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Comets & Asteroids From Earth?

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#61 indydave

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Posted 26 April 2015 - 10:24 AM

>>If it's approaching the Earth-Moon system at only "2-8 mph" as you have suggested, the multiplier approaches 300x. >>

Simple question...would the gravity component (multiplier) also go up for the Moon?  And wouldn't the result be that a slower object has the same chance of being hit by the Moon (vs. the Earth...other than the 13.5x size factor) as a faster object?  And if it is coming from inside 1 AU wouldn't that FIRST have a chance to hit the Moon?...and wouldn't many objects NOT in the 8000 mi. wide "slice" (Earth's diameter) get a chance to hit the Moon within its 500,000 mile wide "slice?"  And wouldn't the SAME object sometimes get SEVERAL chances for the Moon but only ONE for Earth?  Stop evading.

 

>>Any returning material would be able to enter the Earth-Moon system from virtually any direction at virtually any time. >>

Apparently you DON'T "get it" yet...after 23 times.  We are NOT disagreeing about "returning material."  It is ONLY the material being pushed outward by RE/SW.  That stuff NEVER RETURNS.  GET IT?

 

>>There.  It's nailed down.>>

NOT nailed down...that is obvious because you keep referring to "returning material."  Stop being so obstinate.

 

So again (since you neglect to answer this so far) I need to ask:  "DO YOU AGREE that there would be some stuff shot out fast on eccentric orbits which WOULD return to the inner solar system, but other stuff would start out less than 1 AU and get pushed out slowly by RE/SW...NEVER to return.  I am not asking if you agree really...but DO YOU SEE what I mean (and what Brown means) and that we have TWO kinds of things?"  AND (adding this point) there would be some stuff that starts at 1 AU and goes outward SLOWLY (due to RE/SW), which would NOT have any chance to hit E but COULD have several chances to hit the Moon...right?  Most of the stuff would MISS the Moon...so there's no need to repeat the point you made about how many degrees of the sky the Moon occupies.  We agree a lot/most misses the Moon. 



#62 piasan

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Posted 27 April 2015 - 09:49 AM

Indy seems to insist on making this topic about impactors.   Oh well...... 

Pi:

If it's approaching the Earth-Moon system at only "2-8 mph" as you have suggested, the multiplier approaches 300x. >>
 

Indy:
Simple question...would the gravity component (multiplier) also go up for the Moon?  And wouldn't the result be that a slower object has the same chance of being hit by the Moon (vs. the Earth...other than the 13.5x size factor) as a faster object?  And if it is coming from inside 1 AU wouldn't that FIRST have a chance to hit the Moon?...and wouldn't many objects NOT in the 8000 mi. wide "slice" (Earth's diameter) get a chance to hit the Moon within its 500,000 mile wide "slice?"  And wouldn't the SAME object sometimes get SEVERAL chances for the Moon but only ONE for Earth?  Stop evading.

An inability to comprehend on your part does not constitute evasion on my part.

 

We have already discussed slower objects.  The slower the object, the more influence Earth's greater gravitational attraction has.  The minimum factor of 13.5x was for objects with such a high velocity they will escape the solar system.  When the velocity drops down to a more reasonable 6 km/sec (13,400 mph) that goes up to 52x.  You're talking about things as slow as 2 mph.  Objects this slow will approach the maximum of about 300x.

 

If it is coming from inside 1AU (the orbit of the Earth around the Sun) it will only first have a chance to hit the moon if the moon is between the Earth and the Sun.  If the moon is in any other location, the Earth will be hit.  Especially if the object is moving along at the 2-8 mph you suggest.

 

The moon's "slice" is not 500,000 miles wide.  The absolute maximum is 54,300 miles wide and that is at any particular moment.

 

The SAME object will not get SEVERAL chances for the Moon but only ONE for Earth.

 

 

Pi:

>>Any returning material would be able to enter the Earth-Moon system from virtually any direction at virtually any time. >>
 

Indy:
Apparently you DON'T "get it" yet...after 23 times.  We are NOT disagreeing about "returning material."  It is ONLY the material being pushed outward by RE/SW.  That stuff NEVER RETURNS.  GET IT?

If it left Earth....  which is what Brown's model says and what anything launched above escape velocity will do.... then it comes back to Earth, guess what .... it is RETURNING.  GET IT?

 

You were wrong the first 23 times.... are you going for 230?

 

 

Pi:

>>There.  It's nailed down.>>
 

Indy:
NOT nailed down...that is obvious because you keep referring to "returning material."  Stop being so obstinate.

 

So again (since you neglect to answer this so far) I need to ask:  "DO YOU AGREE that there would be some stuff shot out fast on eccentric orbits which WOULD return to the inner solar system, but other stuff would start out less than 1 AU and get pushed out slowly by RE/SW...NEVER to return.  I am not asking if you agree really...but DO YOU SEE what I mean (and what Brown means) and that we have TWO kinds of things?" 

No.  We have only ONE kind of things.... those that have escaped Earth's gravity and are returning to the vicinity of the Earth-Moon system.  You are simply picking a special case of that material that, according to you, has a velocity of only 2-8 mph.

 

 

AND (adding this point) there would be some stuff that starts at 1 AU and goes outward SLOWLY (due to RE/SW), which would NOT have any chance to hit E but COULD have several chances to hit the Moon...right? 

Wrong.  At 2 mph, it will be captured by the gravity of the Earth, not the moon..... as we shall soon see.

 

Using metric units for consistency with the calculators....

The Earth's orbit has a radius of 400,000 km.

The Earth's mass is 6e24 kg.

The moon's mass is 7.35e22 kg

 

Using the gravitational force calculator at http://www.wsanford....calculator.html we find the Earth's gravity at the distance of the Moon's orbit (400,000 km) is 0.00250 N.  This is equivalent to the Moon's gravity of 0.002498 N at a distance of 44,300 km.  This means that at the lunar orbit, anything over 44,300 km from the moon will experience more gravity from Earth than the Moon and will be drawn toward Earth, not the moon.

 

Assuming for purposes of this evaluation that the object is in an orbit matching the speed of the Earth around the Sun and that the relative velocity is only 2 mph.  2 mph is 0.000889 km/sec.  At a distance of 400,000 km (the moon's orbit) the escape velocity from Earth is over 1.4 km/sec.  This is according to the escape velocity calculator at http://www.calctool....escape_velocity .  This means ANYTHING that passes within the orbit of the moon at the speed Indy claims and does not impact the moon will NOT escape Earth's gravity and will be drawn to an impact with Earth.  There are no "several chances to hit the moon."

 

Now, if we assume ALL of this material is on the plane of the lunar orbit around the Earth ... It won't be, but I'm feeling generous.... with a radius of 400,000 km, the Moon's orbit around the Earth totals a distance of over 2,500,000 km.  The moon shields 88,600 km of that.  This means those objects will be a MINIMUM of 28.4x more likely to impact Earth.   More than double the 13.5x MINIMUM.

 

It's STILL nailed down.

 

(Cross posted to the Craters topic where it properly belongs.)



#63 indydave

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Posted 27 April 2015 - 10:01 AM

Indy:

And that "passing water", some of which will not PASS but will COLLIDE with that rock will do WHAT to that rock when it hits the rear end of that rock?

 

Pi>>It will accelerate the rock.  That was never an issue. >>

 

REALLY?  Yes it WAS.  You claimed the rock would NEVER get up to the same speed as water.  I never said it CAUGHT UP to the earlier-launched water.  But it would eventually get to the same speed as LATER-launched water...so then it could indeed capture it, making a "sail."  RIGHT?

 

>>Once it hits the rock, it may even transfer all of its momentum to the rock.... but it will then bounce off and be picked up by that hypersonic stream of vapor that is already racing past the rock and quickly accelerate to those hypersonic velocities and again quickly outdistance that much slower rock.>>

 

Fine.  That would happen for a while.  Just like water in a stream would bash into the back of a log and then be swept along by faster water...UNTIL that log finally gets up to speed.  Eventually the rock would indeed be accelerated (slower) up to the same speed as the faster water molecules...RIGHT?



#64 indydave

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Posted 27 April 2015 - 10:45 AM

>>

We have already discussed slower objects.  The slower the object, the more influence Earth's greater gravitational attraction has.  The minimum factor of 13.5x was for objects with such a high velocity they will escape the solar system.  When the velocity drops down to a more reasonable 6 km/sec (13,400 mph) that goes up to 52x.  You're talking about things as slow as 2 mph.  Objects this slow will approach the maximum of about 300x.>>

 

And so would Moon's gravity go up...by the same factor...RIGHT? 

 

>>If it is coming from inside 1AU (the orbit of the Earth around the Sun) it will only first have a chance to hit the moon if the moon is between the Earth and the Sun.  If the moon is in any other location, the Earth will be hit. >>

 

Untrue.  Picture an object that is behind where the Earth is in it's orbit.  NO chance to hit Earth (if the distance is great enough).  But it still could be hit by the Moon.  Let's go to the extreme...250,000 miles "off" from hitting Earth.  NO chance to hit Earth.  And it still COULD hit the Moon.  It might even get pulled into the Moon if it is a bit further off.  AND ALL the objects sent outward from Earth AWAY from the Sun have ZERO chance to hit Earth...and still have a chance to hit the Moon. RIGHT PI?  YES OR NO. This would apply to the entire 500,000 mile swath or "slice" of the Moon's orbit. 

 

 

 

>>The moon's "slice" is not 500,000 miles wide.  The absolute maximum is 54,300 miles wide and that is at any particular moment.>>

 

I sure don't get that.  And we aren't talking just about "any particular moment."  It's ALL particular moments.

 

>>The SAME object will not get SEVERAL chances for the Moon but only ONE for Earth.>> 

 

We don't have to just consider the SAME object.  But sometimes there would be multiple chances for a single object.  HOWEVER what we really are considering is ALL objects.  MANY which have zero chance to hit Earth (because they aren't in the 8000 mile "slice" of Earth) would indeed get SOME chance to hit the Moon.  This raises the odds from 1/13.5x.  I believe they would be closer to 1/1 than 1/13.5.

 

>>If it left Earth....  which is what Brown's model says and what anything launched above escape velocity will do.... then it comes back to Earth, guess what .... it is RETURNING.  GET IT?>>

Ok...sure.  More tricky and wacky wording from you.  The slow stuff shot INSIDE 1 AU COULD return but it gets ONLY ONE chance to hit Earth.  MANY chances to hit the Moon.  GET IT?  We both agree about high eccentricy fast stuff which comes back MANY TIMES (like comets).  Stop confusing things by constantly speaking about that stuff.

 

>>You were wrong the first 23 times.... are you going for 230?>>

I haven't been wrong, but I may have to say it more than 23...since you still have a problem GETTING IT.

 

>>No.  We have only ONE kind of things.... those that have escaped Earth's gravity and are returning to the vicinity of the Earth-Moon system.  You are simply picking a special case of that material that, according to you, has a velocity of only 2-8 mph.>>

The "specialness" of it is that there is only ONE chance to hit Earth and MANY to hit the Moon while going outward ONCE.  PLUS half of that (that part shot out away from 1 AU) has ZERO chance to hit Earth.  That means odds are NOT 13.5x. 

 

>>

Using metric units for consistency with the calculators....

The Earth's orbit has a radius of 400,000 km.

The Earth's mass is 6e24 kg.

The moon's mass is 7.35e22 kg>>

I assume you mean MOON'S.

 

>>This means that at the lunar orbit, anything over 44,300 km from the moon will experience more gravity from Earth than the Moon and will be drawn toward Earth, not the moon.>>

So this makes my point.  The Moon will first get a chance to pull in stuff in a full 88,600km swath.  And this only "counts" if the Moon were standing still in a direct line with the Earth.  BUT IT MOVES.  And it moves around on a 500,000 MILE SWATH.  Plus half the stuff is shot outward beyond 1 AU so as to NEVER get a chance to hit Earth...while still HAVING a chance to hit the Moon.  You are so stubborn you won't even concede to THAT very SIMPLE point.  DO YOU?

 

>>This means ANYTHING that passes within the orbit of the moon at the speed Indy claims and does not impact the moon will NOT escape Earth's gravity and will be drawn to an impact with Earth.  There are no "several chances to hit the moon.">>

Such an object could take YEARS if ever to be drawn into the Earth and impact the atm.  During that time it would have many chances to hit the Moon. 

 

>>Now, if we assume ALL of this material is on the plane of the lunar orbit around the Earth ... It won't be, but I'm feeling generous>>

You aren't being generous TO ME.  I agree much would be pushed outward and miss BOTH.  Some of this could be above/below the eccliptic.  Stop building straw men...PRETENDING to be generous when you aren't.  Do you also make pretentious claims about your charitable gifts?

 

>>the Moon's orbit around the Earth totals a distance of over 2,500,000 km.  The moon shields 88,600 km of that. >>

 

Only half (the inner half) could have any chance to hit Earth.  If it misses Earth...coming from inside...it still could have a chance...maybe many...to  hit the Moon.  The out half could only hit the Moon.  You should admit that part.  DO YOU?

 

AND even if (as you seem to believe) all the stuff within Moon's orbit radius MUST be pulled into Earth (it wouldn't) if it took many Moon orbits (months) to do that, it means there would be many chances (as E/M orbit the Sun) to hit the Moon before it MIGHT eventually be captured by Earth.  Remember they are MOVING.

 

Did you agree that the increased gravity multiplier would ALSO apply to the Moon's gravity?  You seemed to once again IGNORE that.  Have you conceded I am right about that? 

 

>>the Moon's orbit around the Earth totals a distance of over 2,500,000 km.  The moon shields 88,600 km of that. >>

What you are missing (besides it is only the INNER part of M's orbit) is that the Moon IS MOVING.  So when an object is on a path taking it BEHIND (or far in FRONT OF) the Earth...then it still has a chance to hit the Moon and NO chance to hit Earth.  You want to only think of E/M as if they are standing still.

 

24 times.  NOT nailed down. 



#65 piasan

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Posted 27 April 2015 - 09:18 PM

I'll answer Dave's post #64 on the crater thread where it belongs.  That discussion is about objects that hit the Earth, this one is about objects that do NOT hit the Earth.



#66 indydave

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Posted 28 April 2015 - 06:04 PM

There is a LOT of stuff in THIS thread you need to address and I should NOT have to repeat it again.  I suppose the main question for now would be about the speeding up of the rock by stuff hitting its rear end...so that it would gain speed to be at or about the speed of water...meaning it would then capture it and gain a "sail."  Yes or no...have I shown you that would be true?  Then indeed there would be a RE/SW "push" outward. 

 

Then you should also admit that your mocking about (approx quote) "where did the 17km/s go?"...was wrong and instead of Brown being mocked you should be mocked about that yourself.  It is natural that when stuff would be moved outward, its lateral (orbital) speed diminishes.  It's something shown by one of Kepler's laws.  Please admit that.  Ask one of your experts if you need to. 



#67 piasan

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Posted 28 April 2015 - 10:38 PM

There is a LOT of stuff in THIS thread you need to address and I should NOT have to repeat it again. 

Yes, there is a lot of stuff I need to address in this thread, and the crater thread, and a couple others I'd like to participate in.  I have told you before, you clearly have far more time for this than I do.  Why do I constantly need to explain this to you?

 

Since I was last online, you have made 9 posts that, if I were to answer each of them in detail, would take at least 6-7 hours.... probably more like 9-12.  If I were to stay up and answer all of them, not only would I get no sleep, I'd be late for work.  In addition, I still haven't had a chance to listen to the two hours of Brown you sent me.  Finally, I have no intention of spending all of my available time dealing only with your posts .... though I likely will tonight.  Even then, it will be to respond to a few posts that will take the least time.  Right now, there simply isn't enough time for even one substantive post.

 

 

I suppose the main question for now would be about the speeding up of the rock by stuff hitting its rear end...so that it would gain speed to be at or about the speed of water...meaning it would then capture it and gain a "sail."  Yes or no...have I shown you that would be true?  Then indeed there would be a RE/SW "push" outward. 

Why don't you show how much this "push" outward would be?  My calculation showed it wouldn't work for rock.  IIRC, it was something over 1800 years for one meter.  Extrapolating from that do a density 25x less with a quick SWAG suggests it would still be 72 years (1800/25) per meter.  If so, even the "sail" wouldn't provide enough power.  If you don't like the NASA number for pressure, find your own research on what the number should be.



#68 piasan

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Posted 29 April 2015 - 08:27 AM

OK, I did some calculations for Indy to see if his "sail" will generate enough force to push his "cloud" to beyond the orbit of Mars.

 

He proposes an object with a radius of 5000 m and a density of 100 kg/m3.

 

A sphere with a radius of 5000 m will have a volume of 5.24e11 m3; a mass of 5.24e13 kg; and a geometric area of 7.85e7 m2.

 

NASA says the pressure exerted at 1AU is 4.53e-6 newtons / m2

 

Pressure = force / area.    Solving for force, we get  Force = pressure x area.

 

From Newton's second law,  force = mass x acceleration  substituting into the pressure equation, we get:

 

mass x acceleration = pressure x area   solving for acceleration, we get:

 

Acceleration = pressure x area / mass.  The calculation then becomes:

 

Acceleration = 4.35e-6 x 785e7 / 5.24e13 = 6.52e-12 m/sec2.  It would take1.53e11 seconds or 4866 years for the speed to reach one meter per second.or 2.23 mph.

 

I don't think that's going to work either.

 

(Note:  The reason this larger, less dense obect takes more time is that volume, which is a function of the radius cubed, increases much more rapidly than the area which is a function of the radius squared.  Therefore, the mass increases much faster than the area.)

 

Therefore, the argument about a 5km radius sphere with a density of 100 kg/m3 isn't going to work even if such an object could organize. 



#69 indydave

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Posted 29 April 2015 - 09:38 AM

Yes, there is a lot of stuff I need to address in this thread, and the crater thread, and a couple others I'd like to participate in.  I have told you before, you clearly have far more time for this than I do.  Why do I constantly need to explain this to you?

 

Since I was last online, you have made 9 posts that, if I were to answer each of them in detail, would take at least 6-7 hours.... probably more like 9-12.  If I were to stay up and answer all of them, not only would I get no sleep, I'd be late for work.  In addition, I still haven't had a chance to listen to the two hours of Brown you sent me.  Finally, I have no intention of spending all of my available time dealing only with your posts .... though I likely will tonight.  Even then, it will be to respond to a few posts that will take the least time.  Right now, there simply isn't enough time for even one substantive post.

 

 

Why don't you show how much this "push" outward would be?  My calculation showed it wouldn't work for rock.  IIRC, it was something over 1800 years for one meter.  Extrapolating from that do a density 25x less with a quick SWAG suggests it would still be 72 years (1800/25) per meter.  If so, even the "sail" wouldn't provide enough power.  If you don't like the NASA number for pressure, find your own research on what the number should be.

 

 

I don't know how to do the calc you did a few days ago.  That is why I said you could probably do it easily...since you know the process already.  If you assume there is a cloud that IS ATTACHED BY GRAVITY to the rock...then if the cloud is indeed big enough (maybe figure if it has 10 million x the cross-area of the rock)  then it WILL be pushed outward by SE (which you figured) AND by RE (which you ignored).  I don't know what pressure RE adds...don't think Brown figured it either. 

 

Ah...I just saw your calc. in #68.  It appears to me to be good.  I think I agree that IF it is a sphere then the mass of that sphere would be too great to be pushed to the a-belt by SW alone very fast.  There are several factors to consider which you didn't.  The solar wind varies...it has gone down 20% from the 1990's.  It would begin to push on stuff that is nearer the Sun so it would be harder than the rate you used, and could be harder if the solar wind were stronger at 2500 B.C.  Also as it pushes it would deform the cloud so that it would not be a sphere...so the area would increase...probably doubling it.  Maybe a good bit more than double.  Even WITH your estimate (avg. of .5 m/s over 4866 years) the SW alone would take stuff out to nearly 2 AU.  If the sphere of water vapor is squashed by the push to become double the area (or half the mass)...maybe more...then the SW alone would take it out to 4 AU or more.  But the MAIN thing you neglect is what (if any) effect RE would have.  I don't know.  Maybe (if you think this crashes his HPT) then Brown needs to do some calcs about it.  I think so far he hasn't.  I would guess he believes it to be a harder push than SW is...since he has RE more prominent in his book.



#70 indydave

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Posted 29 April 2015 - 10:59 AM

From NASA here

 

>>

"What we're seeing is a long term trend, a steady decrease in pressure that began sometime in the mid-1990s," explains Arik Posner, NASA's Ulysses Program Scientist in Washington DC.

How unusual is this event?

"It's hard to say. We've only been monitoring solar wind since the early years of the Space Ageā€”from the early 60s to the present," says Posner. "Over that period of time, it's unique. How the event stands out over centuries or millennia, however, is anybody's guess. We don't have data going back that far.">>

 

Pi, how about a simple answer to what I've asked several times that DOES NOT require a bunch of work from you.  Do you agree that if that rock is pounded on the rear end by faster water vapor or droplets of ice...then soon (or maybe after a long time...I think it would be SOON) it would be at the same speed (or close to it) as the water (meaning the water would NOT get stripped away)...and that you were wrong to say the rock would always be slower than the water?  That is a simple question for you.  Instead of taking the TWENTY minutes you took to write #68 as your ATTACK, how about you taking just TWO minutes to write your reply to MY attacking point and we can move on knowing that the rocks CAN capture a cloud for a sail.  I think you just are evading and hope it will go away...so you won't have to admit you were wrong. 



#71 indydave

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Posted 29 April 2015 - 12:45 PM

>>  It would begin to push on stuff that is nearer the Sun so it would be harder than the rate you used, and could be harder if the solar wind were stronger at 2500 B.C.>>

I believe this means if stuff (which was shot out from E retrograde) fell inward and began its trip outward (being pushed by RE/SW) at a position of .5 AU...that means the push force would be 4x as great.  I don't want to redo the calcs but I would guess this could mean (even though it must go an extra 93/2 million miles) that the much higher speed would get it to the a-belt much sooner. 

 

BTW, Brown suggests that the RE push is more lateral...adding speed to the solar orbit, and spiralling the object outward.  He says that this is because the rotation of each cloud is about the same as Earth's rotation, and the greatest heating (or radiation of heat) would happen when the object has turned about 90d from the Sun.  I am not so sure about this, because if asteroids today rotate near the Earth's rotation speed, then that means before they pulled in their cloud, the cloud had to be going a lot slower.

 

Maybe solar wind is pretty minor.  I mean, the cloud may be nearly totally collapsed in just a short time anyway.  I think I figured maybe 20 years.  So RE could be the bigger "player."



#72 indydave

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Posted 29 April 2015 - 05:58 PM

I wrote:>>I believe this means if stuff (which was shot out from E retrograde) fell inward and began its trip outward (being pushed by RE/SW) at a position of .5 AU...that means the push force would be 4x as great.>>

 

I found confirmation here where it said that at .25 AU it would be 16x.  I believe (if I read correctly) that some solar sail missions involved going INWARD toward the sun first to get more acceleration than what would be available at 1 AU. 

 

>>An exciting application of solar sailing is enabled by the greatly increased photon pressure at close solar distances. The sail can spiral down to as close as 0.25 AU with conventional coating and then pitch face on to effect a manoeuvre known as the solar photonic assist. The acceleration at 0.25 AU is 16 times higher than at 1 AU, enabling a considerable velocity change. The solar sail can then rapidly escape the solar system.>>

 

That same source said this: 

 

For a perfectly reflecting surface the observed pressure is twice the value provided by [2.9] due to the momentum transferred to the surface by the incident photons and the reaction provided by reflected photons. Using [2.9] the solar radiation pressure exerted on a solar sail at the Earth's distance from the Sun (1 AU) can be calculated. Since the orbit of the Earth about the Sun is slightly elliptical the energy flux received at the Earth varies by approximately 3.5% during the year. An accepted mean value of the solar constant WE is 1368 J s-1 m-2. Thus, the pressure exerted on a perfectly reflecting solar sail at 1 AU is taken to be
9.12 x 10
-6 N m-2.  

 

The water vapor "sail" would NOT be perfectly reflecting...but it might have SOME additional force than the amount Pi reported...so he probably underestimated the acceleration of SW alone when he reported NASA's number of 4.53e-6 newtons / m2

 

I was also pretty surprised to learn that they can adjust the sail so it can RETURN and go TOWARD the Sun!



#73 indydave

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Posted 01 May 2015 - 11:17 AM

>>Wrong.  At 2 mph, it will be captured by the gravity of the Earth, not the moon..... as we shall soon see.>>

I do not agree with you here, but Pi, would you agree this is NOT true for retrograde stuff going 60km/s?  As it is moving out at 2mph, it still has a closing speed of 60km/s as it comes into the influence of the Earth.  Just a slight tug from Earth would make it be slung outward to an orbit beyond 1 AU.  It would be going so fast that it could not be CAPTURED by Earth at all, and it would have a hyperbolic shape (of its Earth-orbit) and immediately exit the E/M area.  And each time that happened it would get 2 chances to hit the Moon.  That would NOT occur for ANY possible Moon hits.  The moon could not tug on the stuff much at all to change the object's solar orbit, since it is moving and has so much less mass.  This would happen to the vast majority of the retro stuff.  This means that until it moves outward 250,000 mi from the point where it could first hit the Moon, it has ZERO chance to hit E.  It would be moving too fast to be pulled into the Earth (unless lined up perfectly...very rare).  Then it has 8000 mi (diameter of E) where (if it is not 8000 mi above or below the ecliptic) it could NOT be slung hyperbolically around Earth...and then WOULD hit Earth.  And then as it moves outward past 1 AU it has another 250,000 mi (as it keeps moving outward at 2mph) to again possibly hit the Moon, again twice per pass.  The Moon is moving and is smaller...but the swath is wider (500,000 mi or more) AND there are two chances with each "pass" to hit the Moon, which happens each 6 months as E and the object travel in opposite directions around the Sun.  AND as I've said...this only applies for stuff launched INSIDE 1 AU.  For stuff just beyond 1AU, that can ONLY hit the Moon (unless it is perturbed...a rare event). 

 

SO...picture retro stuff moving outward at 2mph.  In six months that stuff would travel 2 x 24 x 180 = 8640 miles.  During that time it has NO chance to hit Earth and it has TWO chances to hit the Moon.  In another 6 months it still has NO chance to hit the Earth, and it has TWO MORE to hit the Moon.  This happens 29 TIMES...over 14.5 years.  As the object gets closer to Earth, there are VERY good odds that the slow movement outward gets perturbed by the gravity of Earth, so that it gets slung out to beyond 1 AU...perhaps even beyond the SS.  There would be only a SLIGHT chance that it could continue UNPERTURBED on its slow march outward at 2mph to then have a good chance to hit Earth on its 30th pass, in the 15th year.  Almost none would hit Earth.  If the stuff is not slung away, there would then be another 29.5 years where it gets 2 chances each 6 months to hit the Moon. 

 

Ok, this seems to fit better with the Craters/Earth topic so I will cross-post it there and reply to replies there.  Also, I believe I will be asking numerous time for Pi to respond to this so for short I will refer to it as the retrograde-sling argument (RGS) for short.  IMO it will be the killer argument on this topic.



#74 indydave

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Posted 01 May 2015 - 07:11 PM

Please check out the drawings I made of the RGS...on the Cratering/Earth topic, pg 17.



#75 piasan

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Posted 03 May 2015 - 11:13 AM

Ok, this seems to fit better with the Craters/Earth topic so I will cross-post it there and reply to replies there.  Also, I believe I will be asking numerous time for Pi to respond to this so for short I will refer to it as the retrograde-sling argument (RGS) for short.  IMO it will be the killer argument on this topic.

It has been answered on the Crater topic. I have allowed Dave ALL of the material that would be launched retrograde such that it would be along the path of the Earth in it's orbit and within the orbital distance of the moon.  That amounted to   With relation to THIS topic, anything that impacts the Earth will no longer be a comet or asteroid and is "out of the game."  In other words, it is irrelevant.

 

BTW, Dave's "killer argument" deals with only about 0.002% of the launched material.  Here is the comment that establishes that:

Let's also put Dave's 500,000 mile wide "zone" into perspective.  The Earth's orbit around the Sun is 58.4 million miles.  In six months, the Earth will travel about 29.2 million miles.  Simply keeping this in two dimensions, we can find the "angle" of the "target.... including the entire distance... 500,000 miles.  The angle is arctan (500,000 / 29,200,000) or about 1o.  The moon occupies 0.5o and that is 0.0005% of the celestial sphere.  Without getting into the trig, the radius of this "target" is twice that of the moon, so it has 4x the area.  That would mean it occupies 0.002% of the celestial sphere.  Frankly, I see no point in arguing over the 0.002% any longer.  You can have the whole 0.002% Dave.... that reduces the minimum multiplier to 13.47.

 

Dave has a long history with me of straining at gnats (0.002%) while ignoring the elephants (99.998%).

 

The outward force imparted by the radiometer effect has a lot more bearing on this discussion than objects that remove themselves from consideration by impacting Earth.



#76 indydave

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Posted 03 May 2015 - 11:52 AM

>>Let's also put Dave's 500,000 mile wide "zone" into perspective.  The Earth's orbit around the Sun is 58.4 million miles.  In six months, the Earth will travel about 29.2 million miles. >>

 

You gotta rework your numbers...although they are totally irrelevant.  "R" is 93 million.  2pi (6.23) x 93 is 584 million. 

 

>>That would mean it occupies 0.002% of the celestial sphere.  Frankly, I see no point in arguing over the 0.002% any longer.  You can have the whole 0.002% Dave.... that reduces the minimum multiplier to 13.47.>>

I think you have another decimal to add, and I suppose you'd think it makes your case stronger, but it is totally irrelevant.  The object is moving about the Sun so it sweeps the full 360o of the solar orbit, moving out by 17,200 mi each year.  So it eventually intersects everything at any location on the orbit.  The Retrograde Sling argument has not been TOUCHED yet by you. 

 

>>BTW, Dave's "killer argument" deals with only about 0.002% of the launched material.>>

It doesn't matter what % is that misses Earth.  100% of the stuff above/below the ecliptic would miss.  Stuff sent out ahead of E (pointed in the direction of the solar orbit) would have no chance to hit E and minimal chance to hit the Moon.  Stuff sent out to the side (at right angles to the solar orbit) also has little chance to hit E/M.  It is mainly the retrograde stuff that has some potential to hit either M or E.  The RGS argument shows it cannot hit Earth.  This is the stuff that made the big craters on the Moon. 

 

All the stuff that misses both E and M can be pushed outward to form the a-belt and TNO's. 

 

AND BTW (acc. to your own persnickety-ness) this belongs on Crater/Earth...since it is about impacting.



#77 piasan

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Posted 03 May 2015 - 12:34 PM

>>Let's also put Dave's 500,000 mile wide "zone" into perspective.  The Earth's orbit around the Sun is 58.4 million miles.  In six months, the Earth will travel about 29.2 million miles. >>

 

You gotta rework your numbers...although they are totally irrelevant.  "R" is 93 million.  2pi (6.23) x 93 is 584 million. 

 

>>That would mean it occupies 0.002% of the celestial sphere.  Frankly, I see no point in arguing over the 0.002% any longer.  You can have the whole 0.002% Dave.... that reduces the minimum multiplier to 13.47.>>

I think you have another decimal to add, and I suppose you'd think it makes your case stronger, but it is totally irrelevant.  The object is moving about the Sun so it sweeps the full 360o of the solar orbit, moving out by 17,200 mi each year.  So it eventually intersects everything at any location on the orbit.  The Retrograde Sling argument has not been TOUCHED yet by you. 

 

.....

AND BTW (acc. to your own persnickety-ness) this belongs on Crater/Earth...since it is about impacting.

Thank you for the correction, it does change the numbers

 

I shouldn't need to point out my comments were in RESPONSE to the off-topic remarks you made... .in other words, YOU, not I initiated them.



#78 indydave

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Posted 03 May 2015 - 11:08 PM

I meant to type 2pi is 6.28.  Sorry.  My multiplying was right, but not the typing. 







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