Inspired by these articles, what would have to be the smallest mass a black hole placed at the center of the Earth would have to have to destroy the Earth or at least wreak some noticeable havoc before it perishes from Hawking radiation? 93.136.4.186 (talk) 02:25, 19 November 2017 (UTC)[reply]

I don't know the numbers, but any black hole which would suck matter in faster than it evaporates would eventually destroy all Earth if placed at the center thereof. 2601:646:8E01:7E0B:404:F3D3:C557:159A (talk) 11:03, 19 November 2017 (UTC)[reply]

If you look at your links, you will see that the Earth will be heated into a plasma, before being swallowed. Is vapourising the Earth counted as destroying it? Graeme Bartlett (talk) 00:34, 20 November 2017 (UTC)[reply]

Stellar_black_hole#Properties cite: "There are no known processes that can produce black holes with mass less than a few times the mass of the Sun.[...].As of April 2008, XTE J1650-500 was reported by NASA[6] and others to be the smallest-mass black hole currently known to science, with a mass 3.8 solar masses and a diameter of only 15 miles (24 kilometers). However, this claim was subsequently retracted. The more likely mass is 5–10 solar masses.". --Kharon (talk) 01:13, 20 November 2017 (UTC)[reply]

Hmmm, what if you shot two black holes at each other at particle-accelerator speeds? One might speculate the black hole merger should be absolute, yet the angular momentum this imposes would make the resulting hole a naked singularity. Is there any theory that suggests they break back apart, perhaps into some kind of compex black hole shrapnel, so the pieces keep their event horizons? Wnt (talk) 01:21, 20 November 2017 (UTC)[reply]

If you believe in Hawkins theories of "Micro black holes" read Micro black hole#Minimum mass of a black hole. In that case also please try to answer the Question what process should produce and apply enough force to make one when even whole suns, much much bigger than ours (a multiple of the resulting smallest black hole atleast - lets say minimum ~20 times our sun's mass), only have a small theoretical chance to produce a black hole. Looks to me like an ant thinking if it could push a planet on another Orbit by keep jumping in one place. --Kharon (talk) 01:40, 20 November 2017 (UTC)[reply]

I was originally thinking of collisions among multiple black holes that may pile up near the galactic center. But regarding your reply... that article suggests that large accelerators might produce black holes. Now if what is needed is 22 micrograms in a Planck radius, the first question that comes to my mind regards the Compton radius of ridiculously relativistic particles with that combined mass, which presumably one collides from opposite directions. The particles are surely foreshortened to a very narrow distance, but do they remain as fuzzy laterally as when they are at rest? I would think they'd have to, in which case they usually pass through each other without forming a singularity. (Of course, you also need a ridiculously accurate means of alignment...) Wnt (talk) 04:04, 20 November 2017 (UTC)[reply]

Odd... Compton radius directs to classical electron radius, not Compton wavelength. I'll have to look later and see if there's a plausible reason for that... Wnt (talk) 16:20, 20 November 2017 (UTC)[reply]

OP here, @Graeme Bartlett, vaporizing Earth counts, I'm aware that complete digestion is impossible. Basically, I'm assuming that there is a certain mass m, where black holes with smaller mass will tend to lose mass to Hawking radiation faster than they can accrete and thus evaporate, while black holes with a higher mass will tend to accrete material fast enough to grow. @Kharon this is just a "what if" scenario, I'm aware that there's p=0 probability of this happening. 93.139.55.105 (talk) 04:16, 20 November 2017 (UTC)[reply]

This might be a stupid question but why isn't yeast used as a meat replacement dietary protein source?

I've been reading about world protein shortages and about how farming animals for meat won't be sustainable in the long term to feed the growing population. Various alternative protein sources like farming insects and plant based proteins are discussed. There doesn't seem to be any attention paid to yeast though.

Yeast is easy to grow and is a compete protein. It requires no light so can be grown pretty much anywhere. Is there a reason that it has been overlooked as a dietary protein source? How much space would be required to grow enough yeast to feed one person?

Thanks Temic3300 (talk) 08:15, 19 November 2017 (UTC)[reply]

No, this is not a stupid question. And yes, yeast is used for food, see yeast extract and vegemite. There's also some yeast in leavened bread and unfiltered beer. Dr Dima (talk) 09:12, 19 November 2017 (UTC)[reply]

and we also have an article on Nutritional yeast. Dr Dima (talk) 09:18, 19 November 2017 (UTC)[reply]

As well as the actual yeast in beer (at least in Real ale, as Dr Dima alludes), the alcohol produced by the yeast also has nutritional value, per Alcoholic drink#Food energy, hence it sometimes being called "liquid bread". {The poster formerly known as 87.81.230.195} 94.0.37.45 (talk) 10:37, 19 November 2017 (UTC)[reply]

Have a look at the article on Quorn which is a meat substitute derived from a type of fungus. The advantage of the particular fusarium fungus is that it produces hyphae with a similar structure to muscle fibres, and can therefore be processed to give a more meat-like texture than would be possible with yeast. Wymspen (talk) 14:36, 19 November 2017 (UTC)[reply]

See also Single-cell protein. Various yeast are mentioned, but it's they're just some of several suggestions. Of course some suggest other alternatives like Entomophagy or plant proteins for various reasons (including the issues with keeping a sterile culture and yield). See also [1]. Of course you also have to convince people to eat the thing, that's one of the reasons why in the short term at least, most of these are ending up as animal feed. Nil Einne (talk) 15:13, 19 November 2017 (UTC)[reply]

• There is also the issue that people like the taste of meat. --Jayron32 12:19, 20 November 2017 (UTC)[reply]
  • Except for vegetarians maybe. To cite the old saying, "All the more for us!" ←Baseball Bugs ^{What's up, Doc?} carrots→ 12:49, 20 November 2017 (UTC)[reply]

One reason for not eating large amounts of yeast is that it is high in purines which can bring on attacks of gout in people who are susceptible. RJFJR (talk) 22:19, 22 November 2017 (UTC)[reply]

In previous discussion if we take into account the conservation of momentum law then there is no a paradox with energy needed for acceleration from 0 to 1 m/s and from 100 to 101 m/s.

${\displaystyle T_{\text{probe 1}}+T_{\text{earth 1}}+U_{\text{probe -earth 1}}=T_{\text{probe 2}}+T_{\text{earth 2}}+U_{\text{probe -earth 2}}}$

${\displaystyle {\tfrac {m{v_{1}}^{2}}{2}}+{\tfrac {M{V_{1}}^{2}}{2}}-{\tfrac {GMm}{R}}={\tfrac {m{v_{2}}^{2}}{2}}+{\tfrac {M{V_{2}}^{2}}{2}}-{\tfrac {GMm}{\infty }}}$

${\displaystyle MV_{1}+mv_{1}=MV_{2}+mv_{2}=V_{0}(m+M)\Rightarrow }$
${\displaystyle V_{1}=V_{0}(1+m/M)-v_{1}m/M;}$
${\displaystyle V_{2}=V_{0}(1+m/M)-v_{2}m/M}$
${\displaystyle \therefore {\tfrac {m{v_{1}}^{2}}{2}}+{\tfrac {M}{2}}{\Big [}{\Big (}V_{0}(1+m/M){\Big )}^{2}+(v_{1}m/M)^{2}-2V_{0}v_{1}{\Big (}m/M+(m/M)^{2}{\Big )}{\Big ]}-{\tfrac {GMm}{R}}={\tfrac {m{v_{2}}^{2}}{2}}+{\tfrac {M}{2}}{\Big [}{\Big (}V_{0}(1+m/M){\Big )}^{2}+(v_{2}m/M)^{2}-2V_{0}v_{2}{\Big (}m/M+(m/M)^{2}{\Big )}{\Big ]}}$

${\displaystyle (m/M)^{2}\to 0}$
${\displaystyle {\tfrac {m{v_{1}}^{2}}{2}}+{\tfrac {M}{2}}{\Big [}{\cancel {{\Big (}V_{0}(1+m/M){\Big )}^{2}}}+0-2V_{0}v_{1}{\Big (}m/M+0{\Big )}{\Big ]}-{\tfrac {GMm}{R}}={\tfrac {m{v_{2}}^{2}}{2}}+{\tfrac {M}{2}}{\Big [}{\cancel {{\Big (}V_{0}(1+m/M){\Big )}^{2}}}+0-2V_{0}v_{2}{\Big (}m/M+0{\Big )}{\Big ]}}$
${\displaystyle {\tfrac {m{v_{1}}^{2}}{2}}-V_{0}v_{1}m-{\tfrac {GMm}{R}}={\tfrac {m{v_{2}}^{2}}{2}}-V_{0}v_{2}m}$
Last equation is a law relating probe start and final velocities (${\displaystyle v_{1},v_{2}}$) for sun reference frame. E.g. for ${\displaystyle v_{2}=42{\text{km/sec}}}$ (which is the escape velocity for the sun from 1 AU distance) it gives ${\displaystyle v_{1}=46.3{\text{km/sec}}}$, which coincides with 16.3 km/sec in earth reference frame.

Why can't we solve an equation like this : ${\displaystyle T_{\text{probe 1}}+T_{\text{earth 1}}+T_{\text{sun 1}}+U_{\text{probe-earth 1}}+U_{\text{probe-sun 1}}+U_{\text{sun-earth 1}}=T_{\text{probe 2}}+T_{\text{earth 2}}+T_{\text{sun 2}}+U_{\text{probe-earth 2}}+U_{\text{probe-sun 2}}+U_{\text{sun-earth 2}}}$
Username160611000000 (talk) 20:11, 19 November 2017 (UTC)[reply]

Isn't this just the three-body problem? Rmhermen (talk) 00:48, 20 November 2017 (UTC)[reply]

No, because no need to know a trajectory. We know the start position and the final position. To solve a problem of escape velocity from solar system we use 2-step method (1st step is an overcoming the Earth gravitation, 2nd step is an overcoming the Sun gravitation from Earth orbit), shown in article. I wonder is it possible to solve the problem directly. Username160611000000 (talk) 05:08, 20 November 2017 (UTC)[reply]

I wouldn't usually bother trying to answer such a malformed question, but I think the OP is trying to calculate the escape velocity from the Earth's surface to interstellar space. I don't know why it is done in two steps, perhaps https://en.wikipedia.org/wiki/Sphere_of_influence_(astrodynamics) explains in enough detail. Greglocock (talk) 06:01, 20 November 2017 (UTC)[reply]

No, the article Sphere_of_influence is not about my question.Username160611000000 (talk) 06:32, 20 November 2017 (UTC)[reply]

As you like. I've just done a MOOC in Space Mission Design and Operations and I can assure you that SOI was fundamental to the three stages of planning the delta V needed for an interplanetary mission. Greglocock (talk) 09:19, 20 November 2017 (UTC)[reply]

As I see from the article the Sphere of influence (SOI) is an approximate imaginary surface, Feynman said nothing like that. And again I do not care about space dynamics. The exercise is to calculate initial speed to make the probe guaranteed to move at infinity with a residual speed v. I simplify the exercise to zero residual speed .
All we know and all we should use is lectures 1-14. And 2 -step method was explained by ToE here and here. The 2-step method is next. When the probe starts from the Earth with speed 11 km/sec it then overcomes Earth gravity and is flying in solar system with speed 30 km/sec in Sun ref.frame and with 0 km/sec in Earth ref.frame. To escape it should have 42 km/sec. So the excess speed = 42 - 30 = 12 km/sec and excess energy = 0.5m(12 km/sec)².
It was not clear why excess energy isn't calculated like 0.5m(42 km/sec)² - 0.5m(30 km/sec)². But when I have counted the conservation of momentum law and got formula ${\displaystyle {\tfrac {m{v_{1}}^{2}}{2}}-V_{0}v_{1}m-{\tfrac {GMm}{R}}={\tfrac {m{v_{2}}^{2}}{2}}-V_{0}v_{2}m}$, this confirmed that at an initial speed of 46 km/sec the final speed would be 42 km/sec.
Username160611000000 (talk) 12:26, 20 November 2017 (UTC)[reply]

Feynman didn't mention SOI because he was only considering two bodies. Duh. You want three bodies (probe earth sun), so SOI becomes an important concept. Out. Greglocock (talk) 18:44, 20 November 2017 (UTC)[reply]

Feynman has mentioned all planets in lecture 9, sec. 7 and showed a way to calculate positions at any moment with any wished accuracy. But the exercise 14.21 is to lectures 13, 14 on Energy. Feynman never proposed exercises that go beyond the material of lectures. In this exercise it is not asked to find all coordinates of the probe. I will use SOI and the numerical methods when in an exercise it will be asked. Username160611000000 (talk) 19:02, 20 November 2017 (UTC)[reply]

I don't want to go on scribd, just on account of not liking the site. But the equation shown seems unnecessarily complicated. The sun and the earth aren't changing - their relative kinetic and potential energy will add up to some constant. Using the sun as our fixed frame of reference, the probe will have a positive kinetic energy because it moves with the Earth's surface, and a negative potential energy relative to the sun, and a negative potential energy relative to the earth. Oddly all of these change during a day before launch because the surface spins with or against the orbital velocity, the probe gets nearer and further from the sun, and the tides increase and decrease the Earth's gravity. But just pick a launch window and stick with it -- I assume the best time is roughly when the probe, if released, would spin out away from the sun rather than toward it? It makes sense that the increasing and decreasing tide would do work on the probe as it stands on the launch pad, I think... Anyway, if you can tot all that crap up you ought to get an answer for pure kinetic energy given a certain distance from the sun. Wnt (talk) 00:01, 22 November 2017 (UTC)[reply]