $T_o$ be the original period of the source and $T$ the received period, then you have found:

$T = T_o - \frac{v T_o}{c} = (1 - \frac{v}{c}) T_o = \left( \frac{c - v}{c} \right) T_o$

If you take the reciprocal to find the new frequency then you get:

$f = \left( \frac{c}{c - v} \right) f_o$

This doesn't look like your desired form of $f = f_o - f_o \frac{v}{c}$ . To get there you'll have to consider the given assumption: Assume v << c where c is the speed of light. Usually this sort of hint implies that some term or terms will become negligible and can be ignored, especially if some form of series approximation is involved.

ah binomial expansion I get it now thanks