<p>Gravity is an acceleration, and force = mass x acceleration.  Thus, if we want to detect the Moon's gravitational pull, we need either a giant mass or a really precise instrument.  Since the oceans are a giant mass, the force on the ocean is thus correspondingly large.  The fluidness of the ocean also allows it to move, even if the force per volume is small.</p>
<p>To expand on this, the tidal acceleration due to the Moon on the surface of the Earth is about 1 μm/s<span class="math-container">$^2$</span> (<a href="https://en.wikipedia.org/wiki/Tide" rel="nofollow noreferrer">1</a>), which means that the force due to the weight of a 1000 kg object will vary by about 0.001 N over the course of a day.  1000 kg of water is about 1 cubic meter, so every cubic meter of the ocean experiences that tiny variation in force as well.  Since the ocean has about 1.3 x 10<span class="math-container">$^{18}$</span> cubic meters (<a href="https://hypertextbook.com/facts/2001/SyedQadri.shtml" rel="nofollow noreferrer">2</a>) of water in it, you could say the tidal force is 1.4 x 10<span class="math-container">$^{15}$</span> N, which is an absolutely enormous force - translated into weight, it's more than the weight of all living things on the Earth (<a href="https://www.wolframalpha.com/input/?i=1.4e14%20kilograms" rel="nofollow noreferrer">3</a>).</p>
<p>There is also a corresponding acceleration from the Sun, about half as strong as the Moon, which also has some effect on the tides, resulting in a monthly very high and very low tide, known as the spring tide, and a somewhat averaged tide, known as the neap tide.  It might also help to realize that this 1/2 as strong force is what is also responsible for keeping the whole Earth in orbit around the Sun.</p>

CC BY-SA 4.0
2020-09-29T02:01:46.203Z
2020-09-29T02:22:10.803Z
2020-09-29T02:22:10.803Z
IronEagle
<p>While gravity itself is weak relative to the other three known fundamental forces, that should not be construed to mean that the moon's gravitational pull on earth is necessarily weak.  To say something is &quot;weak,&quot; you really need some sort of benchmark against which to make that determination (e.g., I'm weak because I can only benchpress 50 kg, compared with my friend who can benchpress 100 kg, but I'm <em>not</em> weak relative to my other friend who can only benchpress 20 kg).</p>
<p>That aside, it helps to realize what gives rise (ha!) to a tidal force: Tidal force is the <em>difference</em> in gravitational pull from one side of an object to another.  As opposed to gravity itself which is just the generic pull.</p>
<p>Think of it this way:  I'm traveling in a car, in the back seat, and I really don't know how fast I'm going.  I would need specialized equipment to do that, like a speedometer.  But, I can certainly tell you that I'm traveling faster than a car I pass, or that a car that passes me is traveling faster than I am.  So it's easier to measure that difference.  (This isn't a perfect analogy, but the point is to illustrate measuring something from a reference frame you are in versus the difference between reference frames.)</p>
<p>Taking this to tides between Earth and Moon, it might be difficult to get an absolute gravitational measurement from the Moon at any given laboratory location on Earth.  You're in the presence of Earth's much more dominating gravitational field.  But, very roughly speaking, the side of Earth closest to the moon is about 109 lunar diameters away.  The side of Earth farthest from the moon is about 113 lunar diameters away.</p>
<p>That difference is not trivial.  The part of Earth closest to the moon is pulled more towards it than the part of Earth farthest from it, and the center of Earth is pulled towards the moon a &quot;medium&quot; amount.  Since water is much more easily deformed than rock, we see this effect as tides in the oceans:  The water facing the moon is pulled towards it (high tide), Earth is pulled towards the moon a little less, and the water facing away, on the other side of the planet, is pulled even less and so, conceptually, it lags behind in space, creating another high tide.</p>
<p>To FAKE put in numbers, the water closest to the Moon would be pulled 100 m towards the moon, the rocky part of Earth 50 m towards the Moon, and the water farthest from the Moon only 10 m towards the Moon, meaning that for people on the far side of Earth, the water appears to rise away from them.  Again, those are not the real numbers, the point is to convey what's going on conceptually.</p>

CC BY-SA 4.0
2020-09-29T02:51:31.967Z
2020-09-29T02:51:31.967Z
Stuart Robbins
<p>That the moon’s gravitational pull is responsible for the tides is established. However, if we want to detect the moon’s gravitational pull on earth, this is very challenging. This seems to be a paradox considering the enormous force needed to drive the tide.</p>

CC BY-SA 4.0
2020-09-29T04:24:14.630Z
`+1` for a a great question! It's quite possible that someone will answer and provide a way to make such a laboratory measurement, though one has to account for the laboratory's rotation around the center of the Earth and about the Earth-Moon center of mass. If you think about it, the Moon's gravitational pull on Earth" is quite strong, it's just *the difference between* it's pull on the Earth and it's pull on an object in the laboratory is small and hard to measure.
uhoh
CC BY-SA 4.0
2020-10-01T12:24:58.623Z
@uhoh - It's the difference between the Moon's gravitational acceleration of the Earth as a whole and Moon's gravitational acceleration on a patch of water on the surface of the Earth that drives the oceanic tides. Even worse, it's just the horizontal component of this acceleration vector that drives the tides.
David Hammen
CC BY-SA 4.0
2020-10-01T15:18:28.983Z
It's also the difference between measuring a constant value and measuring a delta.  What you're really asking, I think, is how to measure the delta-g force in one location as the Earth rotates under the moon.
Carl Witthoft
<the-moon><newtonian-gravity><tidal-forces>
Considering that the moon’s gravitational pull on earth is so weak- we need special instruments to detect it- how does it cause the tides?
James
# Considering that the moon’s gravitational pull on earth is so weak- we need special instruments to detect it- how does it cause the tides?

(Asked by: username_0 on 2020-09-29)

That the moon’s gravitational pull is responsible for the tides is established. However, if we want to detect the moon’s gravitational pull on earth, this is very challenging. This seems to be a paradox considering the enormous force needed to drive the tide.

**Tags:** the-moon, newtonian-gravity, tidal-forces

> **Comments:**
> **username_1**: `+1` for a a great question! It's quite possible that someone will answer and provide a way to make such a laboratory measurement, though one has to account for the laboratory's rotation around the center of the Earth and about the Earth-Moon center of mass. If you think about it, the Moon's gravitational pull on Earth" is quite strong, it's just *the difference between* it's pull on the Earth and it's pull on an object in the laboratory is small and hard to measure.  
> **username_2**: @username_1 - It's the difference between the Moon's gravitational acceleration of the Earth as a whole and Moon's gravitational acceleration on a patch of water on the surface of the Earth that drives the oceanic tides. Even worse, it's just the horizontal component of this acceleration vector that drives the tides.  
> **username_3**: It's also the difference between measuring a constant value and measuring a delta.  What you're really asking, I think, is how to measure the delta-g force in one location as the Earth rotates under the moon.  

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## 2 answers

### Answer by username_4 on 2020-09-29. Score: 2

Gravity is an acceleration, and force = mass x acceleration. Thus, if we want to detect the Moon's gravitational pull, we need either a giant mass or a really precise instrument. Since the oceans are a giant mass, the force on the ocean is thus correspondingly large. The fluidness of the ocean also allows it to move, even if the force per volume is small.

To expand on this, the tidal acceleration due to the Moon on the surface of the Earth is about 1 μm/s$^2$ ([1](https://en.wikipedia.org/wiki/Tide)), which means that the force due to the weight of a 1000 kg object will vary by about 0.001 N over the course of a day. 1000 kg of water is about 1 cubic meter, so every cubic meter of the ocean experiences that tiny variation in force as well. Since the ocean has about 1.3 x 10$^{18}$ cubic meters ([2](https://hypertextbook.com/facts/2001/SyedQadri.shtml)) of water in it, you could say the tidal force is 1.4 x 10$^{15}$ N, which is an absolutely enormous force - translated into weight, it's more than the weight of all living things on the Earth ([3](https://www.wolframalpha.com/input/?i=1.4e14%20kilograms)).

There is also a corresponding acceleration from the Sun, about half as strong as the Moon, which also has some effect on the tides, resulting in a monthly very high and very low tide, known as the spring tide, and a somewhat averaged tide, known as the neap tide. It might also help to realize that this 1/2 as strong force is what is also responsible for keeping the whole Earth in orbit around the Sun.


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### Answer by username_5 on 2020-09-29. Score: 1

While gravity itself is weak relative to the other three known fundamental forces, that should not be construed to mean that the moon's gravitational pull on earth is necessarily weak. To say something is "weak," you really need some sort of benchmark against which to make that determination (e.g., I'm weak because I can only benchpress 50 kg, compared with my friend who can benchpress 100 kg, but I'm *not* weak relative to my other friend who can only benchpress 20 kg).

That aside, it helps to realize what gives rise (ha!) to a tidal force: Tidal force is the *difference* in gravitational pull from one side of an object to another. As opposed to gravity itself which is just the generic pull.

Think of it this way: I'm traveling in a car, in the back seat, and I really don't know how fast I'm going. I would need specialized equipment to do that, like a speedometer. But, I can certainly tell you that I'm traveling faster than a car I pass, or that a car that passes me is traveling faster than I am. So it's easier to measure that difference. (This isn't a perfect analogy, but the point is to illustrate measuring something from a reference frame you are in versus the difference between reference frames.)

Taking this to tides between Earth and Moon, it might be difficult to get an absolute gravitational measurement from the Moon at any given laboratory location on Earth. You're in the presence of Earth's much more dominating gravitational field. But, very roughly speaking, the side of Earth closest to the moon is about 109 lunar diameters away. The side of Earth farthest from the moon is about 113 lunar diameters away.

That difference is not trivial. The part of Earth closest to the moon is pulled more towards it than the part of Earth farthest from it, and the center of Earth is pulled towards the moon a "medium" amount. Since water is much more easily deformed than rock, we see this effect as tides in the oceans: The water facing the moon is pulled towards it (high tide), Earth is pulled towards the moon a little less, and the water facing away, on the other side of the planet, is pulled even less and so, conceptually, it lags behind in space, creating another high tide.

To FAKE put in numbers, the water closest to the Moon would be pulled 100 m towards the moon, the rocky part of Earth 50 m towards the Moon, and the water farthest from the Moon only 10 m towards the Moon, meaning that for people on the far side of Earth, the water appears to rise away from them. Again, those are not the real numbers, the point is to convey what's going on conceptually.


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