EME path loss

Last Updated on February 19, 2025 by John Berry

Achieving a useful EME path loss relies on the Moon being an effective reflector. The transmitted power incident on the Moon is initially captured and then re-radiated. It’s known as the Huygens-Fresnel principle dating from 1818. The Moon’s ability to reflect energy in this way is defined by its relative cross sectional area or echo area. In the radar world, this is termed the radar cross section.

Diagram of the Earth and Moon. The losses and gains of the EME path loss at 144MHz are shown.
The losses and gains of the EME path loss at 144MHz

The scenario is like that used in billboard reflectors or passive reflectors in microwave radio. Until recently, large reflectors were installed on hilltops to reflect the microwave signal down into the next valley. This avoided costly relay stations.  

In such a scenario, the path loss between transmitter and receiver comprises the loss in the first leg, the loss or gain of the reflection and the loss in the second or return leg.   I’ve described the scenario at the apogee in the figure below and described it further in the following text.

The Moon as a reflector

The Earth to Moon path is modelled using the Free Space Loss (FSL) equation to give a one-way loss to the Moon at 144MHz of 187.5dBi at the apogee. Since the distance is less at the perigee, the loss is a little lower, but there’s only 1.2dB in it so I’ll proceed with the worst case. The return path is likewise 187.5dBi. That gives a total of 375dBi for go and return at the apogee.  

In coming to that big number, I’ve assumed that there’s an isotropic radiator at the Moon end, doing the capture and re-radiation. But that’s overly pessimistic. In fact, because the Moon is huge, it is a better reflector that this. If it wasn’t, the huge path loss would render EME communications impossible.

Path loss model

The model below illustrates the gains and losses in simple spatial terms.

Path loss model. Figures in square brackets refer to parameters on the path budget page.
Path loss model. Figures in square brackets refer to parameters on the path budget page.

Working left to right, the TX power is increased by the TX antenna gain. It is then attenuated by the Free Space Loss to the Moon. As noted above, the Moon is a big reflector and increases the signal by the passive reflector gain (called target gain factor in the radar world). But the Moon is an inefficient reflector, and so the signal must be reduced by the Moon’s reflection coefficient. It’s then attenuated again by the FSL on the return leg. It’s finally increased by the RX antenna gain to give the RX input power.

As the budget suggests, if the RX input power is bigger than the receiver threshold (hence having a positive margin), communications will be possible.

Billboard calculation

There are two parts to this reflection. First, I’ve modelled the Moon as a giant billboard of 3,475km diameter. The ‘gain’ of this billboard is the ratio of the power density from the passive repeater, to the power density which would exist at the same point if the passive repeater was replaced by an isotropic antenna.    

It’s well established that the Moon reflects primarily from the area near the centre facing Earth. I can approximate this by assuming that the effective diameter is 0.7*3,475km, or 2,432km. There is much discussion in the ham literature about how much of the Moon is an effective reflector. The 0.7 figure above comes from a paper given at an EME conference back in 2010. This does seem generous and a lower value may more appropriate. 

I’ve calculated this gain using the effective area of the billboard in the equation Gain (dB) = 42.9 + 40log f + 20log Ae. This is taken from microwave engineering. Variable f is the frequency and Ae is the effective area of the Moon. This gives me an effective gain of 143dB.

Second, I need to consider that the Moon is not a perfect reflector, even in the area in which the bulk of the reflections occur. Past measurements have concluded a reflection coefficient of about 6.5%.  So, just 6.5% of the energy incident is reflected. There is some dispute in the literature suggesting a better refection coefficient (of up to 10%) at VHF frequencies, but a pessimistic 6.5% will be assumed here. This equates to a loss of about 12dB.

Round trip path loss

So, I’ve concluded that the total effective gain is therefore 143dB – 12dB or about 131dB.   The EME path loss considering only the Free Space Loss and the reflective capability of the Moon at 144MHz is therefore -375dBi + 131dBi or about 245dBi allowing for rounding. I’ve used this figure to calculate the path budget in an adjacent page. I’ve also calculated the same at 432MHz and 1296MHz.