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Thread: Meteorite that doesn't burn?

  1. #1 Meteorite that doesn't burn? 
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    Is there a conceivable speed and angle for a rock to enter the atmosphere without getting very hot? In other words, it should come in being almost geo-stationary and with little enough vertical speed to fall down to earth in the same manner as Baumgartner.


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    Universal Mind John Galt's Avatar
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    Unlikely. However meteorites generally do not get very hot, except on the outside. There transit through the atmosphere is over quickly.


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    Quote Originally Posted by John Galt View Post
    Unlikely. However meteorites generally do not get very hot, except on the outside. There transit through the atmosphere is over quickly.
    Thanks, but let me slightly re-phrase my question. Is there a conceivable angle and speed for a human astronaut to enter the atmosphere and parachute down safely without being in a re-entry vehicle? I know it would have been impossible for any human astronaut up to now to have obtained that speed and angle, but, just theoretically, is it conceivable?
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  5. #4  
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    Baumgartner travelled through only 39 km.

    Atmospheric effects on the re-entry of spacecraft become noticeable at 120 km above the earth's surface according to wiki.

    Needs a bit more thought.
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    Quote Originally Posted by adelady View Post
    Baumgartner travelled through only 39 km.

    Atmospheric effects on the re-entry of spacecraft become noticeable at 120 km above the earth's surface according to wiki.

    Needs a bit more thought.
    But that thought is exactly what I'm trying to do here ! The reason that spacecraft notice the atmosphere so high up is that they are moving at thousands of kilometers an hour relative to it. If an astronaut somehow managed to be moving in such a way that he is geo-stationary 120km up, there would be no severe friction and he would be able to fall all the way down to the surface without burning up. I'm therefore wondering if it is possible for an object to be almost in that position.
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    For instance, if a spacecraft doesn't only do a decelerating burn that slows it enough to attain the correct angle to enter the atmosphere as intended, but did a far longer burn, would it be possible to slow down to such an extent that it would fall "straight down" through the atmosphere without heating up significantly?
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    Geocentric orbit - Wikipedia, the free encyclopedia
    (LEO) - Geocentric orbits ranging in altitude from 160 kilometeres (100 statue miles) to 2,000 kilometres (1,200 mi) above mean sea level. At 160 km, one revolution takes approximately 90 minutes, and the circular orbital speed is 8,000 metres per second (26,000 ft/s).
    Baumgartner's speed was only 833.9mph (1,342km/h) which is about 373 meters per second.
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    Quote Originally Posted by Harold14370 View Post

    Baumgartner's speed was only 833.9mph (1,342km/h) which is about 373 meters per second.
    The speed you refer to for Baumgartner was his vertical speed, which is not what I'm interested in. I'm talking about horizontal speed. Baumgartner's horizontal speed was 0 km/h, which is the same for a geostationary satellite. The other speed of 26 000 ft/s you mention is a horizontal speed, but not for a geostationary object.

    The reason why horizontal speed is most important here is that that is the number that mainly influences the friction and heat, not vertical speeds (except if the vertical speed is super fast and not only gravity driven, as I'm imagining here).
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    i think we have just about measure of speed, or is that velocity, in this thread. ft/s, m/s, mph, km/h, amazing.

    Sometimes it is better not knowing than having an answer that may be wrong.
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    I think 0-angle entry is not a 'no-burning' angle of entry. At such angle any object drop from 200km in space will come down at 1000meter/second, which will definitely burn even for just 1 minute.

    But if the entry angle is greater than 0, then such object will be traveling faster than the '0-angle-object' (above) but it could use the upper-thinner-atmosphere to skid and 'fly', and thus can gradually reduce speed to the terminal velocity (by skidding alot) before going down to the denser atmosphere.
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    Quote Originally Posted by msafwan View Post
    I think 0-angle entry is not a 'no-burning' angle of entry. At such angle any object drop from 200km in space will come down at 1000meter/second, which will definitely burn even for just 1 minute.

    But if the entry angle is greater than 0, then such object will be traveling faster than the '0-angle-object' (above) but it could use the upper-thinner-atmosphere to skid and 'fly', and thus can gradually reduce speed to the terminal velocity (by skidding alot) before going down to the denser atmosphere.
    Now this is getting closer to an answer to my question! What you're saying makes sense, but is there evidence about it? For instance, are you sure that a 0-angle will result in the object burning? Wouldn't it also just slow down gradually enough as the atmosphere gets thicker to avoid burning? Also, if what you're saying is correct (I'm pretty sure it is), is it therefore possible to enter a spacecraft into the atmosphere like that without a heat shield being used (even if it takes an un-economical amount of fuel to slow it down that much) or for a meteorite to come down like that?
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    Quote Originally Posted by Gerdagewig View Post
    Now this is getting closer to an answer to my question! What you're saying makes sense, but is there evidence about it? For instance, are you sure that a 0-angle will result in the object burning?
    Ok, I was wrong. 0-angle reentry won't burn your stuff. I got this from Mercury1 report (here: http://ntrs.nasa.gov/archive/nasa/ca...1964052551.pdf linked by Wikipedia), the height of Freedom-7 is 187km, horizontal speed about 400m/s (insignificant), and encounter atmosphere at about 90km above ground.

    So 0-angle drop & not burning is absolutely 100% possible.

    Quote Originally Posted by Gerdagewig View Post
    Wouldn't it also just slow down gradually enough as the atmosphere gets thicker to avoid burning? Also, if what you're saying is correct (I'm pretty sure it is), is it therefore possible to enter a spacecraft into the atmosphere like that without a heat shield being used (even if it takes an un-economical amount of fuel to slow it down that much) or for a meteorite to come down like that?
    You don't need much fuel anymore. You can use inflatable heat shield. This shield is soo big: it cause aerodynamic lift during reentry, thus causing sloooower (and thus: less heat) reentry. (ie: A Look at What's Ahead: HIAD - YouTube )
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    Quote Originally Posted by Gerdagewig View Post
    Quote Originally Posted by Harold14370 View Post

    Baumgartner's speed was only 833.9mph (1,342km/h) which is about 373 meters per second.
    The speed you refer to for Baumgartner was his vertical speed, which is not what I'm interested in. I'm talking about horizontal speed. Baumgartner's horizontal speed was 0 km/h, which is the same for a geostationary satellite. The other speed of 26 000 ft/s you mention is a horizontal speed, but not for a geostationary object.

    The reason why horizontal speed is most important here is that that is the number that mainly influences the friction and heat, not vertical speeds (except if the vertical speed is super fast and not only gravity driven, as I'm imagining here).
    An astronaut in geosynchronous orbit does not have a horizontal speed of zero with respect to the surface of the earth. He is actually going 3.1 kilometers per second, whereas a point on the surface has a speed of 0.46 kilometers per second. If he steps out of his space ship wearing a parachute, he does not fall to the earth, but simply orbits right along next to the space ship.
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    Quote Originally Posted by msafwan View Post
    So 0-angle drop & not burning is absolutely 100% possible.

    You don't need much fuel anymore. You can use inflatable heat shield. This shield is soo big: it cause aerodynamic lift during reentry, thus causing sloooower (and thus: less heat) reentry. (ie: A Look at What's Ahead: HIAD - YouTube )
    Thanks a lot for this Msafwan! Great info.
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    Quote Originally Posted by Harold14370 View Post

    An astronaut in geosynchronous orbit does not have a horizontal speed of zero with respect to the surface of the earth. He is actually going 3.1 kilometers per second, whereas a point on the surface has a speed of 0.46 kilometers per second. If he steps out of his space ship wearing a parachute, he does not fall to the earth, but simply orbits right along next to the space ship.
    Harold, I think we're talking past each other here slightly. Your point that an astronaut stepping out of a craft in geosynchronous orbit doesn't fall to earth but keeps orbiting is obviously true, but again misses the hypothetical scenario I'm sketching. First, I used the term geostationary, which is a specific type of geosynchronous orbit that does in fact have 0 horizontal speed as measured against the earths surface. Second, my hypothetical scenario would be that our intrepid explorer isn't in a real geostationary orbit way out at something like 36 000 km above the equator, but rather in a decaying orbit with a nearly geostationary position against the earth as measured horizontally, but with a probably somewhat significant vertical speed down towards the earth. The info supplied my msafwan above seems to suggest that it is conceivable that my astronaut might somehow find herself in such a position and would be able to parachute down safely to the adoration of myself and the rest of humanity!
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    I suppose for a random rock to be able to enter like that, the biggest problem is, how does it get there in the first place?

    Average orbital speed of the earth ->
    Equatorial rotation velocity of the earth ->

    I would think the closest something would be able to come, would be to be in about the same orbit as earth around the sun, but slower or faster by around the amount of the Equatorial rotation velocity of the earth. Then either by catching up to or being caught up to by the earth and approaching form the appropriate side, the only proper motion of the rock relative to the earth surface would be due to the axial tilt, if I am not mistaken.

    Can't think of anything else.
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    Quote Originally Posted by KALSTER View Post
    I suppose for a random rock to be able to enter like that, the biggest problem is, how does it get there in the first place?

    Average orbital speed of the earth ->
    Equatorial rotation velocity of the earth ->

    I would think the closest something would be able to come, would be to be in about the same orbit as earth around the sun, but slower or faster by around the amount of the Equatorial rotation velocity of the earth. Then either by catching up to or being caught up to by the earth and approaching form the appropriate side, the only proper motion of the rock relative to the earth surface would be due to the axial tilt, if I am not mistaken.

    Can't think of anything else.
    Kalster, you came up with a good proposition for how something could end up in a position to fall down to earth in this way. Your suggestion sounds plausible to me, but then it must be remembered that I don't know much about orbital physics. Also, if you're right, any such object would have fallen to earth a very long time ago and there would certainly not be any of them left today. However, I guess it might also be possible (at an extremely low probability) that an extra-solar object could come close to the sun and be caught in an orbit that roughly matches the earths. It would be more probable though that it gets caught in a much more elliptical orbit like comets. In a comet-like orbit I guess it could again be possible at an extremely low probability that the object might almost match the earth's orbit for a little while as it makes its closest pass around the sun and if the earth happens to be in that exact spot at that time it could just fall down as discussed above. But then again I don't think it would be possible that it could be in that position of it's orbit and move as "slowly" as the earth?

    Another possible scenario that I hinted to above would be that a human is deliberately put into that position. I guess it could either be from low earth with an extended deceleration burn that slows it down to almost geostationary but decaying orbit or with the HIAD that Msafwan mentioned.
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