Time Delay Between Skywave and Ground-Wave Arrivals: 04-BS-9, December 2015, Q8

Question 8: Time Delay Between Skywave and Ground-Wave Arrivals (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Horizontal separation $L$ (A to B)	$500$ km
Ionosphere height $h$	$100$ km
Skywave speed	$c=3\times10^{5}$ km/s
Ground-wave speed	$0.95\,c$

Find. The time interval between the ground-wave and skywave arrivals at B.

Ground wave travels straight A→B; skywave reflects off the flat ionosphere. By the image method, the reflected path length equals the straight-line distance from A's mirror image (height $2h$) to B.

Approach. Use the image-source method: the flat-ionosphere reflection (equal angles of incidence and reflection) makes the two-segment skywave path length equal to a single straight line from A's mirror image — reflected through the ionosphere, at height $2h$ — to B. The ground wave is simply the straight distance $L$.

Skywave path length via the image source. A's image sits at height $2h=200$ km directly above A; the straight line from the image to B has length: $$d_{\text{sky}}=\sqrt{L^2+(2h)^2}=\sqrt{500^2+200^2}$$ $$d_{\text{sky}}=\boxed{538.5\ \text{km}}$$
Skywave travel time. $$t_{\text{sky}}=\frac{d_{\text{sky}}}{c}=\frac{538.5}{3\times10^{5}}$$ $$t_{\text{sky}}=\boxed{1.7951\times10^{-3}\ \text{s}}$$
Ground-wave travel time. $$t_{\text{ground}}=\frac{L}{0.95c}=\frac{500}{(0.95)(3\times10^{5})}$$ $$t_{\text{ground}}=\boxed{1.7544\times10^{-3}\ \text{s}}$$
Time interval between arrivals. $$\Delta t=t_{\text{sky}}-t_{\text{ground}}=1.7951\times10^{-3}-1.7544\times10^{-3}$$ $$\Delta t=\boxed{40.7\ \mu\text{s, skywave arrives later}}$$

Quantity	Result
$d_{\text{sky}}$ (reflected path)	$538.5$ km
$t_{\text{sky}}$	$1.7951$ ms
$t_{\text{ground}}$	$1.7544$ ms
$\Delta t$	$40.7\ \mu$s (skywave later)

Concept & Theory

Topic. Two-path (ground-wave vs. sky-wave) radio propagation delay over a flat earth.

Key relations. Image-source reflected path length $d_{\text{sky}}=\sqrt{L^2+(2h)^2}$ for a flat reflector at height $h$ with both endpoints on the ground; $\Delta t=t_{\text{sky}}-t_{\text{ground}}$.

Why this works. The flat-earth, flat-ionosphere assumption reduces the "angle of incidence = angle of reflection" condition to the same image-mirroring trick used for light or sound off a plane mirror: reflecting the source through the mirror plane turns the two-segment reflected ray into an equivalent single straight line, whose length is unaffected by exactly where the reflection point falls.

Common pitfall. Using the ionosphere height $h$ directly instead of $2h$ in the image path length; or computing the ground-wave speed as $1.05c$ instead of $0.95c$ (5% slower, not faster).

Source. Sadiku, Elements of Electromagnetics, Ch. 10 (plane-wave propagation); Young & Freedman, University Physics with Modern Physics, Ch. 33 (reflection geometry, image method).

Question 8 of 8: Time Delay Between Skywave and Ground-Wave Arrivals

Question 8: Time Delay Between Skywave and Ground-Wave Arrivals (20 marks)