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Chapter 5: Problem 1

A satellite passes across the sky sending out radio signals at a constant but inaccurately known frequency \(\omega_{s}\). Assume that the satellite altitude is small compared to the earth's radius and hence that the trajectory can be considered to be a straight line at constant altitude above a flat earth. By beating the signals against a known standard frequency and measuring the difference frequency, an observer on the earth can accurately measure the received frequency, with its Doppler shift, as a function of time, with \(\cos \theta_{s}\) in \((5.8 .3)\) [or \((5.8 .5)\), with \(\gamma=1\), since \(v \ll c]\) ranging from 1 to \(-1\). Show how the observations can be made to yield a value of \(v\), the velocity of the satellite, and a value of \(D\), the closest distance of approach of the satellite. \(\dagger\)

Short Answer

Expert verified
Question: Using the Doppler shift of the received frequency, determine expressions for the velocity of a satellite and its closest distance of approach to an observer on Earth. Answer: The expressions for the velocity of the satellite (v) and the closest distance of approach (D) are given by: $$v = c \left( \frac{\omega_r - \omega_s}{\omega_{s}} \right) \cdot \frac{1}{\cos \theta_{s}} $$ $$D = c \, \textrm{ln} \left(\frac{2\omega_s}{\omega_s - \omega_r}\right)$$

Step by step solution

01

Writing down the Doppler shift formula

The classical Doppler formula (with \(\gamma = 1\) as \(v \ll c\)) for the change in frequency (\(\Delta \omega_{s}\)) is given by: $$\frac{\Delta \omega_{s}}{\omega_{s}} = \frac{v}{c} \cos \theta_{s}$$
02

Rewrite the equation to find the velocity of the satellite

We know from the exercise that the received frequency \(\omega_{r}\) is equal to \(\omega_{s} + \Delta \omega_{s}\). To find the velocity of the satellite, we first have to get the expression for \(v\). We can do this by isolating the term \(v\) from the Doppler shift equation. By substituting \(\Delta \omega_{s} = \omega_r - \omega_s\), we get: $$\frac{\omega_r - \omega_s}{\omega_{s}} = \frac{v}{c} \cos \theta_{s}$$ From this equation, we can isolate \(v\): $$v = c \left( \frac{\omega_r - \omega_s}{\omega_{s}} \right) \cdot \frac{1}{\cos \theta_{s}} $$
03

Find expression for closest distance of approach

The closest distance of approach, \(D\), is related to \(v\) by: $$D = \frac{v}{2} * \textrm{time period of received frequency change}_{\textrm{from} \, 1 \, \textrm{to}\, -1}$$ In the given case, as \(\cos \theta_{s}\) changes continuously from 1 to \(-1\), we can use the derivative of \(\cos \theta_{s}\) with respect to time (\(t\)) and obtain the time period: $$\frac{d(\cos \theta_{s})}{dt} = -\frac{v}{c}\cdot \frac{(\omega_r - \omega_s)}{\omega_{s}}$$
04

Calculate time period from \(\frac{d(\cos \theta_{s})}{dt}\)

Integrate both sides from \(\cos \theta_{s} = 1\) to \(\cos \theta_{s} = -1\) to find the time period \(T\): $$\int_{1}^{-1} \frac{d(\cos \theta_{s})}{\cos \theta_{s}\,(\omega_r - \omega_s)} = -\int_{0}^{T} \frac{v}{c} dt$$ Solve the integral, and we get the time period \(T\): $$T = \frac{2c}{v} \textrm{ln} \left(\frac{2\omega_s}{\omega_s - \omega_r}\right)$$
05

Compute the expression for \(D\)

Now, plug the expression for \(T\) back into the equation for \(D\): $$D = \frac{v}{2} * \frac{2c}{v} \textrm{ln} \left(\frac{2\omega_s}{\omega_s - \omega_r}\right)$$ Simplifying, we get: $$D = c \, \textrm{ln} \left(\frac{2\omega_s}{\omega_s - \omega_r}\right)$$
06

Conclusion

By analyzing the Doppler shift of the received frequency as a function of time and applying the given conditions in the exercise, we have found expressions for the velocity of the satellite (\(v\)) and the closest distance of approach (\(D\)), given by: $$v = c \left( \frac{\omega_r - \omega_s}{\omega_{s}} \right) \cdot \frac{1}{\cos \theta_{s}} $$ $$D = c \, \textrm{ln} \left(\frac{2\omega_s}{\omega_s - \omega_r}\right)$$

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