Janus

You may "always hear that", but it isn't what model actually says. All it says is that the Universe, in it's earliest stages, was extremely hot and dense, and says nothing about it's size. One of the still unknowns about the universe is whether it is finite or infinite. The references to being "smaller than a proton" likely are referring with the "observable universe", which, for all we know, could be a tiny corner of a much vaster, or even infinite universe.

That hasn't been my experience. Both my thumbnails look perfectly normal, and you'd never guess that anything out of the ordinary had ever happened to them.

Unfortunately, it isn't the fungus that causes the regrowth. I talk from experience. Over my life I've lost two thumbnails, due to trauma, that then grew back.

The 1927 Solvay conference.

There is a neat story about Einstein concerning this. This occurred during some scientific conference. Einstein and a few others were at a table debating QM. Einstein would come up with some problem that he felt disputed QM, and the others would go over it until they found a flaw with his argument. Einstein finally came up with on that no one could find the flaw in. It started to get late, so they decided to call it a night. They met up again the next day. Whereupon Einstein stated that he'd been thinking about it, and had found the problem with his own argument.

I fail to see where this would matter, other than Einstein taking credit for that which he was not due. It has nothing to do the validity of the conclusions. And as pointed in the previous post, the theory has survived every test thrown at it, and much of the equipment we use today would not function if it were incorrect to any large degree.

Mine would most likely come through my paternal grandfather. But he was born nearly 170 years ago so it's a bit hard to verify.

I came across some new info that solidifies the idea that I have at least some Sámi in my ancestry. It was a YouTube video which broke down the DNA of a Sámi male. It used results from the same test I took. It identified one of the genetic groups as matching one of mine. It returned a result of 7% Inuit, which adds weight to my suspicion of where my 2% result came from. The icing on the cake was in another video on the Sámi, which had a photo of a Sámi girl. I saw a strong resemblance to my sister. I called my wife over and asked her if the photo reminded her of anyone( without giving her any other context), and she said my sister. So while still not 100% conclusive, I'd say that the odds are pretty high.

The issue I have with these polls is that I doubt that hey have changed their methods since the '22 primary. For example, they do phone polls. But these polls rely on people actually answering their phone. There is a generation gap in this factor. Younger people, who tend to vote more liberal, are much less likely, in general, to answer their phone when they get a call from an unrecognized number. Whereas people in the older age demographic, who tend to vote more conservative, in general, are more likely to do so. As a result, phone polls tend to over sample in favor of conservatives, even if they don't intend to.

The galaxies that show blue shift are those that are part of our group or cluster. Galaxies are not evenly spread out but segregated into collections bound by gravity. First you have local groups of 50 or so galaxies, then these isolated groups form a cluster, then you get super clusters. Collections up to clusters are held together by mutual gravity and don't separate with the universal expansion. As a result, galaxies within them can have various velocities with respect to each other

Gravitational red-shift is due to light going from one gravitational potential to a higher one. Imagine it like a hill. Climbing the hill takes energy. Light gives up the energy it needs to "climb the hill" by through a decrease in it frequency. For our galaxy to see a red-shift from all directions would mean we would be at the "peak of the hill". and everything else lower down the slope. But that means everything else would see our galaxy as higher up the slope, and see light coming from our galaxy as being blue-shifted. For your idea to be correct, our galaxy would have to have a unique and special position in the universe. The second part, as pointed out by others is an old hypothesis, and has been discounted.

In order to reach LEO, a fully loaded shuttle needs all the fuel in that large external tank, plus two solid fuel boosters. As I pointed out in an earlier post, you need about 2.2 km/sec of delta v to enter a trans-lunar orbit from LEO. Even if the entirety of the cargo capacity of the shuttle was extra fuel, the SSME's wouldn't be capable of getting the shuttle up to this speed. Once at the Moon, your shuttle would be moving ~ 0.8 km/sec slower than the Moon, and would need to do another burn in order to be able match speeds and enter orbit around it. Then to return, another burn is needed to re-enter another trans-lunar orbit in order to get back to Earth. Once back, the shuttle will be moving at ~2.2 km/sec more than LEO orbital speed. The shuttle can't hit the atmosphere at this speed, so it will need to do another burn to shed enough for re-entry. So, this works out to at least 6 km/sec total delta v for the trip. This jumps the fuel requirements to 2.8 times the mass of the empty shuttle. Adding a lander would increase this fuel requirement. (some thing the equivalent of the Apollo LEM, would require ~20% more fuel. A good part of the Shuttle's mass is there for re-entry and landing, and would be dead weight as far as the majority of the trip is concerned, so you'd be burning a lot of fuel to get something to the Moon that is of no use when you get there.

Those circular rings in the image were a choice made by the artist. He could have easily chosen to use a rectangular grid instead.

Halc's original statement dealt with the difficulty of even putting a craft into orbit "near" the Sun. Let's take the Parker Solar probe, on its closest approach it is ~7 million km from the Sun in a highly elliptical orbit. This took several orbits and 7 gravitational assists from Venus to achieve. But let's look at its first perihelion, which was at ~22 million km. What would it have taken to circularize the orbit at that distance? A delta V of ~ 56.6 km/sec. This is 5 times the escape velocity from Earth's surface. Now, given a typical launch vehicle engine, it would take roughly 11 kg of fuel for every kg of payload to reach escape velocity. For 56.6 km/sec, that jumps to 232245 kg per kg of payload.

The mass of the oxygen used per person for a trip to the Moon and back is pretty insignificant. compared to the mass of the person themselves. The real savings in not sending a person is in the mass of the person themselves, not in the oxygen they would use.

From Earth orbit, escape velocity from the Sun is ~ 42 km/sec. The Earth in it orbit is already moving at ~30 km/sec, so it would only take roughly 12 more km/sec in delta V to escape the Sun. However, to "drop" an object into a close pass around the Sun, you first have to shed most of the Earth's orbital velocity. Then if you want to insert it into a circular orbit at that distance, you'll have to shed a good part of the velocity it picked up falling in towards the Sun.

From the normal LEO orbit for the Space shuttle, you would need ~ 2.2 km/sec additional delta V to attain a Moon intercept trajectory (~1 km/sec less than needed to attain escape velocity). Using the SSMEs, this would require ~ 50,000 kg of additional propellant, which is about 7% of the mass of propellant in the main external tank at launch. The shuttle cargo capacity is 24,000 kg. So, even if you were to fit an additional fuel tank into the cargo hold, you'd come up short. Now, if you were to put up a secondary fuel tank in a separate launch, and had the Shuttle meet up with it in orbit, you might, with some retrofitting, be able to attach it to the shuttle in much the same way as the external tank was, and give you that extra needed fuel. Of course, this just gets you to the Moon. If you want to come back and put the shuttle into the proper re-entry trajectory on return, you'll need to up this fuel requirement considerably.

The graviton would be a quantum of gravitational radiation ( gravitational waves), playing the same role that photons do for electromagnetic radiation. Virtual photons act as the mediator for electromagnetic forces.( Which they also do with a black hole. Thus a black hole can have a electric charge and field, even though photons cannot escape the event horizon of a black hole.) So, in the same way, Virtual gravitons would act as the mediator for the gravitational force giving the Black hole a gravitational field even though no actual gravitons leave the EH. *

Others have already basically answered your question in that it takes fuel to accelerate the fuel you'll use during the acceleration. mathematically it works out to MR= edV/Ve MR = the mass ration (payload+fuel)/payload dV is the total change in velocity Ve is the exhaust velocity As far as fuel usage goes, accelerating up to 2V is no different than accelerating up to V, then decelerating back to 0. Before we ever reach the point of sending a generation ship out to the stars, we will likely already have spent a good deal of time learning how to build and maintain artificial environments in the form of orbital space colonies in our own system. This is turn means we will already have populations more attuned to this type of life. It will be these people that crew these generation ships.

If Relativity were not true, then the Kinetic energy of the Muons would have to follow Newtonian rules, and the measured energies of the Muons are not large enough. Example: For a muon traveling at 0.98c, Relativity gives a time dilation factor of ~0.2, meaning, for the muon to reach the detector within its lifetime under Newtonian conditions it would have to travel ~5 times as fast or at 4.9c. However, The KE of the muon moving at that speed under Newtonian rules would be almost 3 times greater than that predicted by Relativity and moving at 0.98c, and more importantly, Almost 3 times that measured in the experiments.

The issue would be how would you make that bomb focus all it's released energy towards accelerating the probe? The closest example we have in this respect is an underground nuclear bomb test from 1957. The bomb was placed at the bottom of a shaft with a iron cap. When the bomb was detonated, it blew the cap off. Estimates have put the speed of the cap at 5 times the escape velocity from the Earth. Now, given the size of the cap and the density of Iron, you can get an estimate of how much KE it had. If you then take that KE and apply it to something with the mass of a cellphone, you can get its equivalent speed. It works out to ~ 2% of light speed. And this was using a nuclear devise many times more powerful than the Hiroshima bomb. To reach 50% of c, it would have had to had more than 625 times more energy than that (At this velocity you'd need to use the relativistic KE formula to get an accurate value)

The gravity "slingshot" uses the planet's gravity to alter the trajectory in such a matter that part of the planet's momentum/orbital velocity is transferred to the craft. The theoretical maximum gain from such a maneuver is twice the orbital velocity of the planet. However, this would require placing the craft ahead of the planet in it's orbit, at just the right spot and at rest with respect to the Sun. In practice, this is not practical ( and you'd likely end up wasting more fuel trying to do so than you'd save with the slingshot). In practice, you will always end up in a scenario where you get a smaller boost. This is further complicated by the fact that if you have a final destination in mind, you are limited as to which types of trajectory/boost you can use. My calculations using fission was assuming 100% efficiency as far as the released energy being converted into propulsion. Using a bomb would waste a good percentage of the energy and be less efficient. Also, it is important not to confuse thrust with engine efficiency. For example, Chemical rockets tend to be high thrust and lower efficiency, while ION engines are low thrust and high efficiency. Higher exhaust velocity equals greater delta v for the fuel used, but lower exhaust velocities give you better thrust for the energy used.

Nuclear fission converts ~ 0.1% of the mass into energy. If all of that energy was convert into KE for the remaining mass (acting as the reaction mass), then you might get a exhaust velocity of ~.045c. So let's say that you want to reach 10% of c.(43 yrs to Alpha Centauri). Using the rocket equation gives us an answer of needing over 8kg of fissile fuel per kg of payload you want to get to Alpha C. If you want the trip to end with you being at rest with respect to your destination, this jumps to 75 kg of fuel per kg of payload. This is impractical. Fusion is the better option since it converts a larger percentage of the mass into energy, thus giving you a higher exhaust velocity, which decreases the fuel to payload ratio needed to reach any given velocity.

It is important to distinguish between the acceleration of each object involved and the total closing rate between the two objects. When you drop an object near the Earth, it accelerates towards the Earth at a rate that is solely dependent on its distance from the center of the Earth and the mass of the Earth. Simultaneously, the Earth accelerates towards the object at a rate that is determined by the same distance, and the mass of the object. For the everyday type of objects you are likely to drop, the Earth out-masses them by so many magnitudes that its acceleration is imperceptibly small, and can be ignored for all practical purposes, meaning we can treat the acceleration of the object towards the Earth and the closing rate between the two as being one and same. As you move to larger and larger objects, the acceleration of the Earth starts to make up a greater proportion of the closing rate, and at some point cannot be ignored( where this point is depends on how accurate you need your solution to be) So all objects dropped from a given height from the Earth do accelerate at the same rate regardless of their mass, but the Earth's acceleration towards them does change according to their mass.

So I did a bit of research, and it turns out that the Sámi, which are the indigenous people of the Nordic countries, are basically indistinguishable from the Inuit. This map show the present homeland of the Sámi (Sámpi) https://en.wikipedia.org/wiki/Sámi#/media/File:LocationSapmi.png Here is a map showing the regions my genes come from(Which matches what I know from my family tree)The small Northern regions show some overlap with the Sámi regions, and I know my ancestors were from the Northern parts of those regions So it is not a stretch to assume that the 2% tagged as Inuit is Sámi, and with that bit of info, the results make a bit more sense.