1

Physiology & Medicine / Re: What Is Junk Food?

« on: 28/06/2024 04:37:17 »

I would also add mechanically recovered "meat" dosed with various chemicals to give it some flavour.

2

The Environment / Re: How are global warming and world population expansion linked?

« on: 08/04/2024 10:24:00 »

Yes i get the idea, but in 2 generations for a couple of  new born child rearing age  there would be 8 grand parents who would be retired and 4 parents who although not retired would be most probably 50 plus. That would be 6 people of working age for 9 non workers. (they could all possibly be unemployed)

2 generations, say 40 years. Average age of the working population is 40 so the average worker will already have lost half of his grandparents  assuming a life expectancy of 80.

The simple fact is that there are roughly twice as many people aged 20 - 60 than aged 60 - 100. That's the meaning of "life expectancy", and the reason that tax-funded pensions work at all!

Properly organised, an MRI scan should cost the NHS less than ?150, but thanks to Thatcher and Blair, it's about twice that.

3

Physics, Astronomy & Cosmology / Re: If Andromeda is red shifted how can we collide with it?

« on: 03/02/2024 20:49:03 »

Hi.

Doesnt the red/blue shift  result  partially from  the space itself expanding.?

    Not an exact quote, the word result was added to improve the readability.

    Yes, where it is expanding.   Andromeda isn't all that far away on Astronomical scales and it's still in the local group where space may not be expanding to the same extent that space between us and distant galaxies is assumed to.

Actually there's two things that could cause a red shift:
(i)   If the radiation was emitted from a place that is at a lower gravitational potential than the receivers location.
(ii)  Expansion of space that the light may have travelled through.

The first one is easier to consider.  It is easy to imagine a situation where an object is travelling toward the receiver but the gravitational red-shift effect is overwhelming and you would still obtain a net red-shift.

Best Wishes.

4

Technology / Re: How much abuse can a domestic microwave oven tolerate?

« on: 29/01/2024 17:59:41 »

Most common thing is that the reflected energy in the cavity will, due to the turntable, eventually return energy back into the probe end of the magnetron, leading to a high voltage flashover between the metal cap on the end and the waveguide cavity. This results in a burst of ionised air, and a dead short at RF of the magnetron output, with a loud bang. This RF short, with the stored energy in the magnetron cavities, and the capacitor between cathode and anode (the main capacitor outside has little effect, RF wise it is at the end of a really long RF trap by the cathode) results in the internal copper strapping inside the magnetron being subjected to very high currents, as the circulating energy is rapidly dumped via the output short, and those straps overheat and burn open circuit. This then results in the magnetron no longer being able to reach full power, even though it does draw the same current in operation, but the damaged cavities, like a whistle that has had the casing broken, will no longer oscillate with the circular flow of electrons from the hot cathode to anode.

You open the magnetron, and you will, after you saw through the end seal by the RF side, that there will be melted copper rings that strap the cavities together, so all cavities contribute to the RF output, even though only one has the output coupling loop in it.

https://upload.wikimedia.org/wikipedia/commons/1/16/Magnetron2.jpg

https://upload.wikimedia.org/wikipedia/commons/thumb/b/b0/Magnetron_section_transverse_to_axis.JPG/800px-Magnetron_section_transverse_to_axis.JPG

This shows the rings clearly, you will find them melted with the magnetrons that have had arcing over. the heavy copper strap that is leading up is the output loop, coupling off of the one cavity, and the rings around the cathode (silver in the middle) are the straps that connect the cavities, and also act to reduce harmonics as they strap alternate cavities together so that they will all resonate at the same 2.4GHz frequency.

5

General Science / Re: What Is Gray?

« on: 13/01/2024 21:37:19 »

people behave more carefully when they perceive more risk,

Apparently, after cyclones were given both male & female names (back in 1975), it was discovered that female-named cyclones killed more people than male-named cyclones. ("Cyclone" is Australian for hurricane.)
- I saw a speculation this week that female-named cyclones were perceived as less threatening than males, and so people took fewer precautions!
- Of course, if we went the next step and started giving cyclones names from multiple languages, that effect would disappear - and perhaps we may see racism in cyclone deaths? Names in languages that were perceived as more threatening might kill fewer people?

This came up in the context of heat waves - apparently Spain & Portugal have started naming heat waves
- It has been claimed that heat waves have killed more people in Australia than all other natural disasters combined (since modern records began)
- A week ago, I noticed the Bureau Of Meteorology* starting to quote "Dew Point" in weather reports. They said that with a dew point above 24C, some people will start to suffer heat stress (especially the elderly and the young). And that day had a dewpoint of 24C.

*The obvious (and common) nickname for the Australian Bureau Of Meteorology is "BOM" (pronounced "bomb"). Last year they came out with a press release stating that they were only to be called the Bureau Of Meteorology, and never "BOM"... Maybe people had realised that they were dropping too many climate bombshells, after they had to add an extra colour to the weather maps to denote extreme temperatures (purple is higher than red or black).

Australia_Heat_map.png (429.45 kB . 966x606 - viewed 1168 times)

http://www.bom.gov.au/australia/charts/viewer/index.shtml?type=T&level=2m&tz=AEDT&area=Au&model=CG&chartSubmit=Refresh+View

6

Just Chat! / Re: Anyone For Microdosing?

« on: 07/01/2024 13:57:42 »

Might explain the MAX, though.

"Hey, man, let's make the plane unstable, add a widget that overrides the pilot without warning, not tell him about it, and blame him when the plane crashes."

"Cool! And in the next issue we could save weight by eliminating some of the bonding wires, so the instruments, like, go crazy in a thunderstorm."

"Yeah! What a ride!"

7

Physics, Astronomy & Cosmology / Re: What is a Fermion?

« on: 02/01/2024 20:58:41 »

If it looks like a duck, walks like a duck and quacks like a duck, then it just may be a duck

8

Physics, Astronomy & Cosmology / Re: Has E=mc2 been proved practically?

« on: 15/12/2023 13:08:29 »

But how fast fo electrons travel (or positrons) during radioactive decay modes?

Very fast, a large percent of the speed of light.  Beta particles (electrons) from Uranium 235 fission can actually move faster than light through water.

Edit:  The beta radiation is actually from the decay of the fission products not from the fission reaction itself.

9

COVID-19 / Re: What is "long covid" and what different types are there?

« on: 28/11/2023 09:39:29 »

Chronic Fatigue Syndrome is a well-known condition that can last for months after an infection of mononucleosis or some viruses.
- The causes are not well understood
- Some COVID patients seem to suffer something similar.
https://en.wikipedia.org/wiki/Chronic_fatigue_syndrome

COVID attacks cells via the ACE2 receptor on their surface. This receptor is involved in regulating blood pressure.
- This receptor appears on many cell types including lungs (the normal route of infection), lining of blood vessels, kidneys, and nerve-supporting cells.
- Loss of nerve-supporting cells can kill nerve cells, resulting in COVID's most distinctive symptom: Loss of taste and smell.
- If this spreads into the brain, it could be responsible for the death of brain cells - a study of patient MRIs before and after COVID showed a reduction in brain volume (faster than normal shrinkage). This could be responsible for the "brain fog" symptom which is often reported by long COVID sufferers.
- Since every organ depends on a flow of oxygen and glucose, inflammation of the blood vessels (and resulting blood clots) could damage every organ in the body (depending on where the blood clots occur), resulting in a wide array of medical conditions.

Probably, Long COVID symptoms will eventually be grouped into 5 or more clusters of symptoms - where one person may be suffering from 1 or more of these clusters.
https://en.wikipedia.org/wiki/Long_COVID

(Overlap with Paul)

10

COVID-19 / Re: What is "long covid" and what different types are there?

« on: 28/11/2023 09:27:23 »

If I could answer some of the questions about postviral syndromes I could become quite wealthy-very little is understood. It is diagnosed based on a patient's symptoms, there is no test to do this. Look into the history of chronic fatigue syndrome. There are as many theories concerning the cause as there is in the case of ball lightning. It appears to involve subtle changes in the immune system, possibly t1/t2 lymphocyte dominance but as I said it does remain a mystery despite extensive research. It can range from relatively mild symptoms to incapacitating ones and again why this is so remains unexplained. Some patients improve over time and others do not.

11

Physics, Astronomy & Cosmology / Re: Why are some isotopes fissionable?

« on: 13/11/2023 00:53:14 »

Hi.

I have another question for you: u235 on exposure to a thermal neutron what ratio of u236/ u236* would one expect and are there conditions that affect this ratio?

    I'm not really an expert on Nuclear reactions.  I've just been reading some lecture notes and a textbook on it recently  (they weren't my lecture notes, one of the family just kindly made them available to me).

   U236* should be so short lived, you won't find any significant amount of it.  As fast as you may produce it, it will decay and usually by fission.   So you'll just find the later decay products.   You ( @paul cotter ) mentioned it should have a half life of 10^-12 seconds.   I previously mentioned it may not even be a real particle that exists, it's a useful model and/or may be something like U236 with a deformed nucleus  etc.   There should be very little you can do to increase the proportion of U236*,  the main problem would be keeping it alive (undecayed).

   U236*  is what you shoud be producing  (as the first step) from U235 and a thermal neutron.   You shouldn't produce U236 (the ordinary ground state).   However, there probably are some routes where the excited state  U236* can lose its surplus energy (say by emitting a gamma ray) and become U236 (the ground state).   There are probably also some chains (of neutron absorption and/or decay events) that would also lead to the production of U235.   For example, we can already use neutron bombardment to change one element into another, it's a bit random and the yield isn't great - you produce all sorts of stuff and only a small portion was what you wanted - but it happens.

   Nothing goes absolutely as it should in the textbooks inside a nuclear reactor.   Sometimes a contaminant will be present, that might absorb some neutrons and form something unexpected.   There are various things that could happen,  I'll avoid presenting examples but I'm sure you can imagine the situation.  We know that you end up with traces of all sorts of stuff being left in the reaction vessel and spent fuel rods when reactors are finally decommissioned.

    More significantly, you never really start with just 100% U235.    Natural Uranium is a mixture of isotopes and it's difficult to enrich the proportion of the (desirable for reactors) U235.  Getting to 5% U235 is very good and is considered a high quality fuel.   So, although on paper we can write down what should happen to U235 with a thermal neutron,  in practice you never really had pure U235 to start with, there was quite a lot of other Uranium isotopes in it already.

Best Wishes.

12

Physics, Astronomy & Cosmology / Re: Why are some isotopes fissionable?

« on: 11/11/2023 17:20:23 »

Hi.

    That's a good answer from @paul cotter.   I can add a bit more detail and did start writing last night but it got long and the computer crashed.

   To try and keep it short, I'll just take the comments you made in the OP and reply to each.

Why for example is U235 fissionable yet the more unbalanced u238 not?

   Good question.  There is an answer but it's quite long, see later.

U238 has a greater number of neutrons, so I would assume it would be quicker to decay and be more unstable?

    "Stability"  can be considered in two main ways.

1.   Energetic stability:    When some nucleons (protons and neutrons) come together to form a dense nucleus there is some binding energy released.  A nucleus is more stable when the binding energy is higher.   
   Let's say this another way - some protons and neutrons have a particular mass when they are well separated from each other.   When they come very close together (nuclear separation distances), the strong nuclear force really kicks in and takes over, the potential of each nucleon drops as they approach, so that energy is released.  As such the mass of a nucleus is less than the mass of the individual component nucleons  (by Einsteins famous E = mc^2).
   In general, any two nucleons that approach each other to nuclear separation distances will result in some binding energy being released.   So we expect the toal binding energy of a nucleus to be proportional to the number of nucleons in that nucleus.  As such we usually consider the binding energy PER NUCLEON.   If a large nucleus (say U-235) splits into two smaller nucleii then the binding energy per nucleon increases slightly and as a result the total energy (or mass) that remains in the daughter products is less than it was initially.   
   A nucleus can be considered to be more energetically stable when the binding energy per nucleon is increased.  As you may know, the highest binding energy per nucleon occurrs when a nucleus has around 60 nucleons in total (with about half of those protons and half of them neutrons - so an element like Iron on the periodic table).    U-235 of course has 235 total nucleons, so that splitting in half makes energetic sense.   Indeed it would be reasonable to split in half and then split in half again.

2.   Activation Energy:   A nucleus can't just reach a new final state from an initial state.   U-235 doesn't just spontaneously split because there is some activation energy required, to paraphrase this the nucleus must get over a small hill before it can roll down the big slope on the other side to an energetically stable final state.   This is quite different from another type of radioactive decay (alpha decay). 
   We can consider a simple model, where the two daughter nucleii (well, packages of nucleons that will go on to become the individual daughter nucleii once they separate from each other) pre-form inside a region of space that we are still considering as the nucleus of the original parent.   The strong nuclear force more or less switches off entirely once these two packages of nucleons separate from each other enough that we can consider them to be just touching rather than overlapping.  When this happens the Coulomb force (the electrostatic force between the positively charged protons in each package of nucleons) becomes the important dominating force.   There are textbooks and other sources of information that explain this in more detail.  For now we'll just accept that the Coulomb potential at a critical distance = r₁ + r₂   where  r₁ and r₂ are the radii of these daughter packages (the daughter nucleii that will form), is representive of the height of a potential barrier that would confine these two daughter packages to the parent nucleus region.   While the daughter packages are confined, it all stays together as one parent nucleus instead of splitting into the two daughter nucleii.
    Classical physics suggests that the only way to get over that barrier is to have the daughter nucleon packages with enough energy to get over that barrier.   Quantum mechanics presents another option, the daughter nucleon packages could "tunnel" through the barrier even if they didn't have sufficient energy to legitimately get to the top of that barrier.    Tunnelling is extremely dependant on the width of that barrier, which in turn will be dependant on the height of that barrier at the critical separation distance r₁ + r₂  since outside that parent nucleus region that potential is falling off  like  1/r   (the usual Coulomb potential).   For an alpha particle (which is just 2 protons and 2 neutrons, so  "not much"),  the other package of nucleons  is   Z - 4   nucleons   and for large  Z (mass number or nucleon number,  like we do have in Uranium-235  where  Z= 235)   we have that    Z-4  is approximately Z anyway.    The height of the Coulomb potential  is  q₁.q₂ /  (4πε r)    where r = r₁ + r₂ = the critical separation distance   and  q₁ , q₂  are the charges of the two daughter nucleon packages.  The details aren't too important,  we can simplify and assume r = r₁ + r₂ = the critical separation between the centres of charge is much the same for the situation where we have two daughter packages of about  235/2 ≈ 118 nucleons each,   or when we have  one big daughter package with 231 nucleons and a little alpha particle of 4 nucleons.    So that the height of the potential barrier is then only dependant on the product q₁ . q₂  that appears.   

     For the situation where an alpha particle and a big daughter nucleon package form,  we have  q₁ . q₂  =  2 . P     where 2 is the charge on the alpha particle and P is the charge on the big daughter nucleii left.   (So  P ≈ 90  for Uranium and it's large daughter nucleus left after alpha decay).    For simplicity we have  barrier height ~ P.    Tunneling of the alpha particle out of the parent nucleus does sometimes happen.

      For the situation where the Uranium 235 splits into two nucleii of about the same size,  we have that the height of the potential barrier  =  (P/2) . (P/2)     ~   P².        For large nucleii like Uranimum 235 with P ≈ 90,   P²   is a shocking lot more than  P.   The barrier width is MUCH larger and  tunneling probability falls of exponentially with this barrier width rather than just linearly, so this probaility takes another big hit.    As a result the barrier is just too wide and tunneling almost never happens.

   What this means is that, although U-235 will sometimes spontaneously split into two roughly equal sized daughter nucleii (after millions of years - exactly as quantum tunneling would predict),   the more prevalent decay is just to have an alpha particle emission.

   Anyway, let's get back to the main point....   Quantum tunnelling is not going to allow fission into two roughly equal sized daughter nucleii.   The rule of classical physics does almost hold - the nucleon packages must have enough energy to legitimately get over that barrier,  or at least get sufficiently close to the top that the remaining barrier width is now so small that tunneling is again realistic.   This is the origin of the "activation energy" that is required for fission to proceed.  The daughter nucleon packages must have enough energy to get over the potential barrier that confines them to the parent nucleus.

- - - - - - - - - - -
   OK, so that's the preliminaries all done.

   Now U-235 and U-238 are almost the same things as far as nuclear physics is concerned, they differ by just a few neutrons.   Both can decay by nuclear fission, both can produce two daughter nucleii of about half the original size of the parent.   The total change in binding energy when this fission happens is about 200 MeV for both  U-235  and  U-238.   They are almost the same when you consider the relative energetic stability of the parent nucleus compared to the daughter nucleii.   So, energetic stability isn't the issue here.

   That means that the difference in stability or fission-ability (if fission-ability is a word) must be due to activation energy.  This is the energy required to lift the daughter nucleon packages over the barrier that confines them to the parent nucleus.

   Now it would be nice if we could say that U-238 just has a higher activation energy than U-235.  That would end our discussion because we could argue that the activation energy for U-238 is too big and that's why U-238 is more stable.   However, it's NOT quite that simple....

   As previously mentioned, U-238 and U-235 are almost the same things as far as nuclear physics is concerned.   They both have similar activation energies.   Typically (say in a nuclear reactor) we provide that activation energy by supplying a neutron to the nucleus.   We have some high energy neutrons produced by earlier reactions and they crash into another nuclues of U-235 or U-238.   Now this actually makes  U-236 (or respectively U-239) for a brief while so it is really the activation energy of U-236  (or  U-239) that we are considering.   It's the U-236 (or U-239) that we want to undergo fission.
   First of all, the energy is not really supplied by the kinetic energy (or speed) of the neutron when it hits the parent nucleus.  It's natural to think that to smash apart a nucleus you must use a neutron with some speed but it's not really like that.  Most of the energy is supplied to the parent nucleus just because you have supplied it with a neutron.   We have already mentioned binding energy much earlier in this post.   Initially we had a neutron that was not in a nucleus, so it had mass (energy)  ~   the usual mass of an individual neutron.   Once that neutron gets into a nucleus then we have some binding energy that can be released,  the mass of the new nuclues is less than the sum of the initial nucleus plus the individual neutron.   This binding energy that can be released is usually much more important than the kinetic energy of the original inbound neutron.     This is important so we'll say it again another way:  Most of the energy you supply to a Uranium nucleus is just beause you have supplied it another neutron, the amount of kinetic energy that the neutron also carries in is just a bit of "fine tuning".    Indeed, for U-235 we have a better cross-section (a probability of causing a fission event) when we present it with a low speed neutron.  Much of a nuclear reactor is concerned with slowing down (moderating) the high speed neutrons that are produced and presenting a nice slow neutron to the U-235 nucleii.   Ideally, if you could present U-235 with a neutron that had 0 kinetic energy then that would be just great, that's almost exactly what it wants - but the neutron obviously won't move into the nucleus unless it does have some velocity relative to the U-235 nucleus.   Give the neutron too much speed and it tends to go past or "through" the U-235 nucleus and not result in a fission event at all.

   So, the activation energy for U-235 to undergo fission after acquiring a neutron (so that's really the activation energy of U-236) is 6.2 MeV  (that one I know).   The activation for U-238 for fission after acquiring a neutron is also about 6 MeV.   the exact figures for U-238 you can find in .... books and stuff,  they may differ by about 1 MeV.     To keep it simple we'll just say the activation energy is approximately 6 MeV for either the U-238 and U-235 that you use as the fuel.   That is to say, they are almost exactly the same.  As previously mentioned U-235 and U-238 just are much the same from the point of view of nuclear physics, they differ by just a few neutrons.   So it's NOT that the activation energy for U-238 is much higher than U-235,   that isn't the case and it does NOT explain why one is readily fissile after being supplied with a neutron and the other is not.

   The difference in fissionability is mainly explained by the availability of that 6 MeV just from absorbing a slow neutron.  The simple model for fission following neutron absorption is that for a brief while we can imagine that ²³⁵U will form a particle we label as  ²³⁶U* ,  the asterisk * that appears after the symbol is important.  ²³⁶U*  is  ²³⁶U  in an excited state,    while   ²³⁶U  is just the ground state (normal state) of U-236.   You can find information and data about  ²³⁶U  from...... books and stuff.   They got their information by doing experiments with U-236.   For example, you can measure the mass of one mole of U-236 and thus obtain the mass of one atom of U-236.  A little subtraction for the electrons in that atom and you have the mass of the (ground state) U-236 nucleus.   Meanwhile the mass of ²³⁶U* is not something you can measure in practice,  even if a particle like ²³⁶U* does exist,  it doesn't exist for long.  It will undergo fission or revert back to ²³⁵U plus a neutron quickly.   The particle  ²³⁶U*  is really just a model or idealisation that is used.   The idea then is that ²³⁶U* has a mass which is equal to the mass of ²³⁵U  plus the mass of one individual neutron,   i.e.   it has a mass which has not been adjusted by the release of binding energy that would result when the ²³⁵U  picks up a neutron and adds it to its nucleus.

   Now the activation energy available when ²³⁶U is formed from the addition of a (slow) neutron to the ²³⁵U nucleus is equal to the difference in energy (mass)  of  ²³⁶U*  and ²³⁶U    (the difference in energy between the excited state that is briefly formed and the rest state that would be possible after the binding energy from incorporating the neutron into the nucleus is released).    A similar thing applies for  U-238   (where the activation energy is the difference between  ²³⁹U*  and ²³⁹U ).     It turns out that this difference in energy between the excited states and the ground state is just over 6 MeV for the U-235 starting nucleus   and just under  6 MeV for the U-238 starting nucleus.   That is what matters, when U-235  absorbs a slow neutron this is just enough to lift it over the 6 MeV activation energy barrier,  while for the U-238 absorbing a slow neutron it isn't.
We actually have a good explantion for this small difference in excitation energy and it is precisely due to some of the stuff that @paul cotter has alluded to.   We have studied the binding energy of a nucleus and we have a fair model of how it depends on the number of protons and neutrons.   There is a "pairing term" that is important in this binding energy formula.  It turns out that the binding energy gets a boost when neutrons come in pairs and also when protons come in pairs,   so that there is an overall boost when both the neutron number and proton number is even,  there is no net adjustment when only one is even and the other is odd,  there is negative adjustement when both are odd numbers.   So the key thing to note is that with our U-238 and U-235, they are both Uranium and so both have proton number = 92 which is even.   However one has an odd number of neutrons while the other has an even number, this makes a small adjustment to the binding energy.  Indeed U-233  is just like U-235,  it is readily fissile with thermal neutrons.  Meanwhile, U-236 is just like U-238, it is not readily fissile with thermal neutrons.   It's just whether the number of neutrons they have is odd or even which makes a small, but as it turns out very important difference.

   Now, I mentioned that the kinetic energy of the neutron is some fine tuning.  U-238 will undergo fission after absorbing a neutron but that neutron must be carrying in the extra needed in the form of kinetic energy.  It must be fast.   However, for reasons that we won't explain (and I'm not sure if we have a good explanation), fast neutrons tend to have a lower cross-section for interaction with a nucleus, so fast neutrons just do not get absorbed by a nucleus as often as you would like.   In the case of U-238 and U-235 the cross-section for interaction is important for nuclear power generation and it has been studied in detail.  The cross-section for interaction with U-238 is greatest when the neutron has a kinetic energy of about 1 MeV,  below this and there just isn't enough energy to get over that activation barrier and fission to occur,   above that and we have the usual effect where fast neutrons just tend to go through or past the nucleus and not interact.  1 MeV is the sweet point where there is enough energy but the reduction in chance of interaction due to speed hasn't become too important.    Meanwhile,  U-235 has a preference for the slowest neutrons it can get because just having been supplied with the neutron is enough to get it over the activation energy.     Over-all  U-235 has a much higher cross-section for interaction at all neutron energies ranging between  0  (stationary) and 10⁷  eV   =  10 MeV.    I have attached a diagram.


fission of Uranium.jpg (50.9 kB . 809x507 - viewed 427 times)


   I hope that helps a bit.   The fissionability of  U-235  and U-238 is so different just because one will get just over the activation energy for fission by absorbing a neutron,  while the other doesn't quite get there.

Best Wishes.   

13

Physics, Astronomy & Cosmology / Re: Why are some isotopes fissionable?

« on: 10/11/2023 14:33:47 »

U235 is fissile and u238 is fissionable. Fissile refers to an element than can undergo fission and release sufficient neutrons to enable a chain reaction. Fissionable refers to an element than can undergo fission but cannot sustain a chain reaction, due to insufficient neutrons being released. Why some elements are one or the other I can't tell you but I would suspect it is down to the neutron requirements of the favourable daughter nuclei. Most fissile elements have odd mass numbers but this is not absolute. Most of the energy in the big multi-megaton bombs derives from u238.

14

Physics, Astronomy & Cosmology / Re: Which bit of the Shell theorem is not working?

« on: 16/10/2023 16:14:58 »

I see, so it is gravitational potential within the shell, the "net gravitational force?? The spherically symmetrical shell being uniform in density.

Force and potential are not the same thing, having different units. Inside the shell, the scalar potential is the same everywhere, but lower than it is outside the shell. The force vector inside the shell is zero, and greater than zero (pointing towards the shell) outside of it.

The [force] vector has magnitude GMm/r² and direction toward the barycenter of the masses. In the absence of a barycenter it has no direction.

This only works for simple two-body systems. I can easily set up some orbiting masses where the acceleration of the test particle is away from the barycenter of the system, or in any other direction I like. The moon for instance (in say an otherwise empty Sun/Earth/moon system) rarely accelerates directly towards either Earth or Sun, and on occasion it accelerates directly away from Earth. If the moon was closer to Earth (as it was long ago), it would be capable of at some point accelerating away from the Barycenter of the system.

15

Physics, Astronomy & Cosmology / Re: Which bit of the Shell theorem is not working?

« on: 16/10/2023 09:02:09 »

I don't recall Newton ever saying that gravity was the gradient of a potential.   It is just the force that a unit mass would experience when it is put somewhere.

No. It is (and always has been) an attractive force between masses. The vector has magnitude GMm/r² and direction toward the barycenter of the masses. In the absence of a barycenter it has no direction.

If you insert an arbitrary point mass in an infinite homogeneous medium, the barycenter is the point mass itself! 

16

Physics, Astronomy & Cosmology / Re: Which bit of the Shell theorem is not working?

« on: 16/10/2023 07:58:30 »

Hi.

Has my post been deleted, and Paul's?

   I don't know.

If an object within a shell has no gravitational  effect on any object within it, surely that is from the point of view of the outside Observer? One would assume that the shell would indeed have gravitational attraction, if the she'll where 100, 000km thick and 10 million in diameter?

    There may be some minor errors in the English in that question.   I think I know what you're asking.
Wikipedia gives a short definition or explanation of the shell theorem:

1.         A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its center.

2.   If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell.

 - - - - - - - - - -

   So the shell is made of some material and that material does gravitate - it attracts other stuff to it.    When you're outside that shell it behaves exactly as you would expect .   You would have a force of attraction identical to the force of attraction you would have if you imagine all of the shell was just squahed down into one really small ball and placed at the centre position of that shell.
    It's only when you're inside the hollow shell where it may seem you get an unexpected result.   You are still attracted to each part of that shell.   If you happen to be at the centre of the shell then you are equally attracted in all directions and there is no net force, as you may expect.    However, even when you move away from that centre and toward one piece of the shell,  you still feel no net force of attraction in any direction.   Provided you stay inside the shell, you find no net force acts on you.
    There are many articles and YT videos that provide explanation in varying levels of detail and sophistication.   The simplest explanation is merely suggestive or hand-wavy:     When you move toward one piece of the shell,   the distance between you and that piece decreases,   so by Newton's law the force of attraction to that piece (the bit infront of you) should increase.   That does happen.   However,   since you have moved off-centre, there is now a bit more of the shell behind you rather than infront of you.   That tends to increase the attraction to the direction behind you.   It just so happens that both things happen in the right proportion and the net (or total) force stays at 0.
      The acceleration can be determined in various ways,  you could use any inertial reference frame you want   OR   you can just identify that acceleration more directly by giving the person insde the shell an accellerometer.    To get the right result with an accelerometer in the real world you'd need a really, really big shell and a person who is actually floating in outer space so that they weren't experincing any forces other than the gravitational attraction of the shell.

Best Wishes.

17

Just Chat! / Re: Radiesthesia, shape radiation, device for developing paranormal abilities

« on: 11/08/2023 18:10:29 »

What ever happened to family friendly forum? and from someone in a position of trust too. I think Alan should be banned personally, who moderates this forum?

It's an old joke but worth repeating. Psychiatrist adminstering Rorshach ink-blot tests:

What does that remind you of?
Sex
And this?
Sex
And this?
Sex.......
...I think you are obsessed with sex
Not me mate. You did the drawings.
 
I'm not ashamed to discuss nuclear power and trains with children. They need to know about these things before puberty.

18

Just Chat! / Re: Radiesthesia, shape radiation, device for developing paranormal abilities

« on: 10/08/2023 13:05:55 »

Whaaaaatttt......?

19

Plant Sciences, Zoology & Evolution / Re: Has natural selection been nullified in humans ?

« on: 05/07/2023 21:46:30 »

Zer0, one species of ant consumes only a fungus which they grow underground. Another herds aphids for their sugary excretions.

This Thread has taken an interesting turn & seems intriguing.
Hope the @OP is Fine with it?
(btw any preference for reference? like PC or Pete or Petro ?
Plz do lemme know)

A few more ruling outs...
Sonar...Bats.
Night vision...Owls.
Infrared...Snakes.

Gosh!
I heard the Egyptian Dung Beetle tracks positions in moonless nights using the Milky Way Galaxy.

Was thinking, what about the act of commiting Suicide?
& Then i found this...WoW!

https://www.popsci.com/insect-self-sacrifice-exploding/

20

Geology, Palaeontology & Archaeology / Re: Is there enough lithium?

« on: 01/06/2023 10:16:30 »

Apparently, Lithium has no significant new sources in the whole universe...
- Lithium was produced in the early universe, making up 10^-9 of matter from the Big Bang
- When Lithium finds its way into the core of stars, it is rapidly fused into heavier elements (ie the amount is reducing)
- When stars fuse lighter elements, they "skip over" Lithium, so no new Lithium is being produced in stars

So, it is important to wisely use what Lithium we have (and recycle what we no longer use)...
- Lithium is great for mobile applications, due to its low density
- But for stationary power storage, other chemistries are becoming competitive, like flow batteries