Question E3B08:
How does the maximum range of ground-wave propagation change when the signal frequency is increased?  

Choices:  
A: It stays the same  
B: It increases  
C: It decreases  
D: It peaks at roughly 8 MHz    

Correct Answer is  C: It decreases  

To understand ground-wave range, remember it is a surface wave that “hugs” the Earth and suffers loss from the ground’s imperfect conductivity.

Why C (It decreases) is correct:
- Ground losses increase with frequency because the electromagnetic field penetrates a thinner layer of ground at higher frequencies. The skin depth is
  $$\delta = \sqrt{\frac{2}{\omega \mu \sigma}} \propto \frac{1}{\sqrt{f}}$$
  Smaller $\delta$ at higher $f$ means higher effective loss, so the ground-wave attenuates faster and its usable range decreases.
- Diffraction around the Earth is less effective at higher frequencies, further reducing range.

Why the other choices are incorrect:
- A: It stays the same — Incorrect. Attenuation clearly increases with frequency due to reduced skin depth and poorer diffraction.
- B: It increases — Incorrect. The opposite occurs; ground-wave is best at LF/MF and gets worse as frequency rises into HF.
- D: It peaks at roughly 8 MHz — Incorrect. There is no “peak” frequency for ground-wave range; above a few MHz the ground-wave component rapidly becomes negligible except at short distances. The 8 MHz idea is not a ground-wave phenomenon.

Useful formulas and rules of thumb:
- Field strength with ground-wave roughly follows
  $$E(d) \approx \frac{E_0}{d}\, e^{-\alpha(f,\sigma,\epsilon_r)\, d}$$
  where $\alpha$ increases with $f$ (and depends on ground conductivity $\sigma$ and permittivity).
- Skin depth rule: $\delta \propto 1/\sqrt{f}$ → higher $f$ → smaller $\delta$ → higher ground loss → shorter ground-wave range.
- Practical rule: Ground-wave is most effective at LF/MF (e.g., 160/80 meters marginal; AM broadcast is good), and becomes very limited as you move up the HF bands.