More precisely, a black hole five times as massive as the Earth. And it fits on a page. I think I have to print that out and pin it to a wall somewhere...
edit: ok maybe we'd need it to be between us and the sun for that to work, the lensing would be that of other stars I guess. I suppose the important question is, since it's relatively close to us, would it deform space hard enough for the gravitational lensing to be noticeable at such close proximity.
edit2: more train of thought, just because it's fun, I suppose since it's so small and the gravity relatively weak (i.e. not as strong as some huge stars or galaxies) any halo or distortion would not extend more than a couple meter, which wouldn't be noticeable from our distance, right?
F = G * (m1 * m2) / r^2 . G = 6.674 / 10^11 (m^3 / (kg * s^2)) | m1 = 5 * 5.972 * 10^24 kg | m2 = 100 kg
Only r in the equation above is not a constant, so we can compute F(r) - F(r+1) to get a difference in force that being 1 meter closer to the black hole would make at a distance r. Basically, this is the difference between how hard your toes would be pulled down vs your heard if you curl into a ball.
Distance Force Force + 1 Diff in kg
1 1.99286E+17 4.98214E+16 1.49464E+16
1000 1.99286E+11 1.98888E+11 39797421.92
10000 1992856400 1992457888 39851.15023
20000 498214100 498164282.3 4981.767364
30000 221428488.9 221413727.7 1476.11612
40000 124553525 124547297.6 622.744272
50000 79714256 79711067.53 318.8474585
60000 55357122.22 55355277.03 184.5191277
70000 40670538.78 40669376.79 116.1990494This answer: https://physics.stackexchange.com/a/332510/236187 seems to suggest that it takes mere milliseconds once one is within three times the Schwarzschild radius of the black hole. And that dozens of radii outward, one already moves at a significant fraction of the speed of light.