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On a planet with a geostationary moon, how can I determine the position in the sky of the moon depending on the position on the planet surface?

For example, Pluto's Charon. The moon wouldn't move across the sky standing still but if you were to move where you were on the planet, where would the moon appear to be? I imagine someone standing directly in line with it at the equator would see it somewhere else than a person standing in the southern hemisphere to the east. And as you drifted out of its view (east to west or north to south) would it slowly disappear beneath the horizon?

I'm sure this could be figured out studying one of our own satellites but I wasn't sure how to approach that.

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    $\begingroup$ Look up in the sky where the moon is now. Now imagine it was 10x closer and it just stayed there. Geostationary. $\endgroup$ – SurpriseDog Nov 5 '19 at 19:52
  • $\begingroup$ Welcome to the site. Please note that we strongly encourage users to wait at least 24 hours before accepting an answer. This site has users all over the world; accepting early may discourage other, better answers from appearing. $\endgroup$ – Frostfyre Nov 5 '19 at 21:47
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    $\begingroup$ The answer is in the question. It is a geostationary moon, it will always be in the same location above a point on the equator. If you stand under it, you will always be under it. If you move North or South, it will look like it is setting or rising, but staying there. If you move the the other side of the planet, you will never see it. $\endgroup$ – Vogon Poet Nov 5 '19 at 21:48
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    $\begingroup$ It would stay in same place. A related but not duplicate question that I asked awhile back about the changing nature of its appearance... how a moon in geostationary orbit waxes and wanes $\endgroup$ – EveryBitHelps Nov 5 '19 at 21:58
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Simple geometric considerations can help you here.

Imagine, for the sake of simplicity, that the moon is on the same plane of the equator and on the meridian where you take the latitude 0. If you measure your latitude starting from 0 at the equator and the observer is at a latitude $\theta$, the moon will appear on the sky at an angle $90 - \theta$

moon position in the sky

If you look at the picture above, an observer on the equator (plane a) will see the moon at 90 degrees in the sky, in other word at its zenith.

An observer closer to the pole (plane b) will see the moon much lower on the horizon, until at the pole the moon will be exactly on the horizon.

Same goes with the longitude.

Mind that when the angle exceeds the 90 degrees the moon won't be visible at all, because it will be hidden by the planet. This happens on the farthest side of our Moon, which never sees Earth.

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    $\begingroup$ technically, at the poles the moon will be below horizon. For our Earth, geostationary orbit radius is about 5 times the radius of the planet, so above picture has roughly the right scale. $\endgroup$ – Bald Bear Nov 5 '19 at 19:57
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A geostationary satellite orbits in the same plane as the equator with a period equal to the rotational period of the planet. In other words, from anywhere on earth, the satellite appears fixed in one position in the sky, never moving - it is always directly above the same point on the equator. If you happen to be standing on that point, the satellite will be directly overhead. If you move away from that point, the satellite will dip lower and lower, until the planet itself obscures your view of the satellite. If you're directly north of the geostationary point on the surface, the satellite will be in the sky to your south, and conversely, if you're toward the south, the satellite will appear in the north. If you go west, the satellite will set in the east, and if you go east, the satellite will set in the west.

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