3
$\begingroup$

Space Navies A and B are on sufficiently distant orbits around a black hole. They are on similar orbits to each other, and neither side is risking going around the gravity well (for reasons). Lt. Mezpec of Navy A has an idea: use gravitational lensing for the sufficiently powerful lasers to arc around the well.

Assume the laser is effective at necesary range.

How far away from the gravity well or each other do the navies have to be to have this work? (As in, how far is sufficiently distant)

Edit: The black hole is 16km in diameter and 400 billion (4E+12)kg.

$\endgroup$
4
  • 6
    $\begingroup$ That mass would result in a tiny, tiny black hole - 19 orders of magnitude less massive than our sun, 13 orders of magnitude less than Earth itself. It will not be 16 km in diameter and its ability to bend light passing more than a few metres from it will be negligible. $\endgroup$ Mar 23, 2023 at 21:37
  • 3
    $\begingroup$ Research topic: "laser spread over distance". I don't think very long range lasers are as practical as sci-fi shows have lead us to believe. $\endgroup$
    – JamieB
    Mar 24, 2023 at 13:29
  • 1
    $\begingroup$ omnicalculator.com/physics/schwarzschild-radius Black hole radius calculator. Mass of our Sun just under 2E30 kg gives a 3 km radius. Radius is proportional to mass. $\endgroup$
    – Boba Fit
    Mar 24, 2023 at 16:21
  • 1
    $\begingroup$ For cripes sake, just make it 400 billion kg. $\endgroup$
    – Daron
    Mar 24, 2023 at 17:09

4 Answers 4

15
$\begingroup$

I'm maybe not properly up on gravitaional lensing or laser weapons.

I don't understand why lensing the laser to hit would particularly change anything. The black hole is already bending all the light, a laser would surely be bent just as much as all the other light? If you can see their ship then your laser can hit it, if you can't then your laser can't hit it. Like shooting a laser at someones reflection in a mirror - they may be hiding out of direct line, but the light is bent to you by the mirror. Shoot down that path of light and your laser will find them?

$\endgroup$
1
  • 4
    $\begingroup$ This is the correct way to think about things. In GR light does not bend to gravity. Spacetime bends to gravity, and light moves in a straight line on a curved surface. So if you can see them, they can see you, etc (ignoring speed of light). $\endgroup$
    – Aron
    Mar 24, 2023 at 3:07
3
$\begingroup$

After some consideration, I think there are a couple of fundamental flaws in your concept, that will make the laser ineffective or at least greatly reduce its effectiveness.

First flaw: laser are cool weapons because they give out coherent photons, all having the same phase, momentum and frequency, which greatly amplify the effect of their interaction with matter.

However, once you send a beam around a gravity well, the resulting gravitation lensing means that every photon will have followed slightly different path and will focus on the target with a different phases: you have turned a laser into a very bright flashlight!

Second flaw: having an optic to modify your beam is handy because you can move the optic to focus it on your target. I hope you agree that moving a black hole around is not handy at all, and you can't really hope that your enemy stays in the focal point enough time for you to realize it and fire your laser. And for you to move keeping the lens steady, you will suffer of always being behind in knowing your target position of a time proportional to the distance between you and them.

$\endgroup$
2
  • 2
    $\begingroup$ We already have multipath cancellation technology for radio signals,with moving targets even! Distortion of laser beam is just more extreme version of this problem. It is possible to send pre-distorted beam by using laser array which will converge exactly on target using blackhole's gravitational field. It only needs sufficiently precise timing calculation. $\endgroup$
    – Juraj
    Mar 23, 2023 at 21:06
  • $\begingroup$ "followed slightly different path and will focus on the target with a different phases" that ISN'T what "focus" means. By definition, when focused, a beam will be in phase and have the same path length for all paths to the focus point. Or rather the point where all paths are the same IS the focus. $\endgroup$
    – Aron
    Mar 24, 2023 at 3:01
3
$\begingroup$

Blackhole does gravitational lensing at any distance. Better, because it can easily bend light by 180° or even more, you can hit target in any direction, provided you can aim with sufficient precision (see my comment above).

In fact, there's no hiding "behind" black hole. When you look at realistic animations you always see everything that's in the direction "behind" it, albeit distorted.

$\endgroup$
2
$\begingroup$

Your black hole has a horizon radius round about $1\times10^{-15} $ meters. That's round about 100,000th of the width of a Hydrogen atom. The mass you suggest, $3.98 \times 10^{11}$ kg is, on a planetary scale, kind of small. Mount Everest comes in at about $8 \times 10^{14}$ kg, two thousand times as big.

Example calculation of bending of light by the Sun.

Bending is proportional to the mass and inversely proportional to the minimum radius the light passes by. The Sun with $2\times10^{30}$ kg mass, produces at a min approach distance of 700,000 km, $8.5\times 10^{-4}$ degrees of deflection. To get 1 degree of deflection with your mass, the light would have to pass within $1.18\times 10^{-13}$ meters of your black hole. That is round-about one thousandth the width of a Hydrogen atom.

Your blackhole will be more-or-less invisible to your laser beams. The tidal effects could be interesting, but I didn't work them out. For purposes of laser shots it will be utterly useless. Indeed, unless you get close enough for the tides to be meaningful (probably within a couple meters) you'd be hard pressed to find it.

Except! Except for the Hawking radiation. A black hole with this mass has a useful life span in the billions of years. But it has a nominal luminosity of 2 billion Watts. So the BH itself will be putting out this incredible amount of radiation, probably more than any laser beams you could hope to mount on a spaceship. The upper end of particle energy is round about 100 MeV electrons, which is very nasty indeed. It's not in a beam, but still, you don't want to get too close.

Black Hole Hawking Radiation Calculator

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .