<p>First off, the geostationary Moon would have to be 115 times less massive as the actual one to match the gravitational pull.</p>
<p>With that, one effect would be that the <em>tidal</em> pull would be stronger, but also the tides would not move across the face of the Earth as we are used to seeing.</p>
<p>First, consider the strength of the tidal pull. The tidal pull across Earth imparted by any other massive body is proportional not to the overall gravitation acceleration imparted by the other body onto Earth, but to the <em>difference</em> in gravitational acceleration between one part of Earth and another. That difference is inversely proportional to the <em>cube</em> of the distance to the other body, not the square of the distance as the bulk gravitational acceleration would be.</p>
<p>So the geosynchronous Moon, scaled down in mass by 115 times to match the squared distance and overall gravitational acceleration, ends up generating a difference across the Earth more than ten times as large as the actual Moon because that difference contains an extra power of the distance. Hence a larger tidal height.</p>
<p>However, the &quot;tidal wave&quot; would not wash over the Earth as it does with the actual Moon. If the Moon were truly geosynchronous, the high and low tides -- enhanced by the closer distance even if we normalize the Moon's gravitational pull -- would also be locked in place on Earth's surface, meaning wherever there is a high tide there will always be a high tide and vice versa. In effect the whole Earth would be given a permanent tidal stretch. The resulting combination of enhancing the tidal wave combined with freezing it into place might therefore leave a specific portion of the ocean shore either permanently flooded or permanently &quot;high and dry.&quot;</p>
<hr />
<p>We don't really need to move the Moon to see the effect of relative distances on tidal strength. We can also see this effect with the actual Moon and Sun.</p>
<p>In terms of overall gravity exerted on the Earth, the Moon is puny having less than 1% of the gravitational pull exerted by the Sun. Yet our tides are caused mostly by the Moon because the Moon is 400 times closer than the Sun. The cubic rather than quadratic effect of tidal forces with distance enhances the Moon's contribution enough to beat the more distant Sun.</p>

CC BY-SA 4.0
2022-10-02T06:41:49.160Z
That's quite interesting about the square vs cube distinction for tides! So what I'm understanding is that there wouldn't be a variation of tides; the water level at any given point on the shore would be roughly constant. Is that right?
Sarah
CC BY-SA 4.0
2022-09-24T15:59:23.370Z
2022-09-25T09:51:11.920Z
2022-09-25T09:51:11.920Z
Oscar Lanzi
<p>What kinds of differences from our Earth would we see with a moon in a geostationary orbit over the equator? Assuming the gravitational pull is roughly the same as it is with ours (my back of the napkin math says this moon would have to be one order of magnitude smaller to provide the same pull at its new height), what effects would this have on this new earth?</p>

CC BY-SA 4.0
2022-09-22T06:14:47.847Z
moving the moon from the orbit it has now(earth,moon distance 384400 km now)down to geostationary orbit/geosynchronous equatorial orbit at 35,786 km means moving the moon 10X closer to earth.i guess anything earth science related will change a lot.
trond hansen
CC BY-SA 4.0
2022-09-22T20:59:51.460Z
Not everybody could see it, maybe?
Oscar Lanzi
<tides><moon><earth-moon>
What would be the effect of a geostationary moon?
Sarah
# What would be the effect of a geostationary moon?

(Asked by: username_0 on 2022-09-21)

What kinds of differences from our Earth would we see with a moon in a geostationary orbit over the equator? Assuming the gravitational pull is roughly the same as it is with ours (my back of the napkin math says this moon would have to be one order of magnitude smaller to provide the same pull at its new height), what effects would this have on this new earth?

**Tags:** tides, moon, earth-moon

> **Comments:**
> **username_1**: moving the moon from the orbit it has now(earth,moon distance 384400 km now)down to geostationary orbit/geosynchronous equatorial orbit at 35,786 km means moving the moon 10X closer to earth.i guess anything earth science related will change a lot.  
> **username_2**: Not everybody could see it, maybe?  

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## 1 answers

### Answer by username_2 on 2022-09-24. Score: 5 | Accepted answer

First off, the geostationary Moon would have to be 115 times less massive as the actual one to match the gravitational pull.

With that, one effect would be that the *tidal* pull would be stronger, but also the tides would not move across the face of the Earth as we are used to seeing.

First, consider the strength of the tidal pull. The tidal pull across Earth imparted by any other massive body is proportional not to the overall gravitation acceleration imparted by the other body onto Earth, but to the *difference* in gravitational acceleration between one part of Earth and another. That difference is inversely proportional to the *cube* of the distance to the other body, not the square of the distance as the bulk gravitational acceleration would be.

So the geosynchronous Moon, scaled down in mass by 115 times to match the squared distance and overall gravitational acceleration, ends up generating a difference across the Earth more than ten times as large as the actual Moon because that difference contains an extra power of the distance. Hence a larger tidal height.

However, the "tidal wave" would not wash over the Earth as it does with the actual Moon. If the Moon were truly geosynchronous, the high and low tides -- enhanced by the closer distance even if we normalize the Moon's gravitational pull -- would also be locked in place on Earth's surface, meaning wherever there is a high tide there will always be a high tide and vice versa. In effect the whole Earth would be given a permanent tidal stretch. The resulting combination of enhancing the tidal wave combined with freezing it into place might therefore leave a specific portion of the ocean shore either permanently flooded or permanently "high and dry."

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We don't really need to move the Moon to see the effect of relative distances on tidal strength. We can also see this effect with the actual Moon and Sun.

In terms of overall gravity exerted on the Earth, the Moon is puny having less than 1% of the gravitational pull exerted by the Sun. Yet our tides are caused mostly by the Moon because the Moon is 400 times closer than the Sun. The cubic rather than quadratic effect of tidal forces with distance enhances the Moon's contribution enough to beat the more distant Sun.

> **Comments:**
> **username_0**: That's quite interesting about the square vs cube distinction for tides! So what I'm understanding is that there wouldn't be a variation of tides; the water level at any given point on the shore would be roughly constant. Is that right?  

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