# How big a telescope would we need to see the light distortion effects of a black hole the mass of the Moon, at the distance of the Moon?

Fairly straightforward. I suspect that we'd notice the gravitational effects of it long before we'd be able to see it, but at what point in history would people start noticing that stars jitter around in a specific point in the sky every month?

Note: A black hole with an Schwartzchild Radius the size of the Moon would have the mass of 588 Suns. Tidal forces would shred the entire solar system, so that wouldn't make a very good world BUILDING question.

• Right, but I suspect it would be small enough to require a telescope to actually see it. We were guessing that the tides and the moon were related in the 700's, but without the Moon, I think we'd need Isaac Newton's theory of gravity to know where to point our telescopes. Even then, it might be harder to spot than Neptune. Nov 8, 2022 at 21:20
• Could you clarify if it is: A black hole with mass of the moon. A black hole with an event horizon the diameter of the moon. Pretty sure the former due to gravitational affects of the latter. Both both could fit the 'size of the moon' Nov 8, 2022 at 21:30
• With Gault Drakkor that we need clarification. The title suggests a black hole the "size" (interpret: "diameter") of the Moon (and as a result many solar masses worth), while your comments seem to suggest a black hole with just the mass of the Moon. Two very different ideas.
– BMF
Nov 8, 2022 at 22:26
• Relevant XKCD (courtesy Starfish Prime).
– JBH
Nov 8, 2022 at 23:29
• I have a feeling the lensing effects are going to be on the order of the black hole horizon's radius. Considering John Dallman calculated a tenth of a millimeter, it might be quite some time before anyone directly observes that . . .
– BMF
Nov 9, 2022 at 13:15

John Dallman calculated that the Moon-mass black hole will have a Schwarzschild radius of ~50 micrometers. That's pretty small, but will the lensing effects make up for it? We can get specific about it by solving for the angular size of the BH's Einstein ring when lensing the disk of the Sun:

$$\theta_{1}=\sqrt{\frac{4GM}{c^{2}}\cdot\frac{D_{LS}}{D_{S}D_{L}}}$$ Where:

• $$G$$ is the Gravitational constant, $$6.6743\cdot10^{-11}$$ m^3 kg^-1 s^-2,
• $$M$$ is the mass of the lensing object, $$7.348\cdot10^{22}$$ kg,
• $$c$$ is the speed of light, $$3.0\cdot10^{8}$$ m s^-1,
• and the various $$D_{...}$$ are angular diameter distances of the Lens, Source, and distance between them. Angular diameter distance is distance defined in terms of an object's physical size, $$x$$, and its angular size, $$\theta$$, as viewed from Earth, $$d_{A}=\frac{x}{\theta}$$. Angular diameter of the Sun is about $$0.53\frac{\pi}{180\cdot3600}$$ radians. The angular diameter of our black hole however . . . is nearly zero. For small angles, $$\theta \approx \frac{x}{d}$$, where $$x$$ is the transverse size of the object and $$d$$ is our distance from it. Plugging in our values we get: $$\frac{5.0\cdot10^{-5}}{3.84\cdot10^{8}}=1.3\cdot10^{-13}$$ radians.

• $$D_{S}=\frac{1.496\cdot10^{11}}{0.53\frac{\pi}{180\cdot3600}}$$,
• $$D_{L}=\frac{3.84\cdot10^{8}}{1.3\cdot10^{-13}}$$,
• $$D_{LS} \approx D_{S}$$.

Let's see how this pans out . . . Completing the computation, we find the angular diameter of our Einstein ring to be: $$\theta_{1}=2.716\cdot10^{-13}$$ radians, or about $$0.00000008$$ arc seconds. The Hubble Space Telescope has an angular resolution of around $$0.04$$ arc seconds (at 500 nm wavelengths), and the human eye can resolve objects as small as $$40$$ arc seconds.

I'm not sure if a telescope the size of Earth itself could resolve even that . . .

Edit: Starfish Prime calculated the maximum size of the Einstein ring radius to be $$0.15$$ arc seconds when the object being lensed is 2x lunar distance from Earth and situated exactly behind the black hole. Maybe a comet on close approach, or a near-Earth asteroid could give such a large flare. The closest comet ever observed was Comet Tempel–Tuttle in 1366 at around ~9x lunar distance.

Nearby objects (up to 2x lunar distance) will have larger lensing effects, but for background objects like the Sun and stars the Einstein ring radius approaches $$0.00000008$$ arc seconds.

Edit 2: Here's a Desmos calculator of all this to play around with: https://www.desmos.com/calculator/pot7wiymj6
The largest contributing factor to the Einstein ring radius (when all else is constant) is the angular size of the black hole in the sky. A micrometer-scale black hole viewed from hundreds of thousands of kilometers away is really just too small to be seen.

There's an online Hawking Radiation calculator. That tells me that a black hole with the same mass as the moon (0.0123 Earth masses) is only a tenth of a millimetre in diameter and has an effective temperature of only 1.6 degrees Kelvin. The time until it evaporates is extremely long at 5.8E44 years.

Its tidal effects are far more noticeable than its effects on our view of stars; I suspect its presence would not have been deduced until Newtonian gravity was understood, and that might well have been delayed by the absence of a visible Moon.

• the q asks for a bh the size of the moon, not the mass, though.
– ths
Nov 8, 2022 at 23:51
• @ths: Not any more. Nov 9, 2022 at 0:38
• I would actually suspect that it would have more than its share of micro-meteor hits on that side, but that wouldn't be noticeable until well after Newton's time. Nov 9, 2022 at 1:45
• I sort of suspect its presence would be hypothesised early on, in order to predict the tides. You have tides that seem to be influenced by the Sun, but overlaid on top of them you also have another set of tides, which only line up with the Solar tides once every 28 days. From there it's not so hard to conjecture an invisible object that circles the Earth once per month. It might be a long time before you get any other evidence in favour of that conjecture though. Nov 9, 2022 at 3:29
• @N.Virgo I wonder! It's relatively easy to look at the moon, look at the tides, create a few charts about the timing of each and discover the basic relationship. But to actually hypothesize that something must be in the sky? That's an observation in hind sight. Keep in mind, without a moon in the sky, what's a month? I have my doubts even the solar relationship would be obvious. I agree with Robert - I don't think even the hypothesis is possible before Newton. And I have my doubts that a hypothesis would form for at least a hundred years after his discovery.
– JBH
Nov 9, 2022 at 4:55

So, this is slightly awkward to calculate, but I think I've found a nice simple answer that doesn't require anyone to integrate anything, which is nice.

The effect you're interested in is called gravitational lensing whereby light from a distant object is bent around a massive object between the emitter and the observer.

We know the mass of the moon $$M$$ (~7.4x1022kg), and how far away it is $$D$$ (~385000 km). The lensing effect is at its strongest when the lens is exactly mid way between the emitter and the observer, and the strength of the lensing effect can be shown by the Einstein ring radius:

$$\theta_e = \frac{4GM}{Dc^2}$$

where $$G$$ is the gravitational constant and $$c$$ is the speed of light.

$$\theta_e$$ turns out to be ~0.15 seconds of arc.

That's quite small. The smallest feature that can theoretically be resolved by a telescope has an angle of $$\theta \approx 1.22\frac{\lambda}{d}$$ (see Airy disc or diffraction limit) where $$\lambda$$ is the wavelength of the light you're considering and $$d$$ is the aperture of your telescope. In theory, then, a perfect telescope with an aperture diameter of ~80cm observing 500nm light should do the job... unfortunately, atmospheric interference limits resolving power of terrestrial telescopes to no smaller than about 0.3-0.5 seconds of arc, even for great big telescopes on the top of high mountains.

This means that until the advent of fancy toys like space-based telescope (eg. Hubble) or clever interferometry (like the pair of telescopes at Keck observatory) the fuzzy blob that your moon-lens generates is simply indistinguishable from any other fuzzy blob your telescopes can see.

Hubble was launched in 1990, and the Keck interferometer was fired up in 2001. There's not much scope for spotting lensing earlier than that, I suspect, though maybe the sheer weirdness of the invisible tidal source woudl encourage earlier investigation.

• Perfect, thank you! Plenty of grist for alternate history in figuring out what that thing is that causes tides. It would be particularly interesting if the space program got to witness the vaporization of the first mission to visit our dark dance partner. Nov 9, 2022 at 22:52
• $0.15$ arc second radius for an object 100 micrometers across? As I understood it, the Einstein ring radius should be on the order of a few Schwarzschild radii as that's where the lensing effects are most prominent . . .
– BMF
Nov 9, 2022 at 23:02
• Your equation assumes the black hole is half-way between yourself and the light source being lensed. From your source: "Gravitational lensing is most effective (meaning the ring radius is largest) when the lensing object is half way between us the and background source. In that event, the Einstein ring radius is given by this equation: ..."
– BMF
Nov 9, 2022 at 23:20
• So, if there exists a prominent source of light (a comet, maybe?) at 2x lunar distance, you may get a $0.15$ arc second flare.
– BMF
Nov 9, 2022 at 23:23
• @BMF the ring radius reduces for more distant objects, but not enormously. The halfway requirement is where it is at its largest, and represents a "best case". Feel free to read the source material and run the numbers yourself... it isn't my equation ;-) Nov 10, 2022 at 10:15

Maybe no Telescope? Other answers point out the black hole is smaller than a grain of rice. Even if the hole was directly in front of the Sun, you would not see the black spot we are all familiar with.

However! The moon hole's gravity still pulls at things hundreds of thousands of miles away. This is now you see the hole. You watch as it lenses things behind it by bending the gravity around it.

Here is a GIF from the N.A.S.A of lensing when one star passes in front of another. Even though the back star is blocked, some of the light from it bends around and the total brightness is more than the front star alone. Your moon hole does the same thing. For example when it goes near the rim of the sun, it will appear to create a little lump on the circumference, as light that would otherwise shoot out into space gets bent around and hits the planet. Note the black dot in the picture will not be visible. It is there to indicate the hole's position.

How big of a telescope you need will depend on how big is the lump. If anyone would like to do the calculations for lump size be my guest!

• The OP already knows that space-time distortions is how they'll see the black hole, it's written in the title. "How big the telescope" is the question. This answer doesn't really answer anything . . .
– BMF
Nov 9, 2022 at 13:11
• @BMF The question only says "we'd notice the gravitational effects ". The answer says how in particular we would notice them. Nov 9, 2022 at 14:38
• That's cherry-picked. OP said something to the effect 'we'd notice the gravitational effects first before we'd notice the stars "jitter" around the BH.' The title says "light distortion effects". OP clearly knew about gravitational lensing as measuring/detecting it was the focus of their Q.
– BMF
Nov 9, 2022 at 14:44
• You are right. The earlier version talked about the stars jittering and the title is also about light distortion effects D-: Nov 9, 2022 at 16:13
• BMF is correct. I am hoping someone knows the math well enough to identify how many arcseconds of distortion a blackhole a tenth of a mm across at the distance of the moon would cause. My basic research says that the Hubble wouldn't have been able to spot it. Nov 9, 2022 at 17:47