Theoretically, what is the minimum distance between the camera and an object, where the object begins to not be visible ? Suppose that a camera has f as focal distance and the sensor size 2/3" and the object is a square of 2m x 2m and we are filming in the daylight.
This is not a practical answer (it is not a practical question), but it is a precise answer.
Let's define "not visible". If in an image, I will offer a description of "not visible" that the object is not more than one pixel size in the image, which certainly will not be considered visible (probably 5 or 10 pixels works as well ...), but "it depends", on the overall image size (pixels) and the enlargement at which it will be viewed, and of course, the size of the object, and what we mean by "can be seen".
Then the calculator at http://www.scantips.com/lights/subjectdistance.html can be used to fill in the other details, such as sensor size and image size and focal length. All of the details there will affect it.
Then if you specify "size of object in image" to be 1 pixel (or 5 or 10 pixels), then it will compute the necessary distance to the object (in the same units as was specified for the "real size of the object, perhaps feet?) A one pixel dot cannot be recognizable as anything.
I will just give you a glimpse of what are you asking, so you can do your own math.
We need to take in account:
1) What is the color. Diferent colours have diferent wavelengths, so this affects on the sensor reception, difraction, atmospheric absortion, etc.
2) The contrast with the background. This is pretty obvious, a white board on a white surroundings is less visible than a bright softbox flash on a dark surrounding.
The atmospheric conditions
This is obvious too, haze day, desert, refraction of the hot-cold air, water droplets.
Sensor size, resolution, capacity, quantum characteristics...
Probably this has less effect on a daytime shoot than on astronomy, but this affects too.
Lens absorption, sharpness, aberration, difraction.
Are you using your Iphone or did you rented Hubble to take a shoot of a daytime of the moon?
The main factors, on ideal conditions are focal length versus sensor resolution.
Study this: https://en.wikipedia.org/wiki/Angular_resolution
After that we need to take into account the resolution of the sensor.
Under bright sunlight conditions, a young person with 20/20 vision can resolve an object that is approximately 3000 diameters distant. A 2 meter square object has a diagonal measure of 2.8 meters. This object if viewed from 2.8 X 3000 = 8,400 meters, will appear to be a point without discernable dimension.
The 3000 times its diameter rule of thumb is too stringent for photographic purposes. This is because all lenses have unresolved aberrations. Also, the contrast of the photographic system adds difficulty to resolving objects. The accepted limit for viewing a photographic image is: The image of an object will be seen as a point if it sustains 3.4 minutes of arc. This works out to an image viewed from 1000 diameters distant. See C.B. Neblette The Photographic Lens Library of Congress 64-20637.
Doing the math: Object diameter 2.8 meters Lens focal length 50mm Object distance 100 meters Image size 1.4mm Viewing distance 1.4 X 1000 = 1,400mm (seen as a point)
Object diameter 2.8 meters Lens focal length 50mm Object distance 280 meters Image size 0.5mm Viewing distance 0.5 X 1000 = 500mm (seen as a point standard reading distance).
Object diameter 2.8 meters Lens focal length 100mm Object distance 560 meters Image size 0.5mm Viewing distance 0.5 X 1000 = 500mm (seen as a point standard reading distance).
If both the subject and the camera are located on Earth (or any planet), the distance to the horizon is an upper bound. As explained in the link, with the camera at 1.7 meters above the ground and assuming a spherical planet, the curvature of the Earth limits sight to about 2.9 miles not counting any refraction caused by temperature changes in the air column.
There is no "one answer", it depends from too many variables. It depends on the size of the object, the focal lengh of the lens mounted on the camera, the resolving capability of the lens (its optical resolution, so to say), the resolution of the sensor...and the definition of "not visible" for who is watching the final image. Uh, and the type of light hitting your object.
Maybe someone could theoretically come up with some kind of formula to put all those things together, but I seriously doubt.