So long as both stations have a line of sight path to the moon, they can, in principle, communicate. The Earth’s circumference is about 24,000 miles, so half of that (12,000 miles) would have line of sight to the moon. In practice, the enormous path losses mean that high ERP, high gain antennas, low noise receivers and narrow bandwidth signals are required.
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This is caused by interference between the multiple path lengths of a moon bounce signal. The path lengths are constantly changing because the moon is “librating”. Although the moon does appear to always present the same face to the earth there is a small apparent “wobble” due to the fact that its orbit is not exactly circular. This apparent movement is called libration.
Because the moon has a highly irregular surface this rhythmic wobble causes irregular RF reflection.
The correct answer is the only one with "fading" in it.
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EME means Earth-Moon-Earth, or in other words, bouncing radio waves off of the Moon. Perigee means the point in the Moon's orbit where it is closest to the Earth.
When radio waves leave the antenna, they spread out, so when they travel far and spread out a lot, few waves hit someone else's antenna. This is much the same as a light bulb: when you're close to it, it's bright, and when you're far away, it looks dim.
The Moon is quite far away, so radio waves will spread out a lot before reaching the moon. When the Moon is at its closest point to Earth, the waves don't spread out quite as much as when the Moon is farther away. The difference between the perigee and apogee (farthest point) is about 40000 km, so round-trip is 80000 km or about 50000 miles. That means the trip is 50000 miles shorter when attempting a Moon bounce when the Moon is at perigee compared to when the Moon is at apogee.
This isn't necessarily the greatest cause for path loss for EME, but it is a factor.
Hint: Remember that apogee is farthest away from the earth, so perigee (think p for personal which is close) must mean the closest to earth, and thus should have the least loss in Earth-Moon-Earth (EME) communications.
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In Earth-Moon-Earth communications you are dealing with very weak signals.
Exclude wide bandwidth: it is an irrelevant distractor.
Exclude high dynamic range: it is helpful in dealing with weak signals.
Exclude high gain: it is helpful in receiving weak signals.
low noise is the correct answer because a low noise figure is highly desirable in weak signal communications. A receiver's noise figure is the ratio of its noise to that of an ideal receiver.
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JT65 uses time synchronous transmissions alternating and is commonly used in EME communications.
(This method is also used in FT8, which has become very popular for regular non-EME use.)
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At the very bottom of the 2 meter band, 144.000 to 144.100 is the CW portion, which includes Earth-Moon-Earth (EME) operation. EME operators communicate by bouncing their signals off the moon. (from Bob Witte, KØNR)
Memory trick: (for this VHF question and similar UHF question): The EME segment is only 0.1 MHz, since EME is a small niche area of interest (No band plan segments are so tiny as 0.001 Mhz, so "A" is obviously wrong.) Of the two answers left with 0.1 Mhz widths, pick the LOWER answer for the VHF question, and the HIGHER answer for the UHF question, becaise VHF is lower than UHF. Easy!
Then go look it up in a band plan when you actually want to do moon bounce, since you'll have forgotten these freqs by then!
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CW is invariably used for E-M-E contacts and this is a weak signal CW segment of the 70 cm band. (Digital is also often used for EME.)
Memory trick: (for this VHF question and similar UHF question): the segment is only 0.1 MHz, since EME is a small niche area of interest. Of the two answers left with 0.1 Mhz widths, pick the lower answer for the VHF question, and the higher answer for the UHF question, becaise VHF is lower than UHF. Easy!
Then go look it up in a band plan when you actually want to do moon bounce, since you'll have forgotten these freqs by then!
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Meteor scatter propagation occurs via the E-Layer.
Briefly, the explanation of the signal - at least in the vicinity of 20 meters is forward scattering from ionization trails left behind by the myriads of tiny meteors which pepper the E region of the ionosphere at all times. Hence the maximum range for this form of transmission is essentially that for normal one-hop E-layer transmission, or 1500 miles.
Source: QST April 1953 (via NASA)
Memory tip: There are a lot of Es in "meteor" and "free electrons". Pick E-layer!
Another: Electrons go to the E-layer
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The best band for meteor scatter is the 50 MHz band, where contacts lasting for several seconds or even a minute or so can be made. At higher frequencies, the contacts will be of shorter duration.
There is only one range in the answer choices in which 50 MHz falls, and that is 28 MHz - 148 MHz.
Memory Trick: "Meteor" has six letters. The best band for meteor scatter is 6m. The only answer that covers the 6m band is 28-148MHz.
Additional Memory Trick: MSK144 is a meteor scatter mode. 144MHz is only in the correct answer.
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