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Why are Titan's lakes “black†in radar images rather than transparent?
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Searching after reading Titan rover questions (1, 2, 3) lead me to Space.com's Titan Has More Oil Than Earth which says in part:
"Titan is just covered in carbon-bearing material — it's a giant factory of organic chemicals," said Ralph Lorenz, a Cassini radar team member from the Johns Hopkins University Applied Physics Laboratory. "This vast carbon inventory is an important window into the geology and climate history of Titan."
and later
"We also know that some lakes are more than 10 meters or so deep because they appear literally pitch-black to the radar. If they were shallow we'd see the bottom, and we don't," Lorenz said.
In this answer for example I show ground penetrating radar that can see through kilometers of rock and regolith that contains water - a molecule that is quite strongly absorbing at radio and radar frequencies.
Why would a pool of dry and non-polar hydrocarbons, something I assume would be a good dielectric and therefore mostly transparent to radar, be black and hide the bottom even at a depth of only 10 meters?
below: "PIA13164: North Polar Cap Cross Section" MRO Shallow Subsurface Radar (SHARAD) from JPL.
titan radar
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uhoh
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up vote
2
down vote
favorite
Searching after reading Titan rover questions (1, 2, 3) lead me to Space.com's Titan Has More Oil Than Earth which says in part:
"Titan is just covered in carbon-bearing material — it's a giant factory of organic chemicals," said Ralph Lorenz, a Cassini radar team member from the Johns Hopkins University Applied Physics Laboratory. "This vast carbon inventory is an important window into the geology and climate history of Titan."
and later
"We also know that some lakes are more than 10 meters or so deep because they appear literally pitch-black to the radar. If they were shallow we'd see the bottom, and we don't," Lorenz said.
In this answer for example I show ground penetrating radar that can see through kilometers of rock and regolith that contains water - a molecule that is quite strongly absorbing at radio and radar frequencies.
Why would a pool of dry and non-polar hydrocarbons, something I assume would be a good dielectric and therefore mostly transparent to radar, be black and hide the bottom even at a depth of only 10 meters?
below: "PIA13164: North Polar Cap Cross Section" MRO Shallow Subsurface Radar (SHARAD) from JPL.
titan radar
asked 4 hours ago
uhoh
28.6k1392346
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Searching after reading Titan rover questions (1, 2, 3) lead me to Space.com's Titan Has More Oil Than Earth which says in part:
"Titan is just covered in carbon-bearing material — it's a giant factory of organic chemicals," said Ralph Lorenz, a Cassini radar team member from the Johns Hopkins University Applied Physics Laboratory. "This vast carbon inventory is an important window into the geology and climate history of Titan."
and later
"We also know that some lakes are more than 10 meters or so deep because they appear literally pitch-black to the radar. If they were shallow we'd see the bottom, and we don't," Lorenz said.
In this answer for example I show ground penetrating radar that can see through kilometers of rock and regolith that contains water - a molecule that is quite strongly absorbing at radio and radar frequencies.
Why would a pool of dry and non-polar hydrocarbons, something I assume would be a good dielectric and therefore mostly transparent to radar, be black and hide the bottom even at a depth of only 10 meters?
below: "PIA13164: North Polar Cap Cross Section" MRO Shallow Subsurface Radar (SHARAD) from JPL.
titan radar
asked 4 hours ago
uhoh
28.6k1392346
Searching after reading Titan rover questions (1, 2, 3) lead me to Space.com's Titan Has More Oil Than Earth which says in part:
"Titan is just covered in carbon-bearing material — it's a giant factory of organic chemicals," said Ralph Lorenz, a Cassini radar team member from the Johns Hopkins University Applied Physics Laboratory. "This vast carbon inventory is an important window into the geology and climate history of Titan."
and later
"We also know that some lakes are more than 10 meters or so deep because they appear literally pitch-black to the radar. If they were shallow we'd see the bottom, and we don't," Lorenz said.
In this answer for example I show ground penetrating radar that can see through kilometers of rock and regolith that contains water - a molecule that is quite strongly absorbing at radio and radar frequencies.
Why would a pool of dry and non-polar hydrocarbons, something I assume would be a good dielectric and therefore mostly transparent to radar, be black and hide the bottom even at a depth of only 10 meters?
below: "PIA13164: North Polar Cap Cross Section" MRO Shallow Subsurface Radar (SHARAD) from JPL.
titan radar
titan radar
asked 4 hours ago
uhoh
28.6k1392346
asked 4 hours ago
uhoh
28.6k1392346
asked 4 hours ago
uhoh
28.6k1392346
asked 4 hours ago
uhoh
28.6k1392346
asked 4 hours ago
uhoh
28.6k1392346
28.6k1392346
add a comment |Â
add a comment |Â
1 Answer
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This article suggests that the radar can penetrate the lakes and reports them to be hundreds of meters deep. The space.com article referenced seems to be sourced from a Geophysical Research Letters article from 2008 (not paywalled) which gives the radar wavelength (2.2cm) and claims that it would be absorbed in 2-20m of clean hydrocarbon, but also give other less direct ways of estimating lake depth which give higher values.
[9] A final, and completely independent, measure is radiometric. The darkest parts of some lakes, generally the largest ones, are ‘black holes’, offering no measurable radar return down to the instrument noise floor of ∼ −26 dB [Stofan et al., 2007]. This requires not only that the surface reflection be very small (consistent with a smooth surface of a low dielectric constant material, such as a liquid hydrocarbon surface unroughened by waves) but also requires that the liquid be deep and/or lossy enough to suppress a bottom reflection. Lake bottoms with sediment density increasing smoothly with depth could also suppress bottom reflections via gradientâ€Âindex impedance matching: however, there are morphological indications such as dark channels incised in almostâ€Âasâ€Âdark lakes that suggest that at least in some places bottom reflections are seen. Assuming then that such lakebed features are being hidden in ‘black’ areas by column absorption, a minimum depth can be inferred: the lower the assumed loss tangent δ, the deeper the lake must be. Clean liquid hydrocarbons have δ ∼ 10−4 to 10−3 [Sen et al., 1992; Rodriguez et al., 2003] although suspended or dissolved polar molecules such as nitriles and small tholin particles could increase these values. A penetration depth (1/e oneâ€Âway absorption †see e.g. similar calculations elsewhere in the Saturnian system [Ostro et al., 2006]) of λ/2Àδ√ɛ, with λ the radar wavelength of 2.2 cm and ɛ the real part of the dielectric constant (∼2), would therefore be 2–20 m †lakes with nonzero reflectivity or visible lakebed features are therefore likely shallower than this range.
The Mars data comes from the SHARAD instrument which uses much longer wavelengths (15-25 MHz, 10-20m) which may explain the different absorption.
From its 10-meter (33-foot) antenna, SHARAD transmits radar waves in a frequency band swept from 25 to 15 megahertz, obtaining a 15-meter (50-foot) vertical resolution in free space, which reduces to 10 meters (33 feet) or less in the subsurface.
answered 4 hours ago
Steve Linton
3,4241327
This looks great, thanks!
– uhoh
2 hours ago
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
This article suggests that the radar can penetrate the lakes and reports them to be hundreds of meters deep. The space.com article referenced seems to be sourced from a Geophysical Research Letters article from 2008 (not paywalled) which gives the radar wavelength (2.2cm) and claims that it would be absorbed in 2-20m of clean hydrocarbon, but also give other less direct ways of estimating lake depth which give higher values.
[9] A final, and completely independent, measure is radiometric. The darkest parts of some lakes, generally the largest ones, are ‘black holes’, offering no measurable radar return down to the instrument noise floor of ∼ −26 dB [Stofan et al., 2007]. This requires not only that the surface reflection be very small (consistent with a smooth surface of a low dielectric constant material, such as a liquid hydrocarbon surface unroughened by waves) but also requires that the liquid be deep and/or lossy enough to suppress a bottom reflection. Lake bottoms with sediment density increasing smoothly with depth could also suppress bottom reflections via gradientâ€Âindex impedance matching: however, there are morphological indications such as dark channels incised in almostâ€Âasâ€Âdark lakes that suggest that at least in some places bottom reflections are seen. Assuming then that such lakebed features are being hidden in ‘black’ areas by column absorption, a minimum depth can be inferred: the lower the assumed loss tangent δ, the deeper the lake must be. Clean liquid hydrocarbons have δ ∼ 10−4 to 10−3 [Sen et al., 1992; Rodriguez et al., 2003] although suspended or dissolved polar molecules such as nitriles and small tholin particles could increase these values. A penetration depth (1/e oneâ€Âway absorption †see e.g. similar calculations elsewhere in the Saturnian system [Ostro et al., 2006]) of λ/2Àδ√ɛ, with λ the radar wavelength of 2.2 cm and ɛ the real part of the dielectric constant (∼2), would therefore be 2–20 m †lakes with nonzero reflectivity or visible lakebed features are therefore likely shallower than this range.
The Mars data comes from the SHARAD instrument which uses much longer wavelengths (15-25 MHz, 10-20m) which may explain the different absorption.
From its 10-meter (33-foot) antenna, SHARAD transmits radar waves in a frequency band swept from 25 to 15 megahertz, obtaining a 15-meter (50-foot) vertical resolution in free space, which reduces to 10 meters (33 feet) or less in the subsurface.
answered 4 hours ago
Steve Linton
3,4241327
This looks great, thanks!
– uhoh
2 hours ago
add a comment |Â
up vote
4
down vote
This article suggests that the radar can penetrate the lakes and reports them to be hundreds of meters deep. The space.com article referenced seems to be sourced from a Geophysical Research Letters article from 2008 (not paywalled) which gives the radar wavelength (2.2cm) and claims that it would be absorbed in 2-20m of clean hydrocarbon, but also give other less direct ways of estimating lake depth which give higher values.
[9] A final, and completely independent, measure is radiometric. The darkest parts of some lakes, generally the largest ones, are ‘black holes’, offering no measurable radar return down to the instrument noise floor of ∼ −26 dB [Stofan et al., 2007]. This requires not only that the surface reflection be very small (consistent with a smooth surface of a low dielectric constant material, such as a liquid hydrocarbon surface unroughened by waves) but also requires that the liquid be deep and/or lossy enough to suppress a bottom reflection. Lake bottoms with sediment density increasing smoothly with depth could also suppress bottom reflections via gradientâ€Âindex impedance matching: however, there are morphological indications such as dark channels incised in almostâ€Âasâ€Âdark lakes that suggest that at least in some places bottom reflections are seen. Assuming then that such lakebed features are being hidden in ‘black’ areas by column absorption, a minimum depth can be inferred: the lower the assumed loss tangent δ, the deeper the lake must be. Clean liquid hydrocarbons have δ ∼ 10−4 to 10−3 [Sen et al., 1992; Rodriguez et al., 2003] although suspended or dissolved polar molecules such as nitriles and small tholin particles could increase these values. A penetration depth (1/e oneâ€Âway absorption †see e.g. similar calculations elsewhere in the Saturnian system [Ostro et al., 2006]) of λ/2Àδ√ɛ, with λ the radar wavelength of 2.2 cm and ɛ the real part of the dielectric constant (∼2), would therefore be 2–20 m †lakes with nonzero reflectivity or visible lakebed features are therefore likely shallower than this range.
The Mars data comes from the SHARAD instrument which uses much longer wavelengths (15-25 MHz, 10-20m) which may explain the different absorption.
From its 10-meter (33-foot) antenna, SHARAD transmits radar waves in a frequency band swept from 25 to 15 megahertz, obtaining a 15-meter (50-foot) vertical resolution in free space, which reduces to 10 meters (33 feet) or less in the subsurface.
answered 4 hours ago
Steve Linton
3,4241327
This looks great, thanks!
– uhoh
2 hours ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
This article suggests that the radar can penetrate the lakes and reports them to be hundreds of meters deep. The space.com article referenced seems to be sourced from a Geophysical Research Letters article from 2008 (not paywalled) which gives the radar wavelength (2.2cm) and claims that it would be absorbed in 2-20m of clean hydrocarbon, but also give other less direct ways of estimating lake depth which give higher values.
[9] A final, and completely independent, measure is radiometric. The darkest parts of some lakes, generally the largest ones, are ‘black holes’, offering no measurable radar return down to the instrument noise floor of ∼ −26 dB [Stofan et al., 2007]. This requires not only that the surface reflection be very small (consistent with a smooth surface of a low dielectric constant material, such as a liquid hydrocarbon surface unroughened by waves) but also requires that the liquid be deep and/or lossy enough to suppress a bottom reflection. Lake bottoms with sediment density increasing smoothly with depth could also suppress bottom reflections via gradientâ€Âindex impedance matching: however, there are morphological indications such as dark channels incised in almostâ€Âasâ€Âdark lakes that suggest that at least in some places bottom reflections are seen. Assuming then that such lakebed features are being hidden in ‘black’ areas by column absorption, a minimum depth can be inferred: the lower the assumed loss tangent δ, the deeper the lake must be. Clean liquid hydrocarbons have δ ∼ 10−4 to 10−3 [Sen et al., 1992; Rodriguez et al., 2003] although suspended or dissolved polar molecules such as nitriles and small tholin particles could increase these values. A penetration depth (1/e oneâ€Âway absorption †see e.g. similar calculations elsewhere in the Saturnian system [Ostro et al., 2006]) of λ/2Àδ√ɛ, with λ the radar wavelength of 2.2 cm and ɛ the real part of the dielectric constant (∼2), would therefore be 2–20 m †lakes with nonzero reflectivity or visible lakebed features are therefore likely shallower than this range.
The Mars data comes from the SHARAD instrument which uses much longer wavelengths (15-25 MHz, 10-20m) which may explain the different absorption.
From its 10-meter (33-foot) antenna, SHARAD transmits radar waves in a frequency band swept from 25 to 15 megahertz, obtaining a 15-meter (50-foot) vertical resolution in free space, which reduces to 10 meters (33 feet) or less in the subsurface.
answered 4 hours ago
Steve Linton
3,4241327
This article suggests that the radar can penetrate the lakes and reports them to be hundreds of meters deep. The space.com article referenced seems to be sourced from a Geophysical Research Letters article from 2008 (not paywalled) which gives the radar wavelength (2.2cm) and claims that it would be absorbed in 2-20m of clean hydrocarbon, but also give other less direct ways of estimating lake depth which give higher values.
[9] A final, and completely independent, measure is radiometric. The darkest parts of some lakes, generally the largest ones, are ‘black holes’, offering no measurable radar return down to the instrument noise floor of ∼ −26 dB [Stofan et al., 2007]. This requires not only that the surface reflection be very small (consistent with a smooth surface of a low dielectric constant material, such as a liquid hydrocarbon surface unroughened by waves) but also requires that the liquid be deep and/or lossy enough to suppress a bottom reflection. Lake bottoms with sediment density increasing smoothly with depth could also suppress bottom reflections via gradientâ€Âindex impedance matching: however, there are morphological indications such as dark channels incised in almostâ€Âasâ€Âdark lakes that suggest that at least in some places bottom reflections are seen. Assuming then that such lakebed features are being hidden in ‘black’ areas by column absorption, a minimum depth can be inferred: the lower the assumed loss tangent δ, the deeper the lake must be. Clean liquid hydrocarbons have δ ∼ 10−4 to 10−3 [Sen et al., 1992; Rodriguez et al., 2003] although suspended or dissolved polar molecules such as nitriles and small tholin particles could increase these values. A penetration depth (1/e oneâ€Âway absorption †see e.g. similar calculations elsewhere in the Saturnian system [Ostro et al., 2006]) of λ/2Àδ√ɛ, with λ the radar wavelength of 2.2 cm and ɛ the real part of the dielectric constant (∼2), would therefore be 2–20 m †lakes with nonzero reflectivity or visible lakebed features are therefore likely shallower than this range.
The Mars data comes from the SHARAD instrument which uses much longer wavelengths (15-25 MHz, 10-20m) which may explain the different absorption.
From its 10-meter (33-foot) antenna, SHARAD transmits radar waves in a frequency band swept from 25 to 15 megahertz, obtaining a 15-meter (50-foot) vertical resolution in free space, which reduces to 10 meters (33 feet) or less in the subsurface.
answered 4 hours ago
Steve Linton
3,4241327
edited 3 hours ago
answered 4 hours ago
Steve Linton
3,4241327
answered 4 hours ago
Steve Linton
3,4241327
answered 4 hours ago
Steve Linton
3,4241327
3,4241327
This looks great, thanks!
– uhoh
2 hours ago
add a comment |Â
This looks great, thanks!
– uhoh
2 hours ago
This looks great, thanks!
– uhoh
2 hours ago
This looks great, thanks!
– uhoh
2 hours ago
add a comment |Â
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