Pi>>We even had a physics PhD (DaveB) provide us with an equation that allows us to compute how many impacts we should expect. It turned out the number depends on the relative velocity of the approaching body and ranges from a minimum of about 13.5x for objects that are going so fast they will exit the solar system and never return to 200x for those that are just barely approaching Earth at very, very slow relative velocities.>>
I have accepted this 13.5x factor for comets or asteroids RETURNING from outside of 1 AU so as to collide with Earth. That is not what most of the collisions would be, however. The Brown model has stuff being pushed outward SLOWLY due to radiometer and Yarkovsky effects (RY). What this means to any sensible mind is that ONLY stuff being pushed outward from INSIDE OF 1 AU could possibly hit Earth. You don't LIKE this FACT (if Brown is right) but you have no ANSWER for it yet. This means that of all the stuff being pushed slowly outward, the amount that could hit the Moon is twice that which could hit Earth...because the Moon can be hit by stuff coming from inside 1 AU as well as stuff slightly outside of 1 AU. So even ACCEPTING your 19x factor, this takes your number from 800 to 400...before a couple of other FAIR adjustments...one of which YOU AGREE with.
>>Sorry, Indy.... the number of such encounters is an essential point in this topic.>>
Not if the ALL are obliterated high in our atmosphere. There could be 10,000 of them and we still may see very few that would make it entact to hit land.
>>At 17 km/sec, there will be no impacts during the flood year. >>
Once again, you don't even understand the HPT model. Stuff launched (both pro and retro) will have varying speeds. Half would be pro, so these would be immediately outside 1AU. As they are SLOWLY pushed outward for weeks by RY, these COULD hit the Moon, but not Earth. Some stuff would be in a prograde orbit around the Sun, but slightly slower than Earth and some of this COULD hit Earth during the Flood year. There also would be stuff launched fast enough retrograde that Earth could encounter it 6 months later, and at a speed much higher than 17km/s. This stuff also could hit Moon at greater speed too, so it could take a much smaller object to make the larger Moon craters. I don't know if 50% hitting Earth during the Flood year is the correct number, but SOME would. Again, it is not that relevant if the stuff is obliterated by the atm.
>>While the model is not exact, it does clearly show there will be no impacts during the flood year.>>
That model does not have a way to show stuff being pushed outward.
indydave, on 15 Oct 2016 - 4:39 PM, said:
indydave, on 15 Oct 2016 - 4:39 PM, said:
You don't like the 400 part because you don't understand the model, and some of that is due to how I expressed it. I do not mean stuff shot directly AT the Sun, which you say would then cause it to "return to its point of departure". I mean stuff shot rearward (retrograde) so that its Sun-orbit-speed is reduced so that it falls into a lower orbit around the Sun.
Pi>>No. I don't like the 400 part because I do understand the physics. I say it will return to its point of departure because of the Law of Conservation of Energy and orbital mechanics. As demonstrated by the "solar system" model you found.>>
I shouldn't have to keep repeating this...the spinning Earth would shoot stuff outward and some of that would give it speed so it immediately would be beyond 1 AU. And about half would immediately fall inward to a lower orbit...UNTIL it gets pushed outward by RY. Do you or don't you acknowledge the reality of RY?
>>You get 400 by simply dividing the number of expected impactors by 2 to allow for those that impacted the moon during the launch phase. That is not the correct approach. >>
Nope. Those hitting Moon could happen during just a few hours after launch, BUT those would be VERY few. The Moon covers such a tiny part of the sky, so less than (guessing) .01% would be direct hits. The rest (of what Brown suggests) would come from encounters of stuff being slowly pushed outward from 1AU-400,000 km to 1AU+400,000 km.
>>In order to find out how many struck the moon during launch, the best way is to find out how many craters there are on the back side of the moon because there are very few trajectories that will cause an object to narrowly miss the moon, hook around, and strike the moon on the back side during launch.>>
Perhaps that approach could tell us how many were direct hits. Almost none of the large maria are on the backside and THOSE could have been on the front too but the Moon slowly rotated afterwards into a new tidal lock due to a change in its mass, so that the more massive side faces us today.
>>We know there have been impacts since the flood and those impacts will be random because they are objects returning from solar orbit. Since it would be very unusual for an object launched from Earth to hit the back side of the moon, we can consider them to be after the launch phase. We also know that after launch, objects will be approaching the Earth/moon from random directions.>>
HPT has a MUCH better way to explain the fairly even distribution of craters EVEN AT THE POLES of the Moon. Your idea does NOT. It is rather silly for you to argue about objects returning "since the Flood" since the rate is almost NIL so hardly any would be expected to hit the Moon in the past 4500 years. Even in Brown's model...there would be hardly any to hit either Earth or Moon on a return since the Flood.
>>Unfortunately, according to this list the majority of 100 mile (160 km) craters are on the far side of the moon. This very strongly suggests few, if any, impacts during Brown's launch period.
Nope. As I have shown, the THREE crater size sites show us (with no invalidation for .1 density objects) that a cloud of water vapor, ice particles, rocky particles and small rocks (under 200m) could indeed make 100 mi craters. You have given NO reason to NOT accept that EXCEPT your own incredulity. I have a professional astronomer (Faulkner) the instruction video, and THREE crater size sites, which agree with me. I also just recently saw THIS: LINK
Equation to determine Crater Size:
D = Crater Diameter
Cf = Crater Collapse Factor ( this is equal to 1.3 for craters >4km on Earth)
ge = Gravitational Acceleration at the surface of Earth
g = Acceleration at the surface of the body on which the crater is formed
W = Kinetic Energy of the impacting body (in kilotons TNT equivalent)
pa = Density of the impactor (ranging from 1.8g/cm3 for a comet to 7.3g/cm3 for an iron meteorite).
pt = Density of the target rock
If you look closely D is the result and all the others are the type of INPUTS we found on ALL the crater sites. However, this "jumps to the chase" by not asking about size or density or even speed of the object...but it just wants the kinetic energy "W". That is exactly the same as what the instructor's video said. This formula says by implication that an object of differing size or density but which has the same mass and speed as another object...WILL MAKE THE SAME SIZE CRATER. So this means that if you scale down the density from 1 to .1, and then you scale up the volume so it is 10x...you will get the same size crater. That is what the THREE sites showed. And now this equation says the same thing. I cannot explain why one site said a 95km object was needed, another said 70km and a third said 28km...but I have indeed shown scientific support that a .1 density 70km crater (or even a 95km one, which could also fit what Brown describes) could indeed make a 100mi crater on the Moon.
indydave, on 15 Oct 2016 - 4:39 PM, said:
Then the radiometer and Yarkovsky effects would slowly push it outward (adding tangential speed), keeping its orbit circular.
PI:Insignificant. Of course, we've already discussed these influences extensively without any agreement there either.>>
You have to just deny science to say that RY are not going to affect the orbits of proto-asteroids. These effects would be many many times STRONGER on a proto-asteroid than on an asteroid...due to the greater surface/mass ratio. Do you deny RY? I may agree with you that the number of impacts IN THE FLOOD YEAR would be less than 50%. I just don't know what would be implied by the model. But if RY is true, then only 50% could possibly hit Earth while 100% could hit Moon because half of the stuff would be launched prograde and that could NEVER hit Earth because it would immediately be outside 1AU.
>>Also, the orbits of ALL objects launched from Earth will be elliptical, not circular>>
That Solar System model has NO way to demonstrate RY effects on orbits. It would slowly circularize them so that 4500 years later the a-belt would be circular. If they were in elliptical orbits in the time when they could impact Moon or Earth...SO WHAT?
>>Of course, you mean the YEC astronomer who doesn't agree that these things will even get launched in the first place. Without successful launch, none of the discussion of Brown's proposal matters at all, does it? How's that for something on *my* side of the ledger?>>
Faulkner's second paper against Brown spoke of the launching as being "a black box"...meaning in his mind Brown did not explain it (he DID), but that he (Faulkner) would take the launching as a given. THEN he would give his reasoning why he believed IF it happened, then there would be consequences (heat) and different observations in comets than what we see. In this (and other) threads here, we also BOTH accepted "for sake of argument" that the stuff WOULD get launched. NOW, you want to express that you DON'T accept the launching. Fine. We all knew that from the start. The question is whether you want to discuss (like DF did) what the consequences or effects would be IF it were launched. Apparently you have your running-away shoes on again, so now you want to no longer talk about craters...since you failed to make your case.