On a serious note, how heavy does an object have to be for gravitational lensing or bending to be noticeable with a naked eye?
s/visible/noticeable
s/noticeable/noticeable with a naked eye
s/lensing/gravitational lensing
s/visible/noticeable
s/noticeable/noticeable with a naked eye
s/lensing/gravitational lensing
α = 4GM/((c^2)b), where b is the impact parameter[0].
Apparently human visual acuity is 0.3 milli-radians, so if b = 1 meter, that's approximately "the moon" (in at most a 1 meter radius volume)…
…assuming I didn't mix up my units in this formula I never used before, though it feels about right given the Schwarzschild radius of the Earth is ~ centimetres.
[0] never heard of this before just now; I think it's the shortest distance between the central point and the path the light would have taken if it hadn't been deflected?
I don't know how fast the radius of curvature drops off as a function of the Schwarzschild radius, but I'd imagine it's at least R^-1. So assuming a spherical cow^H^H^H oven with gravitational lensing dying out at ~100x the event horizon, we need a Schwarzschild radius the size of a tennis ball in order to still see the curvature a few metres out. The oven needs to weigh ~4x the Earth's mass for that.
https://dcmovies.fandom.com/wiki/Fortress_Key_(All-Star_Supe...
Last I tried to calculate (poorly), if you were to touch it. You liquify and be sucked into it just before contact.
Assuming "dwarf star material" means "neutronium" (which is unstable at any level less than a neutron star) and the "half million tons" mass quote for the key…
That's about the energy content of all the world's fossil fuel reserves being released in a 10 minute half-life.
22 grams. That is how heavy my glasses are.
>The angle of deflection (theta) is:
theta = (4GM)/(cr^2)
toward the mass M at a distance r from the affected radiation, where G is the universal constant of gravitation and c is the speed of light in vacuum.[0]
The best resolution our eyes can offer is about one arcminute (1/21600 of a turn). Depending on your distance from the object, just plug in some numbers.
Say at the earth-moon distance 384400 km the object must be about 24x the mass of the sun to bend the incoming light at one arcminute (~0,0002909rad).
The sun actually bends light at about 2 arcseconds as seen from Earth; the focal point would be about 542x the distance Sun-Earth. [1]
Alternatively the object of say 1m^3 volume at a distance of 10 meters will bend light by 1 arcminute if it weighs 3.27x10^16 kg, the density of about 1/10th of a neutron star.
To conclude: one will be instantly overwhelmed by the gravitational forces before being able to see an object bend light with one own eyes. That's why this kind of extreme bending/lending is reserved for galaxy clusters.
[0]https://en.wikipedia.org/wiki/Gravitational_lens#Explanation...
Now, figuring out this limit is left as an exercise for the reader as it's way beyond my abilities.