# Is there an advantage to the equatorial region of the far side of the moon for a radio telescope or would any crater on the far side work?

If nearly any crater on the far side of the moon will work, and not just equatorial craters, then we could pick craters near the possible future Artemis base at the South pole which would allow easier access. My assumption is as long as the crater is outside the range of Earth and the satellites in orbit it would be essentially the same as an equatorial crater.

tl;dr: The less complicated you want to make your dish, the closer to the equator you want to put it.

Since you are asking about single craters I assume this question is about using a crater as a natural pre-form for a single, large dish antenna like Arecibo Observatory and FAST.

On the Moon with no atmosphere or ionosphere nearly the whole sky is available for good signal, but as you move away from the current zenith position (i.e. "straight up") there are several problems. One is that the effective area of the telescope may be reduced for geometrical reasons at steep angles, and the other is that the optics of getting a good focus is increase.

Below you can see that Arecibo can't look much beyond 18° away from the zenith, and while the much newer and more complex FAST dish with it's changing dish shape articulated by moving cables and actuators can go as far as 60° away, it loses useful collection area beyond 26° not even counting $$\cos(\theta)$$ losses.

You will also notice that each telescope's maximum zenith angle attempts to be larger than the geographical latitude of its location, so that it can see the ecliptic.

Moving your lunar dish near the poles means you will lose access to a lot of the sky and be restricted to a "cap" of declinations either near +90° or -90°.

So the less complicated you want to make your dish, the closer to the equator you want to put it.

## Arecibo

From Overview of the Arecibo Observatory; ALFALFA, Undergraduate Workshop , Gregory Hallenbeck January 14, 2013

Latitude: 18° 20′ 39″ N

Can move dome to zenith angle position of 19.7°, But only to ~18° with good performance

## FAST

From Wikipedia:

Latitude: 25° 39′ 11″ N

The maximum zenith angle is 60-degree when the effective illuminated aperture is reduced to 200 m, while it is 26.4-degree when the effective illuminated aperture is 300 m without loss.33,3

Although the reflector diameter is 500 metres (1,600 ft), only a circle of 300 m diameter is used (held in the correct parabolic shape and "illuminated" by the receiver) at any one time.22 Thus, the name is a misnomer: the aperture is not 500 m, nor is it spherical.

In my humble opinion it might be necessary to build a number of crater based radio telescopes on the far side of the Moon to have to have full coverage of the sky.

A half circle between the lunar north and south poles will cover 180 degrees.

So if a crater based Lunar radio telescope covers X degrees of sky the number of radio telescopes needed would be 180/X.

So if a radio telescope can cover 10 degrees of sky 18 will be needed,

if it can cover 18 degrees of sky 10 will be needed,

if it can cover 30 degrees of sky 6 will be needed, if it can cover 45 degrees of sky 4 will be needed,

if it can cover 60 degrees of sky 3 will be needed, and so on.

The orbital period of the Moon is 27.321661 Earth Days, or 655.719 Earth hours. Since the Moon's orbit is elliptical there is a slight variation in its orbital speed, but the average angular fraction of its orbit traversed by the Moon is 13.176262 degrees per day, or 0.549 degrees per hour, or 0.00915 degrees per minute, etc.

So if a lunar crater radio telescope covers 10 degrees, an object should be visible to it for 0.7589 days or 18.2 hours, or 0.02777 of a lunar orbit of 27.321661 Earth days.

If a lunar crater radio telescope covers 18 degrees, an object should be visible to it for 1.366 days or 32.78 hours, or 0.04999 of a lunar orbit of 27.321661 Earth days.

If a lunar crater radio telescope covers 30 degrees, an object should be visible to it for 2.2768 days or 54.643 hours, or 0.08333 of a lunar orbit of 27.321661 Earth days.

If a lunar crater radio telescope covers 45 degrees, an object should be visible to it for 3.415 days or 81.965 hours, or 0.12499 of a lunar orbit of 27.321661 Earth days.

If a lunar crater radio telescope covers 60 degrees, an object should be visible to it for 4.553 days or 109.28744 hours, or 0.16666 of a lunar orbit of 27.321661 Earth days.

So therefore I can guess that if a lunar crater radio telescope could cover 60 degrees of the sky, one should be built in the exact center of the far side of the Moon, and two other should be built on the lunar equator 60 degrees from it, and two others on the same meridian as the first one, 60 degrees north and south of it. So that would give coverage of 180 degrees east and west and 180 degrees north and south.

But lunar crater radio telescopes might not be able to cover as much as 60 degrees of the sky, and it might be desirable to have some overlap in the sky coverage of the radio telescopes, so it is possible that it might be necessary to build several times the above five lunar crater radio telescopes.