# Angular size of Venus: Is my math correct? (1 Viewer)

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#### nukeman

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 ?

#### nukeman

what should this number be (9.565) - because im sure thats not right.

Anyone?

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#### Ich

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″.

#### nukeman

Ahh, ok.

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

So, lets 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?

#### Ich

9.181x10^-05 is a dimensionless angle. There's a "dimension" defined for angles, called "http://en.wikipedia.org/wiki/Radian" [Broken]".
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 ('').

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#### nukeman

9.181x10^-05 is a dimensionless angle. There's a "dimension" defined for angles, called "http://en.wikipedia.org/wiki/Radian" [Broken]".
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!

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#### nukeman

so does this mean that MARS would be more visable from earth?

#### Ich

Yes, if your numbers are correct.

#### qraal

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.

#### granpa

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
Now the diameter of Venus is 12104km
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

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