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na3than t1_j01vdff wrote

I don't understand the question. The Earth's tides ARE affected by a mass at the distance of the Moon, namely, the Moon. If the Moon had more mass or less mass the tides on the Earth would be affected. A greater change in mass would mean a greater disruption, but ANY change in mass would affect the tides.

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davewh t1_j01xvsd wrote

I suspect the question is about how much mass is required to make a noticeable effect. I mean, technically, if the moon were no more than a baseball, it would still affect tides. We'd just never notice it.

As for "how it works" there are plenty of websites that'll explain it in varying levels of detail. The force of gravity decreases with the square of the distance so even a very large mass that's too far away will only have minimal effect. Such as our Sun. It too affects the tides, but very little since it's so far away.

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Reviewingremy t1_j0204kg wrote

Technically any size will effect the tides but I'm assuming you mean a noticeable effect.

It also depends on what you mean by effect the tides. For instance if an object 1/10 the mass of the moon were to suddenly appear it would be "in competition" with the moon as it were. So although it wouldn't have an effect on the tides in and of itself it would reduce the effect the moon had on the tides by 1/10.

If you wanted the tides to be caused by the new object than it would have to have a mass greater than the moon and significantly so for the effect to be seen since again the moon would still be effecting the tides.

I don't understand what you mean about gravity, do you mean pull the earth out of it's orbit of the sun, or would this hypothetical new object cause the earth to orbit it. Essentially Turing the earth into a moon of this new astronomical body?

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Scott_Abrams t1_j022y7l wrote

What constitutes a disruption? Does it have to exert a secondary tidal effect rivaling Luna, or can any alteration to the current tidal pattern, no matter how marginal, be considered a disruption? Also, how measurable does this effect have to be? Thirdly, does this object have to follow the same trajectory as Luna's natural orbit? Luna's orbit around the Earth is elliptical and not equidistant. At what distance exactly are you referring to?

I'm going to assume you're taking about Newtonian physics here and not general relativity. Taking the literal understanding of your question, any introduction of any kind of mass into Earth's system will disrupt the Earth's tides and also Luna's orbit. This effect could be minuscule to the point where it cannot be measured by our instruments but the introduction of something new in the system will affect the existing system. Earth and Luna's tidal forces are also not constant as celestial objects such as the Sun or other planets also exert forces on each other at non-constant rates. The orbital path of the object and its velocity are also factors of consideration but the exact disruption will have to take place at an exact time and place.

The force of attraction (gravity) is stronger the closer you are to an object so the closer this object is, the stronger it's effect (inverse square law). To calculate the exact disruption, you will need to input the mass and distance but any mass at any distance will alter the system, no matter how small (even if it's not measurable).

We can calculate the approximate disruption of an inputted mass at an inputted distance but to be honest, it's difficult to answer your question in a meaningful way. We can try to calculate the tidal force of an object (to simplify the calculation, we input a new mass, distance, and orbital path into the local Earth-Luna system) and run the calculation from there but basically, any change will affect the system and while we can calculate the approximate change to the system as a result, in the end, we need more specific conditions (ex. assuming the introduction of a new tidally locked mass opposite of Luna's normal trajectory, at a distance of 360,000 km, how much mass is required to alter the tidal force of Luna by +/-1%?). There are also other factors to consider here such as measurement errors or rounding errors so I can't tell you the exact minimum amount of mass needed for the change but them monkeys at NASA are getting REALLY good at running these kinds of calculations (orbital mechanics).

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Astromike23 t1_j02bzo5 wrote

> The force of attraction (gravity) is stronger the closer you are to an object so the closer this object is, the stronger it's effect (inverse square law).

In the case of the Tidal force, though, it actually scales as the inverse cube.

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dukesdj t1_j02o91t wrote

To be pedantic, it scales as the inverse cube to leading order as there are infinitely many higher order terms in the Taylor expansion.

(A more expanded explanation of this in words...) From potential theory the force that results from a potential is simply the gradient of the potential. We can Taylor expand the potential to make our lives easier. The resulting leading order force scales as the inverse square and this term just describes uniform acceleration that results in orbital motion. All higher order terms are the tidal force. The leading order term is usually the dominant and hence we approximate the tidal force as an inverse cube law.

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