<p>To expand on Keith's comment, the equation for the <a href="https://en.wikipedia.org/wiki/Tidal_force#Formulation" rel="nofollow noreferrer">tidal force</a> is based upon the difference between the acceleration due to the orbiting object, and the gravity of the body itself, i.e. the difference between how hard the Moon pulls on you compared to how hard the Earth pulls.  This results in an overall equation of:</p>
<p><span class="math-container">$$|a_{tidal}| = 2\Delta r G \frac{M}{R^3} $$</span></p>
<p>where <span class="math-container">$\Delta r$</span> is the radius of the Earth (6371 km), G is the gravitational constant (6.674 x 10<span class="math-container">$^{-11} m^3kg^{-1}s^{-2}$</span>, M is the mass of the Moon (7.342 x 10<span class="math-container">$^{22}$</span>kg), and R is the distance between the two (here either 22,500km or 384,400km). To check our equation we can calculate the current tidal acceleration: using 384,400km, we arrive at a value of 1.099 μm/s<span class="math-container">$^2$</span>, which is close to the accepted value of 1.10 μm/s<span class="math-container">$^2$</span> (<a href="https://en.wikipedia.org/wiki/Tidal_force#Formulation" rel="nofollow noreferrer">2</a>).  Calculating the tidal force using the distance of 22,500km, we arrive at a value of 0.0055 m/s<span class="math-container">$^2$</span>, or about 5,000 times the current tidal force.  That could make for some massive tides, however, were there oceans?</p>
<p>Current research suggests that the Moon formed as the result of a major impact, around 4.5 billion years ago (<a href="https://en.wikipedia.org/wiki/History_of_Earth" rel="nofollow noreferrer">3</a>), while the first oceans formed around 4.4 billion years ago (<a href="http://www.geology.wisc.edu/zircon/Wilde_et_al.PDF" rel="nofollow noreferrer">4</a>).  This leaves a gap of around 100 million years between that 22,500km distance and the formation of the oceans, but it is close enough that the tidal force is likely not that different.  However, translating the height of the tides is difficult.  As the average tide is about 0.6m (<a href="https://en.wikipedia.org/wiki/Tidal_range" rel="nofollow noreferrer">5</a>), a tide 5,000 times higher would be almost 3km high, which seems impossible.  Additionally, the height of the tide appears to be more strongly affected by geography than by the tidal force, as the current range of tides is from 0-15m in height.  The earliest evidence we have of tides is about 2.5 billion years ago (<a href="https://agupubs.onlinelibrary.wiley.com/doi/pdf/10.1029/97GL00234" rel="nofollow noreferrer">6</a>), but unfortunately, the geologic record only records the pattern of the tides, not the height.  We can tell that the Earth was spinning faster then, with a day 4.5 billion years ago being about 4-5 hours long, but it may have been too cold to melt the oceans.</p>
<p>So, to sum up:  The tidal force with the Moon at that distance would be 5,000 times greater than it is today, but there may not have been oceans, and the tide height is affected more by the shape of the ocean than the tidal force itself.  However, something like the water planet in Interstellar, with massive, 3,000km-high tides sweeping over the Earth every couple of hours is possible, although a boring ice planet is possible too.</p>

CC BY-SA 4.0
2020-10-01T15:21:19.903Z
Good math -- I might suggest adding a *slope* estimate.  We tend to see dramatic tidal effects at landfall (or the Bay of Fundy), but if we assume a large expanse of ocean, even a 10-km difference between high and low tide, spread over 1/4 the Earth's circumference, is not a very steep slope.
Carl Witthoft
CC BY-SA 4.0
2020-09-29T17:02:47.827Z
2020-09-29T17:02:47.827Z
IronEagle
<p>IMO the difference in orbital distance was a minor factor.  With all other parameters held constant, tidal force, (as being a function of the derivative of gravitational force) is proportional to the relative rate of rotation of earth WRT the gravitational fields being exerted on it by the Sun and the Moon. Consider for a moon-less Earth, that if Earth was tidally locked with the Sun (1 year = 1 day), there would be zero tidal force on earth.</p>
<p>Earth's rotational energy is being dissipated as heat from tidal deformation and is the reason Earth's rotation has been slowing down since the time Earth formed. Earth was spinning faster one billion years ago, a day lasted ~19 hours. Per my assertion, the magnitude of ocean tidal extent (discounting differences in continental plate arrangement, etc) would have been much greater even if the Moon had never existed.</p>
<p>Using a simplified model with Earth's rotational axis parallel to its orbital axis and one that doesn't include the Moon, my assertion for what causes tidal force is that <strong>a particle on Earth's surface at the equator is moving slower than Earth's solar orbital speed at astronomical noon and faster than Earth's solar orbital speed at astronomical midnight.</strong> Therefore, solar gravitational attraction exerted on that particle is greater than orbital centripetal force at noon and less than orbital centripetal force at midnight.  This of course would deform Earth, causing it to bulge in two directions at once with the noon side toward the Sun and the midnight side away from the Sun.  Particles along the axis of Earth's rotation (eg. at the poles) and those laying along Earth's orbital path at any instant would match earth's orbital velocity and therefore not experience any tidal force.</p>
<p>I looked at the &quot;numbers&quot; a long time ago and found the force differences to be substantial, (far more than that from the difference in orbital distance via Newton's gravity equation).  I'll leave it to someone else to verify it.</p>

CC BY-SA 4.0
2024-03-29T13:47:09.287Z
Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers [in the help center](/help/how-to-answer).
Community
CC BY-SA 4.0
2024-09-05T08:57:50.987Z
While partially correct, you are confusing tidal force (that is indeed a function of the gravitational gradient) with tidal friction that acts to synchronise the earth's rotation with the orbit that is exerting the tidal force. Contrary to your assertion, orbital distance plays a large role in this, the largest role, in fact. The moon's tidal force on Earth is much larger than that of the Sun - although we do see an increase in the tide when Sun and Moon allign. This is called spring tide.
L. IJspeert
CC BY-SA 4.0
2024-03-28T15:36:11.600Z
2024-03-28T15:36:11.600Z
Mike Mitchell
<p>In this <a href="https://www.bbc.com/news/science-environment-12311119" rel="noreferrer">article</a> it is stated that the Moon was originally much closer to Earth at a distance of 22,500 km rather than the 384,400 km it is today. What was the height of the tides back then in meters? Surely they would've been orders of magnitude higher?</p>

CC BY-SA 4.0
2020-09-29T02:01:01.413Z
The [tidal force](https://en.wikipedia.org/wiki/Tidal_force#Formulation) and the [height](https://hepweb.ucsd.edu/ph110b/110b_notes/node13.html) of the tides are proportional to the inverse cube of the Earth-Moon distance.
Keith McClary
<the-moon><tidal-forces>
How high were the tides back when the Moon was much closer to Earth?
user177107
# How high were the tides back when the Moon was much closer to Earth?

(Asked by: username_0 on 2020-09-28)

In this [article](https://www.bbc.com/news/science-environment-12311119) it is stated that the Moon was originally much closer to Earth at a distance of 22,500 km rather than the 384,400 km it is today. What was the height of the tides back then in meters? Surely they would've been orders of magnitude higher?

**Tags:** the-moon, tidal-forces
> **Comments:**
> **username_1**: The [tidal force](https://en.wikipedia.org/wiki/Tidal_force#Formulation) and the [height](https://hepweb.ucsd.edu/ph110b/110b_notes/node13.html) of the tides are proportional to the inverse cube of the Earth-Moon distance.  

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## 2 answers

### Answer by username_2 on 2020-09-29. Score: 4

To expand on Keith's comment, the equation for the [tidal force](https://en.wikipedia.org/wiki/Tidal_force#Formulation) is based upon the difference between the acceleration due to the orbiting object, and the gravity of the body itself, i.e. the difference between how hard the Moon pulls on you compared to how hard the Earth pulls. This results in an overall equation of:

$$|a\_{tidal}| = 2\Delta r G \frac{M}{R^3} $$

where $\Delta r$ is the radius of the Earth (6371 km), G is the gravitational constant (6.674 x 10$^{-11} m^3kg^{-1}s^{-2}$, M is the mass of the Moon (7.342 x 10$^{22}$kg), and R is the distance between the two (here either 22,500km or 384,400km). To check our equation we can calculate the current tidal acceleration: using 384,400km, we arrive at a value of 1.099 μm/s$^2$, which is close to the accepted value of 1.10 μm/s$^2$ ([2](https://en.wikipedia.org/wiki/Tidal_force#Formulation)). Calculating the tidal force using the distance of 22,500km, we arrive at a value of 0.0055 m/s$^2$, or about 5,000 times the current tidal force. That could make for some massive tides, however, were there oceans?

Current research suggests that the Moon formed as the result of a major impact, around 4.5 billion years ago ([3](https://en.wikipedia.org/wiki/History_of_Earth)), while the first oceans formed around 4.4 billion years ago ([4](http://www.geology.wisc.edu/zircon/Wilde_et_al.PDF)). This leaves a gap of around 100 million years between that 22,500km distance and the formation of the oceans, but it is close enough that the tidal force is likely not that different. However, translating the height of the tides is difficult. As the average tide is about 0.6m ([5](https://en.wikipedia.org/wiki/Tidal_range)), a tide 5,000 times higher would be almost 3km high, which seems impossible. Additionally, the height of the tide appears to be more strongly affected by geography than by the tidal force, as the current range of tides is from 0-15m in height. The earliest evidence we have of tides is about 2.5 billion years ago ([6](https://agupubs.onlinelibrary.wiley.com/doi/pdf/10.1029/97GL00234)), but unfortunately, the geologic record only records the pattern of the tides, not the height. We can tell that the Earth was spinning faster then, with a day 4.5 billion years ago being about 4-5 hours long, but it may have been too cold to melt the oceans.

So, to sum up: The tidal force with the Moon at that distance would be 5,000 times greater than it is today, but there may not have been oceans, and the tide height is affected more by the shape of the ocean than the tidal force itself. However, something like the water planet in Interstellar, with massive, 3,000km-high tides sweeping over the Earth every couple of hours is possible, although a boring ice planet is possible too.

> **Comments:**
> **username_3**: Good math -- I might suggest adding a *slope* estimate.  We tend to see dramatic tidal effects at landfall (or the Bay of Fundy), but if we assume a large expanse of ocean, even a 10-km difference between high and low tide, spread over 1/4 the Earth's circumference, is not a very steep slope.  

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### Answer by username_4 on 2024-03-28. Score: -2

IMO the difference in orbital distance was a minor factor. With all other parameters held constant, tidal force, (as being a function of the derivative of gravitational force) is proportional to the relative rate of rotation of earth WRT the gravitational fields being exerted on it by the Sun and the Moon. Consider for a moon-less Earth, that if Earth was tidally locked with the Sun (1 year = 1 day), there would be zero tidal force on earth.

Earth's rotational energy is being dissipated as heat from tidal deformation and is the reason Earth's rotation has been slowing down since the time Earth formed. Earth was spinning faster one billion years ago, a day lasted ~19 hours. Per my assertion, the magnitude of ocean tidal extent (discounting differences in continental plate arrangement, etc) would have been much greater even if the Moon had never existed.

Using a simplified model with Earth's rotational axis parallel to its orbital axis and one that doesn't include the Moon, my assertion for what causes tidal force is that **a particle on Earth's surface at the equator is moving slower than Earth's solar orbital speed at astronomical noon and faster than Earth's solar orbital speed at astronomical midnight.** Therefore, solar gravitational attraction exerted on that particle is greater than orbital centripetal force at noon and less than orbital centripetal force at midnight. This of course would deform Earth, causing it to bulge in two directions at once with the noon side toward the Sun and the midnight side away from the Sun. Particles along the axis of Earth's rotation (eg. at the poles) and those laying along Earth's orbital path at any instant would match earth's orbital velocity and therefore not experience any tidal force.

I looked at the "numbers" a long time ago and found the force differences to be substantial, (far more than that from the difference in orbital distance via Newton's gravity equation). I'll leave it to someone else to verify it.

> **Comments:**
> **username_5**: Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers [in the help center](/help/how-to-answer).  
> **username_6**: While partially correct, you are confusing tidal force (that is indeed a function of the gravitational gradient) with tidal friction that acts to synchronise the earth's rotation with the orbit that is exerting the tidal force. Contrary to your assertion, orbital distance plays a large role in this, the largest role, in fact. The moon's tidal force on Earth is much larger than that of the Sun - although we do see an increase in the tide when Sun and Moon allign. This is called spring tide.  

---

