Shooting star = grain of sand?

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In summary: I don't remember what the quoted part said specifically, but it wasn't that the object itself is called a meteor, it was that the trail is called a meteoroid. Not sure if that helpsIt says that a typical (so, a small one) meteoroid (so, a rock) will produce a metor (so, a trail) of that size. It also states that brightness and length of a meteor depends on the size of the meteoroid.So, it seems that a 1 microgram meteoroid would produce a 1 m by 20 km trail.
  • #71
NASA has a super rifle they use for such things, it can fire projectiles over 17,000 mph [around 8 km/sec]. They use it, among other things, to test shielding of spacecraft against meteoroid impacts.
 
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  • #73
BenAS said:
According to this website
http://www.amsmeteors.org/meteor-showers/meteor-faq/ most meteoroids are between the size of a grain if sand and a small pebble and weight less than 1-2 grams. The light we see is caused by the KE ionizing atmospheric molecules.
I'm late to the thread, but I was going to say this. The only explanation for why something seemingly incapable of being visible due to its low energy output (that I can think of) is ionization of surrounding particles. This would not only create a larger area of light, but a wider spectrum as well.
 
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  • #74
So far there's a lot of arm waving but no actual math. So here goes. Imagine a speck of rock, density 3000 kg/m3. It's 1mm across, with a volume 10-9 m3 and mass 3 x 10-6 kg. If it hits at Earth's orbital velocity, 30 km/sec (30,000 m/sec) then its kinetic energy is 1/2 mv2 = 1/2 x 3x10-6 x (3x10^4)^2 = 4.5 x 10^2 joules. Now, if it takes a second to flame out (typical for the small meteors I've seen), then it's radiating 450 joules/sec = 450 watts. Crudely, auto headlights are around 50 watts and can be seen many miles away at night. So our meteor is about ten times that.

How do we convert our meteor to actual brightness? Incandescent lights are very inefficient. But then again, our meteor is also emitting due to incandescence, although more efficiently since it's a lot hotter. The Sun delivers 1361 watts per meter squared on earth. The apparent magnitude of the Sun is -26.7. Sirius is just about 25 magnitudes fainter or 5 steps of 5 magnitudes or one ten billionth as bright, so its energy flux on Earth is 1.36 x 10^-7 W/m2. Now, if our meteor is 100 km away (slant range) it will emit 450W/4pi x 100,000 x 100,000)m2 = 3.6 x 10-9 W/m2 at the observer's location. It's about 1/38 as bright as Sirius, or about 4 magnitudes, about M = 2.5. That's roughly as bright as a star in the Big Dipper.
 
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  • #75
Steve Dutch said:
So far there's a lot of arm waving but no actual math.
Did you miss page 3 and 4? We have those calculations already, with the same approach but with more accurate resuts: Comparing the shooting star power to a real star gives an error of a factor ~40 in brightness.
 
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  • #76
Steve Dutch said:
So far there's a lot of arm waving but no actual math. So here goes. Imagine a speck of rock, density 3000 kg/m3. It's 1mm across, with a volume 10-9 m3 and mass 3 x 10-6 kg. If it hits at Earth's orbital velocity, 30 km/sec (30,000 m/sec) then its kinetic energy is 1/2 mv2 = 1/2 x 3x10-6 x (3x10^4)^2 = 4.5 x 10^2 joules. Now, if it takes a second to flame out (typical for the small meteors I've seen), then it's radiating 450 joules/sec = 450 watts. Crudely, auto headlights are around 50 watts and can be seen many miles away at night. So our meteor is about ten times that.

How do we convert our meteor to actual brightness? Incandescent lights are very inefficient. But then again, our meteor is also emitting due to incandescence, although more efficiently since it's a lot hotter. The Sun delivers 1361 watts per meter squared on earth. The apparent magnitude of the Sun is -26.7. Sirius is just about 25 magnitudes fainter or 5 steps of 5 magnitudes or one ten billionth as bright, so its energy flux on Earth is 1.36 x 10^-7 W/m2. Now, if our meteor is 100 km away (slant range) it will emit 450W/4pi x 100,000 x 100,000)m2 = 3.6 x 10-9 W/m2 at the observer's location. It's about 1/38 as bright as Sirius, or about 4 magnitudes, about M = 2.5. That's roughly as bright as a star in the Big Dipper.

Yes we all have done the math. As you applied 3 milligrams of space stuff hitting us perpendicular and at Earth's orbital velocity. Applying our dear KE formula we all love so dearly. Most of us just do it in our heads with nice round numbers as you used, nice indeed.
The arm waving helps us lose some weight so we remain attractive and tan with about 1.4 kw as a tanning agent as you indicated. :cool:
mfb said:
Did you miss page 3 and 4? We have those calculations already, with the same approach but with more accurate resuts: Comparing the shooting star power to a real star gives an error of a factor ~40 in brightness.
With your calc. being one of the complete ones, unlike mine mfb. :redface:
 
  • #77
Gary Weller said:
I'm late to the thread, but I was going to say this. The only explanation for why something seemingly incapable of being visible due to its low energy output (that I can think of) is ionization of surrounding particles. This would not only create a larger area of light, but a wider spectrum as well.

Good point Gary. The wider perceived bandwidth or visible spectrum due to more or added elements ionizing in the surrounding Earth's atmosphere than the original rock/ice chemistry. You just can't beat 'fly by the wire gas spectrometry' (if there is such a thing).
 
  • #78
This is all basic physics, unless I missed something relevant. The KE, and physics, of meteoroids entering the atmosphere is understood well enough to allow to return astronauts to earth. Next question.
 
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  • #79
mfb said:
If the spectrum has more infrared or more UV it needs more energy, but we are still talking about hundreds of Joules.
Gary Weller said:
I'm late to the thread, but I was going to say this. The only explanation for why something seemingly incapable of being visible due to its low energy output (that I can think of) is ionization of surrounding particles. This would not only create a larger area of light, but a wider spectrum as well.
One of the most fascinating talks I had the pleasure of attending in grad school was given by Dudley Herschbach on this very topic. He went through the basic kinematics and worked out the blackbody spectrum for an average meteor and concluded that, if that's all there is, then we shouldn't be able to see meteors. Then he delivered my all-time favorite line: "If it isn't physics, it must be chemistry!" and went on to explain that a good chunk of the light that we see comes from the sodium D line (smack dab in the middle of the visible spectrum). The talk then meandered its way through the existence of sodium and iron layers in the upper atmosphere (http://www.albany.edu/faculty/rgk/atm101/sodium.htm). Research in atmospheric chemistry has been focused for so long on things like ozone/CFCs, carbon dioxide, and acid rain, that his talk was a nice break from the doom and gloom often associated with the field.
 
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  • #80
TeethWhitener,
thanks for adding that fresh info to the conversation from the chemist's perspective.
Would love to have been a fly on the wall to hear Dr.Herschbach's presentation. :wink:
 
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  • #81
megacal said:
TeethWhitener,
thanks for adding that fresh info to the conversation from the chemist's perspective.
Would love to have been a fly on the wall to hear Dr.Herschbach's presentation. :wink:

Me too megacal, me too!
TeethWhitener, See the later portion of post #77 and #73 ... There was some spectro suggested there. :blushing:
 

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