APOD: Meteor over the Bay of Naples (2024 Feb 17)

[APOD Robot]

Meteor over the Bay of Naples

Explanation: A cosmic dust grain plowing through the upper atmosphere much faster than a falling leaf created this brilliant meteor streak. In a serendipitous moment, the sublime night sky view was captured from the resort island of Capri, in the Bay of Naples, on the evening of February 8. Looking across the bay, the camera faces northeast toward the lights of Naples and surrounding cities. Pointing toward the horizon, the meteor streak by chance ends above the silhouette of Mount Vesuvius. One of planet Earth's most famous volcanos, an eruption of Mount Vesuvius destroyed the city of Pompeii in 79 AD.

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[epitalon]

I apreciated the "falling leaf" story

[GoatGuy]

Lovely shot.

I was wondering how big such a meteor could be. Various sources say things like 'less than a centimeter' and 'millimeters or so'. I guess so. Then I decided to use the other "can be brighter than Venus" to work some Excel magic, to figure out how big a bit-o-dust it might have been.

Using drivers of

Venus insolation = 2700 W/m²
… diameter 12,000 km
Bright arc of 30%
Bright distance about 75% of 1 AU from Earth

Determined that Venus must bathe Earth with about 1.3×10⁻⁶ W/m² at its brightest. Then went on to ask mathematically "how much output would a meteor at 45 km altitude have to produce in order to also have that 1.3×10⁻⁶ W/m² where the observer is standing?"

That turned out to be 34,000 W of optical power.

From there and E = ½mv², estimating the velocity at 30 km/s, then the little morsel need only weigh in at 0.07 grams. And, while that's not very much, if the meteor is 'stony' say, with a 6 g/cm³ density, then that's about 0.012 cm³, 0.28 cm diameter, about 3 millimeters.

3 mm is yep … still a grain of sand … but not fine sand. Big coarse sand.

Working that 100% backwards, all the numbers confirmed out pretty well. So … 3 to 10 mm chunk of star stuff! Yay!

⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅

[Chris Peterson]

Lovely shot.

I was wondering how big such a meteor could be. Various sources say things like 'less than a centimeter' and 'millimeters or so'. I guess so. Then I decided to use the other "can be brighter than Venus" to work some Excel magic, to figure out how big a bit-o-dust it might have been.

Using drivers of

Venus insolation = 2700 W/m²
… diameter 12,000 km
Bright arc of 30%
Bright distance about 75% of 1 AU from Earth

Determined that Venus must bathe Earth with about 1.3×10⁻⁶ W/m² at its brightest. Then went on to ask mathematically "how much output would a meteor at 45 km altitude have to produce in order to also have that 1.3×10⁻⁶ W/m² where the observer is standing?"

That turned out to be 34,000 W of optical power.

From there and E = ½mv², estimating the velocity at 30 km/s, then the little morsel need only weigh in at 0.07 grams. And, while that's not very much, if the meteor is 'stony' say, with a 6 g/cm³ density, then that's about 0.012 cm³, 0.28 cm diameter, about 3 millimeters.

3 mm is yep … still a grain of sand … but not fine sand. Big coarse sand.

Working that 100% backwards, all the numbers confirmed out pretty well. So … 3 to 10 mm chunk of star stuff! Yay!

⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅

What's tricky, and generally makes such calculations of limited value, is that the references are parked over the same cluster of pixels, while the meteor is moving. What that means is that the references are exposed for much longer, making their apparent brightness greater than the meteor, which is only dwelling on a pixel for a short time. Consider that an exposure like this is typically 30 seconds long, while the meteor duration is on the order of a second, and only on any one pixel for a fraction of that. So the actual meteor could easily be 100 times brighter than the image would suggest.

[Ann]

I note that the meteor is mostly cyan-colored.

Does anyone recognize the star field beyond the meteor? I note a bright blue star at upper right. Anyone know what star that is?

Ann

[Chris Peterson]

I note that the meteor is mostly cyan-colored.

Does anyone recognize the star field beyond the meteor? I note a bright blue star at upper right. Anyone know what star that is?

Ann

Alkaid... end of the Big Dipper handle.

[johnnydeep]

Lovely shot.

I was wondering how big such a meteor could be. Various sources say things like 'less than a centimeter' and 'millimeters or so'. I guess so. Then I decided to use the other "can be brighter than Venus" to work some Excel magic, to figure out how big a bit-o-dust it might have been.

Using drivers of

Venus insolation = 2700 W/m²
… diameter 12,000 km
Bright arc of 30%
Bright distance about 75% of 1 AU from Earth

Determined that Venus must bathe Earth with about 1.3×10⁻⁶ W/m² at its brightest. Then went on to ask mathematically "how much output would a meteor at 45 km altitude have to produce in order to also have that 1.3×10⁻⁶ W/m² where the observer is standing?"

That turned out to be 34,000 W of optical power.

From there and E = ½mv², estimating the velocity at 30 km/s, then the little morsel need only weigh in at 0.07 grams. And, while that's not very much, if the meteor is 'stony' say, with a 6 g/cm³ density, then that's about 0.012 cm³, 0.28 cm diameter, about 3 millimeters.

3 mm is yep … still a grain of sand … but not fine sand. Big coarse sand.

Working that 100% backwards, all the numbers confirmed out pretty well. So … 3 to 10 mm chunk of star stuff! Yay!

⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅

What's tricky, and generally makes such calculations of limited value, is that the references are parked over the same cluster of pixels, while the meteor is moving. What that means is that the references are exposed for much longer, making their apparent brightness greater than the meteor, which is only dwelling on a pixel for a short time. Consider that an exposure like this is typically 30 seconds long, while the meteor duration is on the order of a second, and only on any one pixel for a fraction of that. So the actual meteor could easily be 100 times brighter than the image would suggest.

What "references"? The calculations should be correct independent of brightness measurements by CCDs , right?

[Chris Peterson]

Lovely shot.

I was wondering how big such a meteor could be. Various sources say things like 'less than a centimeter' and 'millimeters or so'. I guess so. Then I decided to use the other "can be brighter than Venus" to work some Excel magic, to figure out how big a bit-o-dust it might have been.

Using drivers of

Venus insolation = 2700 W/m²
… diameter 12,000 km
Bright arc of 30%
Bright distance about 75% of 1 AU from Earth

Determined that Venus must bathe Earth with about 1.3×10⁻⁶ W/m² at its brightest. Then went on to ask mathematically "how much output would a meteor at 45 km altitude have to produce in order to also have that 1.3×10⁻⁶ W/m² where the observer is standing?"

That turned out to be 34,000 W of optical power.

From there and E = ½mv², estimating the velocity at 30 km/s, then the little morsel need only weigh in at 0.07 grams. And, while that's not very much, if the meteor is 'stony' say, with a 6 g/cm³ density, then that's about 0.012 cm³, 0.28 cm diameter, about 3 millimeters.

3 mm is yep … still a grain of sand … but not fine sand. Big coarse sand.

Working that 100% backwards, all the numbers confirmed out pretty well. So … 3 to 10 mm chunk of star stuff! Yay!

⋅-⋅-⋅ Just saying, ⋅-⋅-⋅
⋅-=≡ GoatGuy ✓ ≡=-⋅

What's tricky, and generally makes such calculations of limited value, is that the references are parked over the same cluster of pixels, while the meteor is moving. What that means is that the references are exposed for much longer, making their apparent brightness greater than the meteor, which is only dwelling on a pixel for a short time. Consider that an exposure like this is typically 30 seconds long, while the meteor duration is on the order of a second, and only on any one pixel for a fraction of that. So the actual meteor could easily be 100 times brighter than the image would suggest.

What "references"? The calculations should be correct independent of brightness measurements by CCDs , right?

The stars or planets with known magnitudes that are being used for reference. In the above calculation, Venus is the reference. If Venus is exposed for 30 seconds, and the meteor for 0.1 second, we have a problem.

[Ann]

I note that the meteor is mostly cyan-colored.

Does anyone recognize the star field beyond the meteor? I note a bright blue star at upper right. Anyone know what star that is?

Ann

Alkaid... end of the Big Dipper handle.

Thanks, Chris!

Ann

[rwlott]

I apreciated the "falling leaf" story

I, too, appreciate this O Henry classic short story. Kudos to the APOD editors for including a link to it. By the way, I’m convinced that there are many persons whose lives are saved, or at least the quality of their lives are significantly improved, by caring acts of friends and neighbors.

Russ

[johnnydeep]

What's tricky, and generally makes such calculations of limited value, is that the references are parked over the same cluster of pixels, while the meteor is moving. What that means is that the references are exposed for much longer, making their apparent brightness greater than the meteor, which is only dwelling on a pixel for a short time. Consider that an exposure like this is typically 30 seconds long, while the meteor duration is on the order of a second, and only on any one pixel for a fraction of that. So the actual meteor could easily be 100 times brighter than the image would suggest.

What "references"? The calculations should be correct independent of brightness measurements by CCDs , right?

The stars or planets with known magnitudes that are being used for reference. In the above calculation, Venus is the reference. If Venus is exposed for 30 seconds, and the meteor for 0.1 second, we have a problem.

Hmm. But the OP's statement:

Determined that Venus must bathe Earth with about 1.3×10⁻⁶ W/m² at its brightest. Then went on to ask mathematically "how much output would a meteor at 45 km altitude have to produce in order to also have that 1.3×10⁻⁶ W/m² where the observer is standing?"

Is using the brightness of Venus as a seemingly observer-independent 1.3×10⁻⁶ W/m², and calculating the attributes of the meteor that would match it. A watt has time factored in already. I'm still not seeing where the problem lies.

[Chris Peterson]

What "references"? The calculations should be correct independent of brightness measurements by CCDs , right?

The stars or planets with known magnitudes that are being used for reference. In the above calculation, Venus is the reference. If Venus is exposed for 30 seconds, and the meteor for 0.1 second, we have a problem.

Hmm. But the OP's statement:

Determined that Venus must bathe Earth with about 1.3×10⁻⁶ W/m² at its brightest. Then went on to ask mathematically "how much output would a meteor at 45 km altitude have to produce in order to also have that 1.3×10⁻⁶ W/m² where the observer is standing?"

Is using the brightness of Venus as a seemingly observer-independent 1.3×10⁻⁶ W/m², and calculating the attributes of the meteor that would match it. A watt has time factored in already. I'm still not seeing where the problem lies.

How does his calculation connect to the image?

[johnnydeep]

The stars or planets with known magnitudes that are being used for reference. In the above calculation, Venus is the reference. If Venus is exposed for 30 seconds, and the meteor for 0.1 second, we have a problem.

Hmm. But the OP's statement:

Determined that Venus must bathe Earth with about 1.3×10⁻⁶ W/m² at its brightest. Then went on to ask mathematically "how much output would a meteor at 45 km altitude have to produce in order to also have that 1.3×10⁻⁶ W/m² where the observer is standing?"

Is using the brightness of Venus as a seemingly observer-independent 1.3×10⁻⁶ W/m², and calculating the attributes of the meteor that would match it. A watt has time factored in already. I'm still not seeing where the problem lies.

How does his calculation connect to the image?

Well, are you saying that brightness comparisons using even the naked eye are highly affected by the duration of the event? I can see that might matter. Maybe a lot.

[Chris Peterson]

Hmm. But the OP's statement:



Is using the brightness of Venus as a seemingly observer-independent 1.3×10⁻⁶ W/m², and calculating the attributes of the meteor that would match it. A watt has time factored in already. I'm still not seeing where the problem lies.

How does his calculation connect to the image?

Well, are you saying that brightness comparisons using even the naked eye are highly affected by the duration of the event? I can see that might matter. Maybe a lot.

Well, yes, estimating the magnitude of a meteor visually is a tricky business, and its speed is definitely a factor. But what I'm talking about here is trying to derive the brightness of a meteor from a still image like this one.

[snak88]

Seems fake to me, the meteor not behind the clouds enough, and just so happens to be pointing to the volcano. AI generated, how will we know?

[Chris Peterson]

Seems fake to me, the meteor not behind the clouds enough, and just so happens to be pointing to the volcano. AI generated, how will we know?

Looks perfectly fine to me. There's some reddening of the trail where it's behind clouds, and a little bit of blue scatter off segments with less cloudiness. Nothing about this looks faked.

Also, the imager has a long history of collecting these sorts of images (predating any AI image generators) as well as other astronomical landscapes.

[johnnydeep]

How does his calculation connect to the image?

Well, are you saying that brightness comparisons using even the naked eye are highly affected by the duration of the event? I can see that might matter. Maybe a lot.

Well, yes, estimating the magnitude of a meteor visually is a tricky business, and its speed is definitely a factor. But what I'm talking about here is trying to derive the brightness of a meteor from a still image like this one.

Ok, got it. But the OP didn't seem to be referencing this image particularly, just the comparison of Venus to a meteor in general. So, the math should be correct regardless (though I actually didn't check it).

[Chris Peterson]

Well, are you saying that brightness comparisons using even the naked eye are highly affected by the duration of the event? I can see that might matter. Maybe a lot.

Well, yes, estimating the magnitude of a meteor visually is a tricky business, and its speed is definitely a factor. But what I'm talking about here is trying to derive the brightness of a meteor from a still image like this one.

Ok, got it. But the OP didn't seem to be referencing this image particularly, just the comparison of Venus to a meteor in general. So, the math should be correct regardless (though I actually didn't check it).

I read the post as referencing this image ("such a meteor", "how big... it might have been", and similar).

I take the calculation to be a sort of back-of-the-envelope analysis, and as such it is quite reasonable (but not really if applied to this image).

More rigorously, we estimate the mass of meteoroids by integrating the luminosity over their path (which typically requires video data, not a long exposure still image like this one) and applying one or more mass-luminosity calculations to that (which also requires the inclusion of a luminous efficiency term... rarely known... since different materials ablate quite differently).

[johnnydeep]

Well, yes, estimating the magnitude of a meteor visually is a tricky business, and its speed is definitely a factor. But what I'm talking about here is trying to derive the brightness of a meteor from a still image like this one.

Ok, got it. But the OP didn't seem to be referencing this image particularly, just the comparison of Venus to a meteor in general. So, the math should be correct regardless (though I actually didn't check it).

I read the post as referencing this image ("such a meteor", "how big... it might have been", and similar).

I take the calculation to be a sort of back-of-the-envelope analysis, and as such it is quite reasonable (but not really if applied to this image).

More rigorously, we estimate the mass of meteoroids by integrating the luminosity over their path (which typically requires video data, not a long exposure still image like this one) and applying one or more mass-luminosity calculations to that (which also requires the inclusion of a luminous efficiency term... rarely known... since different materials ablate quite differently).

Ok, thanks for indulging me. I used to be able to do integrals, maybe even "path integrals". I was a wiz at math...once. But that was a long time ago.