# Talk:55 Cancri Af

I read this in the Daily Mail and wondered if this is new info. As i'm not really into science maybe one of you guys could field this one! See this article for me to see if i'm right, thanks! And please get back to me. From Pafcool2 20:22, 7 November 2007 (UTC)
No, this planet is not an Earth-like planet. It is a gas giant with a mass 190 times as great as the Earth. However it could possibly have a large moon, which NASA suggests might be something like the Earth; using Celestia I've made an image of this planet which can be seen here together with a hypothetical Earth-like planet. Eburacum 87.102.28.2 14:45, 8 November 2007 (UTC)

Looks just like the fictional Pandora from the 2009 film Avatar. 200.168.20.215 (talk) 12:29, 29 December 2009 (UTC)

I'm confused about something.  The 55 Cancri article says that the star's luminosity is .63 solar.    If the average distance is .781 AU, that should make the average insolation slightly higher than Earth (.63/.781² = 1.033), yet this article says that it receives considerably less insolation, even at perastron. Nik42 (talk) 22:00, 13 June 2008 (UTC)

To be honest, I don't even think that the insolation chart belongs in the article.  It's original research (see discussion at Talk:Gliese 581 c, and it also has a bunch of extraneous information unrelated to 55 Cancri f.  Deleting it now. J. Langton (talk) 13:31, 18 August 2008 (UTC)


### Insolation received by 55 Cancri f
Main articles: thermodynamic equilibrium & Climate modeling

This seems incorrect - working from the stellar radius and temperature in the 55 Cancri article gives a bolometric luminosity of 
  
    
      
        
          
            
              L
              
                L
                
                  ⊙
                
              
            
          
          =
          
            
              (
              
                
                  
                    0.96
                     
                    
                      
                        R
                        
                          ⊙
                        
                      
                    
                  
                  
                    
                      R
                      
                        ⊙
                      
                    
                  
                
              
              )
            
            
              2
            
          
          
            
              (
              
                
                  
                    5243
                     
                    
                      K
                    
                  
                  
                    5780
                     
                    
                      K
                    
                  
                
              
              )
            
            
              4
            
          
          =
          0.62
        
      
    
    {\displaystyle \textstyle {\frac {L}{L_{\odot }}}=\left({\frac {0.96\ \mathrm {R_{\odot }} }{\mathrm {R_{\odot }} }}\right)^{2}\left({\frac {5243\ \mathrm {K} }{5780\ \mathrm {K} }}\right)^{4}=0.62}
  

Applying the inverse square law gives 
  
    
      
        
          
            
              f
              
                f
                
                  ⊕
                
              
            
          
          =
          
            (
            
              
                L
                
                  L
                  
                    ⊙
                  
                
              
            
            )
          
          
            
              (
              
                
                  d
                  
                    1
                     
                    
                      A
                      U
                    
                  
                
              
              )
            
            
              −
              2
            
          
        
      
    
    {\displaystyle \textstyle {\frac {f}{f_{\oplus }}}=\left({\frac {L}{L_{\odot }}}\right)\left({\frac {d}{1\ \mathrm {AU} }}\right)^{-2}}
  
 which for apastron gives 71% and for periastron gives 160%. However note that the eccentricity of 55 Cnc f is not well constrained - it is consistent with zero (which would give a factor of 102%) and indeed in the Newtonian fit taking planet-planet interactions into account (table 4 in Fischer et al. 2008 "Five planets orbiting 55 Cancri") the eccentricity is a very low 0.0002. Icalanise (talk) 22:22, 15 September 2008 (UTC)
First, before anyone who thinks they are really smart mentions it, the Insolation article says Flux is a synonym. Second, if you are going to critisize the eccentricity used in the formula then why haven't you changed it (still 0.2) in the article itself? Thirdly, I think the formula in  "FUNDAMENTALS OF ASTRONOMY" (QB43.3.B37 2006) by Dr.Cesare Barbieri, professor of Astronomy at the university of Pauda, Italy is the best planet (complete/unsimplified) Equilibium Temperature formula I have found yet.GabrielVelasquez (talk) 20:45, 20 September 2008 (UTC)
Or you can stick with your fantacy eccentricity and uselessly more complicated formulation when the simple L=4piR2aT4 and L=4pid2f  will do just fine to find f=R2aT4/d2. GabrielVelasquez (talk) 20:07, 1 October 2008 (UTC)
I have no idea why you are calling the eccentricity a fantasy - it is what is written in the discovery paper. How is my formulation I posted above any more complicated than yours, after doing some basic algebra it is exactly the same as what you have written! Icalanise (talk) 15:31, 2 October 2008 (UTC)
Exactly, and you don't even recognise when you have contradicted yourself: "This seems in error" you said of that formula, now when pointed out how simple it is you change your tune to say your formula is "exactly the same as you have written." The eccentricity in this article is 0.2 and it is referenced, and still in the article, so it is your 0.002 that is fantacy. In essesnce you give the same formula you critisized and no reason why it is incorrect.  GabrielVelasquez (talk) 18:48, 5 October 2008 (UTC)
I was criticising the results of the formula. You appear to be getting incorrect numbers out of the formula, I don't know why, perhaps you are putting incorrect numbers in. I also provided a reference for the value 0.0002 is also referenced in this article. I suggest you read the paper in question, particularly section 4.4 and table 4.  Icalanise (talk) 18:54, 5 October 2008 (UTC)

This does not seem correct - firstly 16 Cygni Bb's orbit takes it through the habitable zone, secondly even if we select the group of extrasolar planets whose orbits lie entirely within the habitable zone, HD 28185 b got there first. I'm removing the statement from the article. Icalanise (talk) 22:06, 15 September 2008 (UTC)

The highly eccentric orbit of 16 Cygni Bb would make it unlikely to harbor habitable environments.200.168.20.215 (talk) 12:34, 29 December 2009 (UTC)
I agree with Icalanise - HD 28185 b was first discovered in 2001 and found in 2006 to fully orbit withing its parent stars' habitable zone. See reference below:
a b Jones, Barrie W.; Sleep, P. Nick; Underwood, David R. (2006). "Habitability of Known Exoplanetary Systems Based on Measured Stellar Properties". The Astrophysical Journal 649 (2): 1010 – 1019. doi:10.1086/506557. http://iopscience.iop.org/0004-637X/649/2/1010 -- Kugellager (talk) 18:03, 28 October 2010 (UTC)
I just came across this "Eccentricity (e)  0.2 ± 0.2" in the articlethen the range of possible eccentricity could be zero to 0.4Besides that the Extra Solar planet encyclopeadia lists "Eccentricity 0.0002"     And my own calculations give this a solar constant (irradiance) 93% of the Earth's 1366 watts per meter squared. Maybe not first but definately notable. 24.79.40.48 (talk) 05:06, 7 June 2012 (UTC)

This review is transcluded from Talk:55 Cancri f/GA1. The edit link for this section can be used to add comments to the review.
Hi!  I will be reviewing this article for GA status, and should have the full review up soon. Dana boomer (talk) 19:37, 16 October 2008 (UTC)

GA review (see here for criteria)
It is reasonably well written.
a (prose):  b (MoS): 
The lead is supposed to be a summary of the entire article, and therefore should not contain original information, which means it doesn't need references for anything other than backing up direct quotes.
The first sentence of the Orbit and mass section is "55 Cancri f is located about 0.781 AU away from the star which takes 260 days to complete."  Is there something missing from this sentence?  It takes 260 days to complete what?  (I'm assuming its orbit around Cancri, but other readers may not realize this.)
It is factually accurate and verifiable.
a (references):  b (citations to reliable sources):  c (OR): 
The last half of the discovery section needs a reference.
As does the last half of the second paragraph of the Orbit and mass section.
It is broad in its coverage.
a (major aspects):  b (focused): 

It follows the neutral point of view policy.
Fair representation without bias: 

It is stable.
No edit wars etc.: 

It is illustrated by images, where possible and appropriate.
a (images are tagged and non-free images have fair use rationales):  b (appropriate use with suitable captions): 

Overall:
Pass/Fail: 

Overall, a nice article, very well written.  I have a couple of comments/questions about the prose, MOS and referencing, and so I am putting this article on hold so they can be resolved.  If you have any questions, drop me a note here on the review page or on my talk page. Dana boomer (talk) 19:47, 16 October 2008 (UTC)

I believe all the comments have been solved. Nergaal (talk) 02:32, 17 October 2008 (UTC)

Just to add a quick comment, the first sentence states that it is "approximately 41 light-years away". It seems to me that it should be clarified that this is "41 light-years away from Earth". GaryColemanFan (talk) 19:50, 16 October 2008 (UTC)

Oy, good catch!  I've gone ahead and added this myself. Dana boomer (talk) 19:59, 16 October 2008 (UTC)
Everything looks good, so I am passing the article. Nice work! Dana boomer (talk) 12:23, 17 October 2008 (UTC)

There is a move discussion in progress on Talk:Upsilon Andromedae d which affects this page. Please participate on that page and not in this talk page section. Thank you.  —RMCD bot 02:59, 17 December 2015 (UTC)

The image in the infobox should not be used as it does not follow the usage guide for the template:

This template is part of a group of templates that are used to display information about a specific extrasolar planet.
Images of published planetary properties are preferred where available, especially when they are available from cited publications.
Artist's conception, regardless of the source, should be avoided.
Examples of acceptable images include
* direct images, such as one used for GJ 758 b, in the rare cases where these are available;
* output of a model that is integral to a cited paper, such as the image used in HD 80606 b;
* user-generated images that clearly illustrate published properties, such as the size comparisons currently used in GJ 1214 b or Gliese 436 b.

My edits followed these guidelines but were revered by User:MarioProtIV. I'm opening discussion as to why ...
--EvenGreenerFish (talk) 09:26, 21 April 2017 (UTC)

I would like if the discussion was held here, rather then on all of the other pages. --MarioProtIV (talk/contribs) 11:13, 21 April 2017 (UTC)

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