See the waxing Moon meet the outermost planet on 10 October

By Ade Ashford

The 12-day-old Moon lies in the same low-power binocular field as the outermost planet late into the evening of Thursday, 10 October 2019. At 11pm BST, observers in the UK can find the pair highest in the southern sky against the constellation of Aquarius. The glare of the gibbous lunar disc will present a challenge, but look one-third of the way from magnitude +4.2 star phi (φ) Aquarii to magnitude +5.4 83 Aquarii to find the magnitude +7.8 ‘star’ that is Neptune. This simulated view shows field stars as faint as the planet. AN graphic by Ade Ashford.
In an earlier observing story, I wrote about Neptune’s close encounter with a magnitude +4.2 naked-eye star called phi (φ) Aquarii in the first week of September. The outmost planet’s great distance – currently some 29 times farther away from the Sun than Earth – means that its retrograde (east-to-west) motion against the stars of Aquarius is slow. Hence even by the end of October, Neptune lies no more than 1.3° (or two-and-a-half lunar diameters) from φ Aquarii.

Using φ Aquarii as a stellar stepping stone to Neptune is not too challenging, but if you’re looking for a really convenient celestial marker to the outermost planet from the UK, look no further than the waxing gibbous Moon on the night of 10 October. At 11pm BST, the lunar orb lies just 4½ degrees below Neptune – slightly less than the field of view of a typical 10×50 binocular. However, to comfortably encompass the Moon and planet requires an 8× or preferably a 7× instrument.

Neptune lies a staggering 4,349 million kilometres from Earth on the night of 10 October – almost 10,800 times farther away than our Moon. Even at the speed of light, rays from the Sun reflected from the cloud tops of the outermost known planet take four hours to travel across the Solar System to our eyes. At such a vast distance, it’s perhaps not surprising that Neptune takes almost 165 years to orbit the Sun.