Not sure how that’s relevant to what I wrote above.
We’re talking about different objects with the same apparent size. So let’s say object A is 2x farther away than object B, yet have the same apparent size. This means the actual diameter of object A is 2x larger than that of object B. So it has a surface area 4x larger than B, and its volume is 8x larger than B. If they have the same density, that means object A is 8x more massive. Since tidal force is proportional to mass and inversely proportional to cube of distance, the tidal force from object A is 8x(2^-3)=1 times larger than B (i.e. the same).
The implication is interesting - it means if you pick a tiny pebble and hold it up in the sky so it appears to be the same size as the Moon, the pebble is exerting as much tidal force on your body as the Moon.
It is precisely relevant because the **apparent size ** of an object scales with the square of distance, not the cube. If object A is the same size as object B but twice as far away, object A will appear to be one-quarter the size as object B. An object with linear dimensions one-half as much as object A at the same distance as object B will appear to be the same size as object A, in the same way that the moon and the sun appear to be the same size to an observer on Earth.
We aren’t talking about the apparent mass or volume, we’re talking about the are of your field of view that the object takes up. The moon and the sun don’t appear to have the same volume from earth. They do appear to have the same area as seen from earth. And that area scales per square of distance, not cube.
The actual volume of an object of a fixed angular radius varies with the cube of the distance. Therefore, two objects of the same angular radius will have actual volumes that are proportional to the distance cubed. This will cancel out the inverse cube of distance term in the tidal equation, leaving only the density term.
Statement 1: “Objects with the same apparent size and same density have the same tidal force”
Statement 2: “The tidal force from a given object is proportional to its apparent area.”
The two statements are not equivalent. Statement 1 is correct, and that’s what Lemur866 said. What you’re arguing against is Statement 2, and I agree it is incorrect.
Or to put it another way - if you move 2x as far away from a given object, I agree that the apparent area (solid angle) is reduced to 1/4, and the tidal force is reduced to 1/8. But at that point, how much mass do you need to add to the object to restore its original tidal force? The answer is, you need to make it 8 times more massive. If the density is the same, then you need to make its volume 8 times larger. As a result, its diameter ends up being 2x larger. The apparent size goes back to what it was before.
OK. Lemur866 is correct about the difference between the tidal forces between the moon and sun being mostly due to density. Worked out the math on paper and it seems to hold.
From what I can see a pebble held far enough away to appear the same size as the moon won’t have the same tidal forces however. Tidal forces don’t quite vary with the cube of distance. The cube of distance ratio for tidal forces is an approximation that’s fairly close when the size of the object the forces are being applied to is very small compared to the distance between objects. For the moon and sun this is close enough. For the pebble held at arm’s length the math gets more complex.
It’s not the tidal forces on the pebble, it’s the tidal forces on your body. So you’re looking at about a foot or so from one side of your body to the other, versus maybe twice that to the pebble. The math gets more complex in that case.
That I agree with; the answer for the close field (where the size of the object feeling the tidal force is a significant percentage of the distance between the object feeling the tide and the object causing the tide) is going to be somewhat different than the answer you get with the approximation.
OK, for “tidal force on your body” you’re right. For “tidal force at your location” it’s a valid approximation, where “your location” means an small area near your head.
And you could also use, say, a decent-sized boulder instead of a pebble, which would put it a significant number of body lengths away, and put us back into the “far-field” approximation.
