# Moon is a Black Hole

## Recommended Posts

1 hour ago, YNM said:

But I've also read that "Δν" as is colloquially used is not always the same as "change in velocity".

It is, if it's applied as a sufficiently short impulse. Things get complicated when you have low thrust engines, like ions, that you are using over a long transfer orbit. When you're just making a correction burn while passing periapsis, though, you can assume that it's always going to be equivalent to change in velocity. But yes, there is a reason why relevant engine stat is specific impulse, not specific velocity.

Energy is also conserved, but you have to look at total energy. So you have to account for kinetic, potential, and chemical energy of fuel before, and kinetic, potential, and thermal energy of the exhaust after. Momentum, being a vector quantity, is a lot easier to work with.

##### Share on other sites

Heard something about spinning black hole bending space and giving rockets more push.

##### Share on other sites
1 hour ago, K^2 said:

When﻿ you're just making a correction burn while passing periapsis, though, you can assume t﻿h﻿at it's always going to be equivalent to change in velocity.

Yeah.

I guess what's the problem with tying energy to orbital changes is that it's very variable. Inclination changes (normal burns) and eccentricity changes (radial burns), for instance, doesn't change any of the overall orbital energy, but expends some "energy" and momentum change (impulse).

I could only imagine the same happening with oberth effect, just to be a higher "multiplier".

##### Share on other sites

Well correct me if I'm wrong, but, disregarding kinetic energy for a moment, is this not also simply a matter of... time?

So say you're entering a gravity well with a discrete SOI boundary (for the sake of argument). You fall in, you experience acceleration, you fall back out, you experience the same deceleration. Now the time spent falling in is the same as the time spent falling out, so the change in velocity from acceleration is the same as the change in velocity from deceleration, so you come out going as fast as you went in.

If you fire your engine before going in, well, then you still come out as fast as you came in.

Now if you fire your engine while at the lowest point in the dive (periapsis), this would mean that, you go out faster than you came in. So you reach the far side of the SOI boundary in less time than it took you to reach periapsis from the SOI entry point. So you spend more time experiencing acceleration than you spend experiencing deceleration. So the net change in velocity from accelerating down the well is greater then the net change in velocity you get from decelerating on the way out. So you end up with more speed then just the speed boost you get from firing your engine.

Does it work that way?

##### Share on other sites
15 minutes ago, ChainiaC said:

Does it work that way?

Yes. But this kind of analysis is treacherous. Boost at periapsis changes shape of the trajectory, not just how fast you are moving. And as you coast along, the speed is only altered by component of gravity along direction of motion. As an extreme case, a craft in circular orbit never speeds up or slows down no matter how long it orbits. So in general, one cannot insist that trajectory that lingers longer is guaranteed to have a greater change in speed. I think this will be true for any free coasting trajectory near a single source of gravity, but I'd have to think about how I'd prove it mathematically.

That's not to say that your intuition about it is wrong, just that you have to be careful with using this chain of thought. Energy conservation is a lot easier to work with in this context, since it works regardless of the path taken - at least, for a conservative potential, which gravity from a single body is.

##### Share on other sites

How can you have a flat black hole?

And what's the moon?

Seriously though it would probably be pretty spooky if it just spontanously happened and I'm sure many theories would be thrown out the window.

##### Share on other sites
22 hours ago, A Soviet Tank said:

How can you have a flat black hole? ﻿

Probably rotating it fast.

##### Share on other sites

Assuming humans are not cooked by the x ray,

On 7/5/2018 at 2:15 PM, Xd the great said:

Heard something about spinning black hole bending space and giving rockets more push.

So, black hole can give us some boost

And the moon is rotating, so by general relativity, conservation of angular momentum...CHAOS

##### Share on other sites

Whilst a miniature black hole in itself is probably relatively benign, wouldn't anything falling into its gravity well release a tremendous amount of energy?

##### Share on other sites
16 hours ago, RCgothic said:

Whilst a miniature black hole in itself is probably relatively benign, wouldn't anything falling into its gravity well release a tremendous amount of energy?

Yes, but Moon-mass black hole has such a tiny cross-section that very few things would actually fall in. It'd probably form a tiny little accretion disk, but it won't present a danger.

##### Share on other sites

Rough estimation of tidal force:  F = 2GMmh/R3.

Say, m = 70 kg, h = 1 m, GM = 6.67*10-11 * 7.35*1022 = 4.9*1012(lazy to resolve the units).

F ~= 2 * 4.9*1012 * 70 * 1 / R3 ~= 7*1014/R3.

Say, Fcritical ~= 500 kgf ~= 5000 N.

Critical distance ~= (7*1014 / 5000) 1/3 ~= 5 km.

So, you will feel longer at < 10 km from the center of the moonhole.

##### Share on other sites

Would there be time dilation if you were orbiting above the event horizon?

##### Share on other sites
14 minutes ago, James Kerman said:

Would there be time dilation if you were orbiting above the event horizon?

Yes. In fact, a lot of time dilation. Essencially a time machine that sends you to the future.

##### Share on other sites
18 hours ago, Xd the great said:

Yes. In fact, a lot of time dilation. Essencially a time machine that sends you to the future.

Small black holes tend to have too much tidal force to be practical as such. It becomes overwhelmingly strong long before you start noticing relativistic effects.

If your goal is to travel thousands of years into the future, you want a really, really large black hole. Something like the Sagittarius A*. You can orbit one of these with periapsis really close to event horizon without any ill effects, then utilize the insane Oberth effect to depart without much trouble.

##### Share on other sites
24 minutes ago, K^2 said:

Small black holes tend to have too much tidal force to be practical as such. It becomes overwhelmingly strong long before you start noticing relativistic effects.

If your goal is to travel thousands of years into the future, you want a really, really large black hole. Something like the Sagittarius A*. You can orbit one of these with periapsis really close to event horizon without any ill effects, then utilize the insane Oberth effect to depart without much trouble.

Lol this is true, but if you are crazy enough to get close to a black hole and use it as a time machine, who cares about being torn apart.

Or if you are lying flat on your stomach. Should be better.

##### Share on other sites

Being a mad scientist is about placing big bets on what you've calculated to be winning moves. Experiencing spaghettification has no net-positive outcome under any circumstance. Even sending in henchmen seems like a waste. Maybe the failed henchmen, as a punishment... But I digress. Perhaps, the overwhelming majority of people who would want to use a black hole as a time machine are crazy enough not to care to be torn apart. But once we filter it down to these who'd be able to pull it off in the first place, given an opportunity, we're down to people who have a plan. And that plan doesn't involve being atomized.

##### Share on other sites
12 hours ago, K^2 said:

Small black holes tend to have too much tidal force to be practical as such. It becomes overwhelmingly strong long before you start noticing relativistic effects.

If your goal is to travel thousands of years into the future, you want a really, really large black hole. Something like the Sagittarius A*. You can orbit one of these with periapsis really close to event horizon without any ill effects, then utilize the insane Oberth effect to depart without much trouble.

Just watched the Scott Manley video on this.   https://www.youtube.com/watch?v=IM8HvoaKsBU  summary follows:

SR= schwarzschild radius, i.e. the event horizon
1 to 1.5 SR: any attempt to orbit at this radius is doomed.  I'm having issues with the math, but I think you need to burn a delta-v near the speed of light for each time you cross through this zone.  Avoid at all costs.
1.5SR technically allows an orbit at the speed of light, but you will either zoom off to infinity or fall into the "doomed zone" listed above.
2.0 SR zero energy radius.  Scott didn't begin to explain it.  Also unstable for unexplained reasons.
3.0 SR and beyond - stable orbit.  Also gets at most 30% time dilation (how much from your orbital speed and how much thanks to gravity, I'm not sure).

Scott also mentioned that the numbers above are for non-rotating black holes.  The black holes in the movie Interstellar are rotational and require even more math (and are not covered in said video).

I suspect that to get your Obereth, you would basically get into an elliptical orbit, then burn at Pe for significant boost (just plan ahead as you will presumably get a lot more than 30% time dilation by doing that).

On 7/2/2018 at 10:32 PM, K^2 said:

Almost shocking little. See my signature. Moon is a bit heavier than Rhea, so the output will have effective temperature of a brown dwarf star coming from an absolutely tiny object. It would be entirely undetectable. If it forms a tiny little accretion disk, you'd be able to pick it up on IR cameras of a space probe orbiting nearby.

I'm not quite sure why you chose that mass, did it have something to do with the pressures inside the blackhole being right for fusion (which really wouldn't be valid) or the accretion disc (which would)?  The Sun's would be a 2.95km, the Earth's would be 1cm, and presumably the Moons would be quite a bit less than 64nm.

##### Share on other sites
On 7/16/2018 at 10:47 AM, wumpus said:

1 to 1.5 SR: any attempt to orbit at this radius is doomed.

You certainly can't stay within that region for long. But an orbit can dip into that region, so long as apoapsis is multiple SR away. And you kind of want an elliptical orbit for when it's time to get out. Either way, yes, having an ergosphere makes it all a lot easier.

On 7/16/2018 at 10:47 AM, wumpus said:

I'm not quite sure why you chose that mass, did it have something to do with the pressures inside the blackhole being right for fusion (which really wouldn't be valid) or the accretion disc (which would)?  The Sun's would be a 2.95km, the Earth's would be 1cm, and presumably the Moons would be quite a bit less than 64nm.

At that mass, Hawking Radiation matches output of Sun's photosphere in spectrum, hence the "Black Hole Sun." Moon's HR would be a bit cooler and comparable in intensity. In other words, nothing remotely detectable from Earth.

Edited by K^2

##### Share on other sites

We would not be here

##### Share on other sites
6 hours ago, Cheif Operations Director said:

We would not be here

Why not?

##### Share on other sites
50 minutes ago, Gargamel said:

Why not?

Because the Moon has not mass enough to turn into a blackhole by itself, so something else must mechanically crunch it into the Schwarzschild radius. But the Newton's Third still applies - so whatever gargantuan force is applied to crunch the moon, will  "leak" back to space, and part of it will hit us.

I don't have the slightest idea how to calculate such force, but I think it will at least scorch the part of Earth facing the event.

##### Share on other sites

(Can't find if this was already mentioned).

Mass = const.
Radius= 1.735*106 m → 10-4 m., i.e decreased by 1.7*1010 times.

Original rotation period = 27.3 days = 2.4*106 s.

Solid sphere, I ~ r2.
Iw = const.

So, rough estimation of the new rotation period: 2.4*106  / (1.7*1010)2 ~= 10-14 s, frequency 1014 Hz.

Can it stay intact? Can it appear as a black hole at all? Maybe I missed something?

Edited by kerbiloid

##### Share on other sites
11 hours ago, kerbiloid said:

(Can't find if this was already mentioned).

Mass = const.
Radius= 1.735*106 m → 10-4 m., i.e decreased by 1.7*1010 times.

Original rotation period = 27.3 days = 2.4*106 s.

Solid sphere, I ~ r2.
Iw = const.

So, rough estimation of the new rotation period: 2.4*106  / (1.7*1010)2 ~= 10-14 s, frequency 1014 Hz.

Can it stay intact? Can it appear as a black hole at all? Maybe I missed something?

From here, I got the numbers:

```Python 3.6.6 (default, Jun 28 2018, 05:53:46)
[GCC 4.2.1 Compatible Apple LLVM 9.0.0 (clang-900.0.39.2)] on darwin
>>> import math
>>> M = 7.35 * 10**22
>>> G = 6.67408 * 10**-11
>>> c = 299792458
>>> rs = (2*G*M)/(c**2)
>>> r = 1738.1 * 1000
>>> w = (2*math.pi)/(27.32158 * 24 * 60 * 60) # in rad/s
>>> v = r*w # Tangencial Speed in m/s
>>> L = r * M * v # Angular Momentum
>>> # Angular Momentum is constant, M too. But we want the new V given the new rs
... vs = L/(rs*M)
>>> # from v = r*w, we derive w = v/r
... ws = vs/rs
>>> hs = ws/(2*math.pi)
>>> print("new Frequency Hz: " + str(hs))
new Frequency Hz: 107397776927487.7
>>>```

What I think is slighter faster than your numbers.: 10739777692787.7 Hz, or… ~1.074 * 1014 Hz.. Damn… The fastest known pulsar spins about 716 Hz.

But from here, I learnt that General Relative allows a maximum spinning rate of about 56,000 miles per second (90,000 kps), what appears to contradict this that says that the fastest spinning black hole does near the speed of light. Hummm.

In a way or another, we can calculate the maximum spinning rate for this MoonHole, as we know its rs, so:

```>>> v2 = (c / (2*math.pi*(rs**2))) * (2*math.pi)
>>> w2 = v2/rs
>>> h2 = w2/(2*math.pi)
>>> print(h2)
3745755982473533.5```

The second article give me numbers that leads to the conclusion that Yes, it's possible that the Moon be turned into a MoonHole as the maximum possible Frequency for a body with the MoonHole' radius would be 3745755982473533.5 Hz,  ~34.8 times more than the 107397776927487.7 Hz we have.

And yes, I finally reached the same number as you. I made a confusion about Angular Speed and Angular Momentum on my calculations!  .

Our numbers match right now (with mine being slightly more precise, as I delegated the number munching to the computer).

Note: I'm still trying to figure out the correct numbers for 'v1" on the last batch of numbers. I'm curious now.

Note2: The first article don't give me numbers to work. Doing some research on OJ 287, I found that it's spinning on 28% of the maximum spinning rate allowed by the Kerr metric. We should probably apply it on our MoonHole, but for now it's beyond my understanding.

Edited by Lisias
Fixed a dumb mistake on calculating the new angular velocity for the MoonHole, and forgot about Angular Momentum

##### Share on other sites
Spoiler
13 minutes ago, Lisias said:

What I think is slighter faster than your numbers.: 4.06 * 1016 Hz.

Indeed faster, though this is angular velocity (2pi/T), while 1014 is "linear" frequency (1/T), so iirc 2 pi should decrease the difference a little.
But that doesn't matter.

Just got another idea.
If the lunar pre-hole would be torn apart by rotation, and the lunar equator plane is near the lunar orbit plane...

Then the Earth will be in the rotation plane of the lunar disk saw, when the failed moon pre-hole matter will be spread around.

Spoiler

Even if the moonhole survives, anyway Earth lies in its plane of rotation.
So it will be being bombed by random trash and particles.

Edited by kerbiloid

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.

×   Pasted as rich text.   Paste as plain text instead

Only 75 emoji are allowed.

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.