Starship Asterisk*

MarkBour wrote: ↑Tue Aug 07, 2018 4:47 pm

alter-ego wrote: ↑Mon Aug 06, 2018 2:21 am
... you've significantly underestimated your weight reduction which = Δg/g x Weight = 1.8 x 10^-7 x Weight. Normalizing to 1 pound and converting to gram-force = 454 gm-force/lb, the weight reduction = 82 microgram-force per pound of weight. So for a 175lb person, their weight reduction = 14 "milligrams".

Thanks alter-ego, for clarifying that. How did you know my weight? 14 milligrams actually sounds measurable!

• Not to be cocky but 14 milligrams of generic Viagra just cost me $5.75

MarkBour wrote: ↑Tue Aug 07, 2018 4:47 pm

neufer wrote: ↑Mon Aug 06, 2018 12:03 pm
Eclipses are unnecessary, of course. A Super Moon (or a Super New Moon) is already lined up sufficiently.

• But climb Chimborazo to really lose weight:

https://physics.stackexchange.com/questions/22613/lowest-gravity-on-earths-surface wrote:
<<[Sea level] gravitational acceleration at the equator is 9.8144 m/s², vs 9.8322 m/s² at the poles;

the International Gravity Formula can help, giving 9.761 m/s² atop Chimborazo.>>

– Dave Jarvis Dec 30 '15 at 3:03

I actually would like to go to the summit of Chimborazo some day. Sounds easier than Everest* and it's the point on the surface of the Earth that is farthest from the center (much farther than Everest).

It would make some sense that it would be the point on the surface with the lowest gravity, but I know that mass concentrations in the Earth make this a complex question. I realize you didn't make the claim that it was the *lowest* gravity. Surely it is a good contender. I found this reference, but it is a little older than yours:

https://www.newscientist.com/article/dn ... -extremes/

Mount Nevado Huascarán in Peru has the lowest gravitational acceleration, at 9.7639 m/s2, while the highest is at the surface of the Arctic Ocean, at 9.8337 m/s2.

I have heard that climbers on these peaks have reported feeling light-headed.

* Evidently climbing Chimborazo is still a dangerous climb, the worst danger being avalanches. And I don't want to be cocky, it is likely beyond my capabilities. But the air pressure is far more tolerable. Air pressure is a very complex issue, itself, but there are extensive observations. The Himalayas are definitely the lowest air pressure on the planet, and Chimborazo is not too bad.

• Higher air density buoyancy can't hurt.

https://en.wikipedia.org/wiki/Chimborazo wrote:

........Mass concentrations make up to ±100 m adjustments to effective height....... Note: Sri Lankan BLUE actually represent LOW sea level gravity

---

<<The summit of Chimborazo is widely reported to be the farthest point on the surface from Earth's center, with Huascarán a very close second. The summit of the Chimborazo is the fixed point on Earth that has the utmost distance from the center – because of the oblate spheroid shape of the planet Earth, which is "thicker" around the Equator than measured around the poles.[It has been difficult to resolve this issue definitively because of error margins in summit elevations and geoid data. Application of the formula at Earth radius#Radius at a given geodetic latitude shows that the Earth's radius is 520 metres greater at Chimborazo than at Huascaran, with most recent data showing another 5 metres due to local variations in gravity, for a total of 525 metres. Chimborazo's summit is roughly either 20 metres or 40 metres further from the Earth's center than that of Huascaran.] Despite being 2,585 m lower in elevation above sea level, Chimborazo [at 1.5° south] is 2,163 m farther from the Earth's center than the summit of Everest [nearly 27.6° north].>>

alter-ego wrote: ↑Mon Aug 06, 2018 2:21 am ... you've significantly underestimated your weight reduction which = Δg/g x Weight = 1.8 x 10^-7 x Weight. Normalizing to 1 pound and converting to gram-force = 454 gm-force/lb, the weight reduction = 82 microgram-force per pound of weight. So for a 175lb person, their weight reduction = 14 "milligrams".

neufer wrote: ↑Mon Aug 06, 2018 12:03 pm Eclipses are unnecessary, of course. A Super Moon (or a Super New Moon) is already lined up sufficiently.

• But climb Chimborazo to really lose weight:

https://physics.stackexchange.com/questions/22613/lowest-gravity-on-earths-surface wrote:
<<[Sea level] gravitational acceleration at the equator is 9.8144 m/s², vs 9.8322 m/s² at the poles;

the International Gravity Formula can help, giving 9.761 m/s² atop Chimborazo.>>

– Dave Jarvis Dec 30 '15 at 3:03

https://www.newscientist.com/article/dn ... -extremes/
Mount Nevado Huascarán in Peru has the lowest gravitational acceleration, at 9.7639 m/s2, while the highest is at the surface of the Arctic Ocean, at 9.8337 m/s2.

alter-ego wrote: ↑Mon Aug 06, 2018 2:21 am

MarkBour wrote: ↑Sun Aug 05, 2018 10:47 pm ...
So, the Moon can reduce my weight when it is overhead (if it's a Super Moon) by 1.3 x 10^-7 g and the Sun by 5.0 x 10^-8 g. Then when we get an eclipse, the effect can be as strong as 1.8 x 10^-7 g, summing those together. Then according to Art's calculations, even if everything else lined up perfectly and under unrealistically good conditions for all of them, they would not total more than 8.4 x 10^-12 g, I think. Anyway, so just an eclipse event affects my weight by about 0.2 micro grams, and the others are in the realm of a few pico grams. My guess of a yocto gram was too low.

And thanks for the reminder, Bruce, the folks on both sides of the Earth would feel lighter.

Indeed, yocto is too small a prefix. However, you've significantly underestimated your weight reduction which = Δg/g x Weight = 1.8 x 10^-7 x Weight. Normalizing to 1 pound and converting to gram-force = 454 gm-force/lb, the weight reduction = 82 microgram-force per pound of weight. So for a 175lb person, their weight reduction = 14 "milligrams".

• But climb Chimborazo to really lose weight:

https://physics.stackexchange.com/questions/22613/lowest-gravity-on-earths-surface wrote:
<<[Sea level] gravitational acceleration at the equator is 9.8144 m/s², vs 9.8322 m/s² at the poles;

the International Gravity Formula can help, giving 9.761 m/s² atop Chimborazo.>>

– Dave Jarvis Dec 30 '15 at 3:03

MarkBour wrote: ↑Sun Aug 05, 2018 10:47 pm ...
So, the Moon can reduce my weight when it is overhead (if it's a Super Moon) by 1.3 x 10^-7 g and the Sun by 5.0 x 10^-8 g. Then when we get an eclipse, the effect can be as strong as 1.8 x 10^-7 g, summing those together. Then according to Art's calculations, even if everything else lined up perfectly and under unrealistically good conditions for all of them, they would not total more than 8.4 x 10^-12 g, I think. Anyway, so just an eclipse event affects my weight by about 0.2 micro grams, and the others are in the realm of a few pico grams. My guess of a yocto gram was too low.

And thanks for the reminder, Bruce, the folks on both sides of the Earth would feel lighter.

BDanielMayfield wrote: ↑Sun Aug 05, 2018 2:42 pm

neufer wrote: ↑Sun Aug 05, 2018 4:17 am

MarkBour wrote: ↑Sun Aug 05, 2018 12:12 amWill people on that side of Earth weigh less than usual that night? Maybe by a yocto kilo, or something?

Tidal forces (uncompensated by free fall) are dominated by:

• 1) a close Super Moon plus the Sun at perihelion
  2) a close Venus conjunction

Everything else is under the radar.

Tidal forces scale as the cube of the angular diameter times the density:

Tidal force of a close Super Moon: 132,595 = (34.1')³ x 3.344 g/cm³

Tidal force of Sun at perihelion: 49,232 = (32.7')³ x 1.408 g/cm³
----------------------------------------------------------------------------------------------------
Tidal force of Venus at closest conjunction: 7.223 = (1.1')³ x 5.427 g/cm³
----------------------------------------------------------------------------------------------------
Tidal force of Jupiter at closest opposition: 0.772 = (0.835')³ x 1.326 g/cm³

Tidal force of Mars at closest opposition: 0.287 = (0.418')³ x 3.9335 g/cm³
----------------------------------------------------------------------------------------------------
Tidal force of Mercury at closest conjunction: 0.054 = (0.217')³ x 5.243 g/cm³

Tidal force of Saturn at closest opposition: 0.026 = (0.335')³ x 0.687 g/cm³

Interesting Q & A y'all. But Mark, notice that Art didn't completely answer your question. He left out some units for you or someonelse to finish the calculation. I could, but it was your question. Oh, and the people on the dayside of the Earth would also feel a little lighter, but they'd be in the dark as to why.

Art, interesting that sweet little ole Venus exerts more than 10 times the pull of mighty Jupiter! Girl power!

Bruce

alter-ego wrote: ↑Sun Aug 05, 2018 4:11 am In Stellarium, I created this 40° view from geosynchronous orbit. To accentuate the planets, I limited the stellar magnitude visibility to about 6. Note, a thin lunar crescent would also be visible.
 
Sep 8, 2040 - 5-Planet Grouping from 36000km.JPG

neufer wrote: ↑Sun Aug 05, 2018 4:08 pm

BDanielMayfield wrote: ↑Sun Aug 05, 2018 2:42 pm
Art, interesting that sweet little ole Venus exerts more than 10 times the pull of mighty Jupiter!

Girl power!

And much more tidal pull than her brother Mars.

Just as Selene/Artemis/Diana/Luna exerts more tidal pull than her brother Helios/Apollo:

https://en.wikipedia.org/wiki/Selene wrote:
<<In Greek mythology, Selene (Ancient Greek: Σελήνη "Moon") is the goddess of the moon. She is the daughter of the Titans Hyperion and Theia, and sister of the sun-god Helios, and Eos, goddess of the dawn. She drives her moon chariot across the heavens. In classical times, Selene was often identified with Artemis, much as her brother, Helios, was identified with Apollo. Selene and Artemis were also associated with Hecate, and all three were regarded as lunar goddesses, but only Selene was regarded as the personification of the moon itself. Her Roman equivalent is Luna.>>

BDanielMayfield wrote: ↑Sun Aug 05, 2018 2:42 pm
Art, interesting that sweet little ole Venus exerts more than 10 times the pull of mighty Jupiter!

Girl power!

https://en.wikipedia.org/wiki/Selene wrote:
<<In Greek mythology, Selene (Ancient Greek: Σελήνη "Moon") is the goddess of the moon. She is the daughter of the Titans Hyperion and Theia, and sister of the sun-god Helios, and Eos, goddess of the dawn. She drives her moon chariot across the heavens. In classical times, Selene was often identified with Artemis, much as her brother, Helios, was identified with Apollo. Selene and Artemis were also associated with Hecate, and all three were regarded as lunar goddesses, but only Selene was regarded as the personification of the moon itself. Her Roman equivalent is Luna.>>

neufer wrote: ↑Sun Aug 05, 2018 4:17 am

Also sprach Zarathustra

---

MarkBour wrote: ↑Sun Aug 05, 2018 12:12 amWill people on that side of Earth weigh less than usual that night? Maybe by a yocto kilo, or something?

Tidal forces (uncompensated by free fall) are dominated by:

• 1) a close Super Moon plus the Sun at perihelion
  2) a close Venus conjunction

Everything else is under the radar.

Tidal forces scale as the cube of the angular diameter times the density:

Tidal force of a close Super Moon: 132,595 = (34.1')³ x 3.344 g/cm³

Tidal force of Sun at perihelion: 49,232 = (32.7')³ x 1.408 g/cm³
----------------------------------------------------------------------------------------------------
Tidal force of Venus at closest conjunction: 7.223 = (1.1')³ x 5.427 g/cm³
----------------------------------------------------------------------------------------------------
Tidal force of Jupiter at closest opposition: 0.772 = (0.835')³ x 1.326 g/cm³

Tidal force of Mars at closest opposition: 0.287 = (0.418')³ x 3.9335 g/cm³
----------------------------------------------------------------------------------------------------
Tidal force of Mercury at closest conjunction: 0.054 = (0.217')³ x 5.243 g/cm³

Tidal force of Saturn at closest opposition: 0.026 = (0.335')³ x 0.687 g/cm³

MarkBour wrote: ↑Sun Aug 05, 2018 12:12 am

neufer wrote: ↑Fri Aug 03, 2018 12:19 am Sunset, Sat 2040 Sep 8 will be the thing to see.
(I predict a dozen APODs minimum.)

Thanks for emphasizing that one. It's really not too far out on my calendar, and I see I currently have no bookings (except for Sunday afternoon), so I will put it down to watch.

Actually, I wonder if we could get a space-based observation that included a crescent Earth and everything else in one image. Of course, ground-based imagers will be clever enough to catch a mountain range or something so they can count the Earth in their image, and somebody will capture it reflected in a lake, to make the count artificially higher, but I still wonder if there would be anything an off-planet position and a good scope could do that would be extraordinary.

Will people on that side of Earth weigh less than usual that night? Maybe by a yocto kilo, or something?

• 1) a close Super Moon plus the Sun at perihelion
  2) a close Venus conjunction

MarkBour wrote: ↑Sun Aug 05, 2018 12:12 am ...
Actually, I wonder if we could get a space-based observation that included a crescent Earth and everything else in one image. Of course, ground-based imagers will be clever enough to catch a mountain range or something so they can count the Earth in their image, and somebody will capture it reflected in a lake, to make the count artificially higher, but I still wonder if there would be anything an off-planet position and a good scope could do that would be extraordinary.
...

neufer wrote: ↑Fri Aug 03, 2018 12:19 am Sunset, Sat 2040 Sep 8 will be the thing to see.
(I predict a dozen APODs minimum.)

neufer wrote: ↑Thu Aug 02, 2018 12:36 pm

https://web.archive.org/web/20051214013307/http://www.sunspot.noao.edu/PR/alignment.html wrote: ...
For example, the alignment of September 2040 is closest on 8 September 2040. It is the 6th closest alignment in the 8213-year period at an alignment width of 7.7 degrees, and the central point is about 24 degrees to the east of the Sun, i.e., the planets are all to the east of the Sun (that is, visible during the early evening).>>

MarkBour wrote: ↑Thu Aug 02, 2018 10:49 pm
5:10 am May 6, 2492. No heavy drinking on Cinco de Mayo, or I'll sleep through it.

https://science.nasa.gov/science-news/science-at-nasa/2000/ast30mar_1m wrote:
<<The five naked-eye planets cluster together in the sky within a circle 25 degrees or less in diameter once every 57 years, on average. The next time it will happen is September 8, 2040. The 2040 grouping will include Mercury, Venus, Mars, Jupiter, Saturn and the crescent Moon. Clustered well to the east of the Sun, the planets will stage a spectacular show at 7:30 p.m. in the evening. (Mark your calendar now!)>>

alter-ego wrote: ↑Thu Aug 02, 2018 4:32 am

neufer wrote: ↑Wed Aug 01, 2018 6:29 pm
If one limits oneself to just the five visible planets:

Just after sunset on Sat 2040 Sep 8 23:25 UTC:

New York View toward western horizon : Sat 2040 Sep 8 23:25 UTC

Such a close alignment happen only about once ever 1,000 years

The millennia estimate is a bit long.

The 2040, 5-planet grouping separation actually is ~9.4°. Referencing the 1994 paper by De Meis and Meeus, between the years -3101 TDT and +2735 TDT, there are ten 5-planet groupings within ≤ 10°. The average then (over 5700 years) for such a grouping ≈ 570 years.

• I was ball-parking the somewhat longer (more accurate ) study below :

• 8213 years / ~6 close alignments ≈ a millenium.

https://web.archive.org/web/20051214013307/http://www.sunspot.noao.edu/PR/alignment.html wrote:
National Solar Observatory
Sacramento Peak
Planetary Alignments

<<I've studied the alignment (as seen from the Earth) of the planets Mercury through Saturn for a period of 3,000,000 days between 4713 BC and AD 3501. The alignment width shows periodic behavior with main periods of about 10.9, 20.1, and 200 years.

For example, the alignment of September 2040 is closest on 8 September 2040. It is the 6th closest alignment in the 8213-year period at an alignment width of 7.7 degrees, and the central point is about 24 degrees to the east of the Sun, i.e., the planets are all to the east of the Sun (that is, visible during the early evening).>>

neufer wrote: ↑Wed Aug 01, 2018 6:29 pm
If one limits oneself to just the five visible planets:

Just after sunset on Sat 2040 Sep 8 23:25 UTC:

New York View toward western horizon : Sat 2040 Sep 8 23:25 UTC

Such a close alignment happen only about once ever 1,000 years

The next time all 8 planets line up such that they can all be viewed at once
(rather than just sometime over the entire night) will be May 6, 2492:

http://wtamu.edu/~cbaird/sq/2013/08/28/when-do-the-planets-in-our-solar-system-all-line-up/ wrote:
When do the planets in our solar system all line up?
By Dr. Christopher S. Baird: August 28, 2013

South from NYC at 5:10 am on May 6, 2492

---

<<The closest the eight planets will come to being aligned will be on May 6, 2492. This image shows what you would see if you were off the coast of New York City and you looked due south at 5:10 am on May 6, 2492, as calculated by the software program Stellarium 0.13.3. As you can see, the planets are not visually sitting on top of each other, which would be the case if they were all located on a line emanating from the earth (as would be needed to maximize their net gravitational force on the earth). Furthermore, they are not even visually sitting on a line in the sky. They are simply in the same general region of the sky. Public Domain Image, source: Christopher S. Baird.>>

MarkBour wrote: ↑Wed Aug 01, 2018 5:29 pm
I'm wondering how often [planetary alignment] happens:

Once every billion years or so?

Never? (Maybe orbital resonances make it impossible.)

If one limits oneself to just the five visible planets:

Just after sunset on Sat 2040 Sep 8 23:25 UTC:

New York View toward western horizon : Sat 2040 Sep 8 23:25 UTC

Such a close alignment happen only about once ever 1,000 years

For both Uranus & Neptune to be in these 7.7º wide groupings
happens only once every (360/7.7)² ~ 2,000 millennia

http://wtamu.edu/~cbaird/sq/2013/08/28/when-do-the-planets-in-our-solar-system-all-line-up/ wrote:
When do the planets in our solar system all line up?
By Dr. Christopher S. Baird: August 28, 2013

South from NYC at 5:10 am on May 6, 2492

---

http://www.fourmilab.ch/cgi-bin/Solar

---

<<The closest the eight planets will come to being aligned will be on May 6, 2492. This image shows what you would see if you were off the coast of New York City and you looked due south at 5:10 am on May 6, 2492, as calculated by the software program Stellarium 0.13.3. As you can see, the planets are not visually sitting on top of each other, which would be the case if they were all located on a line emanating from the earth (as would be needed to maximize their net gravitational force on the earth). Furthermore, they are not even visually sitting on a line in the sky. They are simply in the same general region of the sky. Public Domain Image, source: Christopher S. Baird.>>

Chris Peterson wrote: ↑Tue Jul 31, 2018 5:52 pm (Saturn and Jupiter are currently getting closer together in the sky, and will remain close for a number of years. They'll be very close at the end of 2020, just six arcminutes apart. A truly spectacular sight and imaging opportunity that will be- Saturn separated from Jupiter by the same amount as the Galilean moons! Mars and the inner planets go through much quicker cycles. So there will be many opportunites in the next years to see all the bright planets either together or over a single evening.)

MarkBour wrote: ↑Tue Jul 31, 2018 5:26 pm I'm wondering if this month has been an especially good time for planet viewing, or if I'm just becoming more aware.

Chris Peterson wrote: ↑Sat Jul 28, 2018 6:26 pm A couple of weeks ago I simultaneously observed Mercury, Venus, Jupiter, and Saturn. About an hour later I saw Venus, Jupiter, Saturn, and Mars. That was pretty cool. With a fisheye I could have easily imaged either group of four. That frame would also have contained Pluto, although it wouldn't have been bright enough to show up.

With current practical technology (and the right date) I think the best we could hope to capture in a single image would be all the planets from Venus to Neptune. So, six. Mercury would probably be out because the sky would still be too bright for the dimmer planets to show. And Pluto would be out because no very wide angle lens system is likely to be able to record such a dim object given a reasonable exposure time.

ta152h0 wrote: ↑Sun Jul 29, 2018 3:06 am
I took a photo of the Moon on July 26, 2018 at 12:57 AM and I think I caught a collision at about 2=11 o;clock position on the face.

https://en.wikipedia.org/wiki/Alpha_Capricornids wrote:
<<Alpha Capricornids is a meteor shower that takes place as early as 15 July and continues until around 10 August. The meteor shower was discovered by Hungarian astronomer Miklos von Konkoly-Thege in 1871. This shower has infrequent but relatively bright meteors, with some fireballs. Peter Jenniskens and Jeremie Vaubaillon identified the parent body as asteroid 2002 EX12, which in the return of 2005 was found weakly active near perihelion. This object is now called comet 169P/NEAT.

According to Jenniskens and Vaubaillon, the meteor shower was created about 3,500 to 5,000 years ago, when about half of the parent body disintegrated and fell into dust. The dust cloud evolved into Earth's orbit recently, causing a shower with peak rates of 2-5/h, sometimes having outbursts of bright flaring meteors with rates up to 5-9/h. The bulk of the dust will not be in Earth's path until the 24th century. The Alpha Capricornids are expected to become a major annual storm in 2220–2420 A.D., one that will be "stronger than any current annual shower.">>

• Did it look something like this:

http://www.esa.int/spaceinimages/Images/2018/07/Two_lunar_flashes_GIF wrote:

Click to play embedded YouTube video.

Two lunar flashes GIF
Released 27/07/2018 2:06 pm
Copyright Moon Impacts Detection and Analysis System (MIDAS)/Jose Maria Madiedo

<<On 17 July 2018, an ancient lump from space thwacked into the Moon with enough energy to produce a brilliant flash of light. With another space rock seemingly in pursuit, a second flash lit up a different region of the Moon almost exactly 24 hours later.

This GIF shows both lunar flashes, edited to within just moments of each other.>>

Chris Peterson wrote: ↑Sun Jul 29, 2018 4:15 am

alter-ego wrote: ↑Sun Jul 29, 2018 4:03 am

Chris Peterson wrote: ↑Sat Jul 28, 2018 6:26 pm

A couple of weeks ago I simultaneously observed Mercury, Venus, Jupiter, and Saturn. About an hour later I saw Venus, Jupiter, Saturn, and Mars. That was pretty cool. With a fisheye I could have easily imaged either group of four. That frame would also have contained Pluto, although it wouldn't have been bright enough to show up.

With current practical technology (and the right date) I think the best we could hope to capture in a single image would be all the planets from Venus to Neptune. So, six. Mercury would probably be out because the sky would still be too bright for the dimmer planets to show. And Pluto would be out because no very wide angle lens system is likely to be able to record such a dim object given a reasonable exposure time.

I did some searching for simultaneous, 8-"planet" visibility within a fisheye view. I located an example where the fainter planet was Mars, and Mercury has a good chance for being imaged. These 8-planet examples are expectedly very rare. The example below is May 5, -1972 TDT. The sun is 10° below the horizon for an observer on the equator. Mercury has an extincted magnitude of about +0.7, while Mars +2.2 with extinction. Jupiter (lower altitude than Mercury) has about -0.9 with extinction. Interestingly, Neptune might not be imaged as comes in at 9.1 mag w/extinction and Uranus is 5.6. I'd say for this example, Pluto definitely won't be imaged, Neptune is questionable, but Mercury is has a good chance.
 

Mercury is easy. Uranus/Neptune are easy. The tradeoff would seem to be between those.

alter-ego wrote: ↑Sun Jul 29, 2018 4:03 am

Chris Peterson wrote: ↑Sat Jul 28, 2018 6:26 pm

MarkBour wrote: ↑Sat Jul 28, 2018 6:00 am Fantastic! Too bad Mercury was uncooperative. Not long ago, I asked what the record was for number of planets in one image.

A couple of weeks ago I simultaneously observed Mercury, Venus, Jupiter, and Saturn. About an hour later I saw Venus, Jupiter, Saturn, and Mars. That was pretty cool. With a fisheye I could have easily imaged either group of four. That frame would also have contained Pluto, although it wouldn't have been bright enough to show up.

With current practical technology (and the right date) I think the best we could hope to capture in a single image would be all the planets from Venus to Neptune. So, six. Mercury would probably be out because the sky would still be too bright for the dimmer planets to show. And Pluto would be out because no very wide angle lens system is likely to be able to record such a dim object given a reasonable exposure time.

I did some searching for simultaneous, 8-"planet" visibility within a fisheye view. I located an example where the fainter planet was Mars, and Mercury has a good chance for being imaged. These 8-planet examples are expectedly very rare. The example below is May 5, -1972 TDT. The sun is 10° below the horizon for an observer on the equator. Mercury has an extincted magnitude of about +0.7, while Mars +2.2 with extinction. Jupiter (lower altitude than Mercury) has about -0.9 with extinction. Interestingly, Neptune might not be imaged as comes in at 9.1 mag w/extinction and Uranus is 5.6. I'd say for this example, Pluto definitely won't be imaged, Neptune is questionable, but Mercury is has a good chance.