Angular size of Venus: Is my math correct?

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Discussion Overview

The discussion revolves around calculating the angular size of Venus and Mars from Earth, including the necessary mathematical formulas and conversions between different units of angular measurement. Participants explore the implications of their calculations and the visibility of the planets based on their distances and sizes.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Exploratory
  • Debate/contested

Main Points Raised

  • One participant presents an initial calculation for Venus's angular size, suggesting a value of 9.565 and later 7.738, but questions the correctness of the initial number.
  • Another participant challenges the use of the number 9.565, clarifying that it represents a range in arc seconds and not a direct input for calculations.
  • A different participant provides a formula for angular diameter, calculating it as 8.091E-5 radians, which converts to 16.69 arc seconds.
  • Further, a participant inquires about calculating the angular size for Mars, using its diameter and distance from Earth, arriving at a dimensionless angle of 9.181x10^-05.
  • Another participant explains the conversion of this dimensionless angle into radians, degrees, arc minutes, and arc seconds, confirming the calculations as correct.
  • There is a question about whether Mars would be more visible from Earth compared to Venus, with one participant affirming this if the calculations are accurate.
  • A participant discusses the orbital distances of Venus and Mars, noting that despite Venus being closer, its visibility is affected by its position relative to the Sun.
  • Another participant presents a different calculation for Venus's angular size, incorporating additional details about the diameter of the human eye and retinal cell spacing.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of initial calculations and the interpretation of angular measurements. There is no consensus on the accuracy of the initial values presented, and multiple approaches to the calculations are discussed without resolution.

Contextual Notes

Some calculations depend on specific assumptions about distances and definitions of angular measurements. The discussion includes various methods for converting between units, which may not be universally agreed upon.

Who May Find This Useful

Individuals interested in astronomy, particularly those exploring planetary visibility and angular measurements, as well as those studying mathematical conversions in scientific contexts.

nukeman
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Ok, just trying to figure out the angular size of venus. I have charted and figured out the distance between Earth and Venus at this certain time is: 149,600,000.00km

Now the diameter of Venus is 12104km.

venus angular diameter is: 9.565

so is it 9.565 x 12104/149,600,000

Is that correct?

So the answer would be: 7.738 ?
 
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what should this number be (9.565) - because I am sure that's not right.

Anyone?
 
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Sorry, that makes no sense.

nukeman said:
venus angular diameter is: 9.565
I see you copied this number from Wikipedia.
Look at the complete figure:
Wikipedia said:
9.565″ – 66.012″
First, there's a ″. It means arc seconds.
Second, that's a range, with 9.565 being the minimum value.
Third, that's actually the result you're asking for, not something to plug into a formula w/o motivation.

Here's how to do it:
Angular diameter = diameter/distance = 12104/149,600,000 =8.091E-5 = 16.69″.
 
Ahh, ok.

How did you get 8.091E-5 turned into = 16.69

So, let's say for mars. I need to find the Angular Size

Mars diameter is 6794km, and distance from Earth is: 74,000,000km

It would be: 9.181x10^-05 ? How do I get the angular size from that?
 
9.181x10^-05 is a dimensionless angle. There's a "dimension" defined for angles, called "http://en.wikipedia.org/wiki/Radian" ".
Look up the conversion tables.
So you have 9.181x10^-05 rad = 0,00526° = 0.31562' = 18.937''.
That's rad, degrees (°), arc minutes ('), arc seconds ('').
 
Last edited by a moderator:
Ich said:
9.181x10^-05 is a dimensionless angle. There's a "dimension" defined for angles, called "http://en.wikipedia.org/wiki/Radian" ".
Look up the conversion tables.
So you have 9.181x10^-05 rad = 0,00526° = 0.31562' = 18.937''.
That's rad, degrees (°), arc minutes ('), arc seconds ('').

Great thanks!

so this
9.181x10^-05 rad = 0,00526° = 0.31562' = 18.937''

is that correct calculation (on my part?)

Great, thanks!
 
Last edited by a moderator:
so does this mean that MARS would be more visable from earth?
 
Yes, if your numbers are correct.
 
Venus has an average orbital radius from the Sun of 0.72 AU while Mars' is 1.52 AU, which means the minimum distance between Earth and Venus is just 0.28 vs 0.52 AU between Earth and Mars. The very closest approaches are a little bit less less due to their elliptical orbits, but their perihelia don't perfectly coincide so that figure would be a bit of an underestimate. But, as you can see, Venus should look bigger than Mars - BUT because Venus is an interior planet all we see is it's shaded back when the two planets are at their closest. A thin crescent at best. So when it is biggest it is also at it's dimmest and very close to the Sun in the sky, so hard to see.
 
  • #10
nukeman said:
Ok, just trying to figure out the angular size of venus.
I have charted and figured out the distance between Earth and Venus at this certain time is: 149,600,000.00 km[/color]
Now the diameter of Venus is 12104km[/color]

12,104 / 149,600,000 = 0.00008 radians

the diameter of an eyeball is 2.5 cm (25,000 micrometers)
retinal cone cells are packed 50 per 100 micrometres in the most central fovea
thats 2 micrometer apart
2 / 25,000 = 0.00008 radians

according to wolfram alpha the current distance to venus is 0.3343 au = 5 * 10^7 km
http://www.wolframalpha.com/input/?i=venus

12,104 / 50,000,000 = 0.000242 radians
 
Last edited:

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