Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Problem 6.1.5: The Doppler sweep of a satellite pass

Problem 6.1.5

The Doppler sweep of a satellite pass

A small satellite in a low earth orbit moves with $v = 7.5 km/s$ and transmits a beacon at $f = 437 MHz$ . During a pass the angle $θ$ between its velocity and the line from the satellite to a ground station changes from nearly $0^{\circ}$ (rising, approaching) to nearly $18 0^{\circ}$ (setting, receding).

Find the received frequency shift $f^{'} - f$ for $θ = 0^{\circ}$ , $6 0^{\circ}$ , $9 0^{\circ}$ , $12 0^{\circ}$ and $18 0^{\circ}$ .
Over what frequency range must the receiver be tuned during a pass?
How large is the error of the first-order formula compared with $f^{'} = f / (1 - \frac{v}{c} cos θ)$ ?

Try it yourself first.

Strategy

The satellite is a moving source: $f^{'} \approx f (1 + \frac{v}{c} cos θ)$ by Eq. (6.1.69), so the shift is the fixed amount $f v / c$ times $cos θ$ . The exact classical formula $f / (1 - x)$ , $x = \frac{v}{c} cos θ$ , differs by the series terms $x^{2}, x^{3}, \dots$

The scale of the shift is

f \frac{v}{c} = 437 \times 1 0^{6} \times \frac{7.5 \times 1 0 ^{3}}{3 \times 1 0 ^{8}} = 10.925 kHz,

and $f^{'} - f = 10.925 cos θ kHz$ :

$θ$	$f^{'} - f$ (kHz)
$0^{\circ}$	$+ 10.93$
$6 0^{\circ}$	$+ 5.46$
$9 0^{\circ}$	$0$
$12 0^{\circ}$	$- 5.46$
$18 0^{\circ}$	$- 10.93$

From $f - 10.93 kHz$ to $f + 10.93 kHz$ : a range of about $21.9 kHz$ , wider than the channel of a narrow-band beacon (a few kHz). The station has to track the frequency during the pass.

$\frac{1}{1 - x} = 1 + x + x^{2} + \dots$ , so the first-order formula misses at most

f (\frac{v}{c})^{2} = 437 \times 1 0^{6} \times (2.5 \times 1 0^{- 5})^{2} = 0.27 Hz .

Exactly: $+ 10 925.27 Hz$ at $0^{\circ}$ and $- 10 924.73 Hz$ at $18 0^{\circ}$ .

Answer

f^{'} - f = 10.93 cos θ kHz, tuning range \approx 21.9 kHz, second-order error \leq 0.27 Hz

Check

The shift is positive while the satellite approaches ( $θ < 9 0^{\circ}$ ) and changes sign at closest approach, as it must. A car at $30 m/s$ heard on $437 MHz$ would shift by only $\pm 44 Hz$ , a GPS satellite, whose radial speed reaches about $0.9 km/s$ , by $\pm 4.7 kHz$ at $1575.42 MHz$ . At $v / c = 2.5 \times 1 0^{- 5}$ the relativistic corrections are of the same negligible size as $x^{2}$ .

The sweep in time

Seen in time, a pass gives an S-shaped curve: flat near the horizons and steepest at closest approach (Figure 6.1.28). For an overhead pass at the height $h$ , with the satellite moving on a straight line, the distance is $r = h^{2} + (v t)^{2}$ with $t$ counted from closest approach, and $cos θ = - v t / r$ , so near $t = 0$

f^{'} - f \frac{d f ^{'}}{d t}_{t = 0} \approx - \frac{f v}{c} \frac{v t}{h}, = \frac{f v ^{2}}{c h} = \frac{437 \times 1 0 ^{6} \times ( 7.5 \times 1 0 ^{3} ) ^{2}}{3 \times 1 0 ^{8} \times 5 \times 1 0 ^{5}} Hz/s = 164 Hz/s

for $h = 500 km$ . A real orbit curves round the Earth: the satellite moves on a circle of radius $R + h$ about the Earth's centre ( $R = 6371 km$ ), which reduces the rate at closest approach by the factor $R / (R + h) = 0.927$ , to $152 Hz/s$ , and limits the shift at the horizon to $10.13 kHz$ (the Earth's rotation is neglected here). Either way the beacon slides through $f$ at about $150$ to $160 Hz$ per second. Measuring $f^{'} (t)$ during one pass gives the time of closest approach and the distance at that instant: this is how the Transit system, the first satellite navigation, located ships.

**Figure 6.1.28.** Problem 6.1.5. Left: the first-order shift $f^{'} - f = 10.93 cos θ kHz$ with the five values of part (a). Right: the shift against time during an overhead pass at $h = 500 km$ , from horizon to horizon. The straight-track model gives the slope $164 Hz/s$ at closest approach, the circular orbit $152 Hz/s$ and a smaller shift at the horizon, because the line of sight there is never along the velocity.