Which of course reduces to [itex]\theta = 2r / d[/itex] (where [itex]\theta[/itex] comes out in radians) in the small-angle approximation, which surely holds for Jupiter viewed from the Earth (or even the moon or sun viewed from the earth).Janus said:[tex]\theta = 2 \arctan{\frac{r}{d}}[/tex]
The number you gave is the diameter, which is already twice the radius, so all you should need to do is divide those two numbers and then convert to arcseconds:Vast said:I’m getting about 19.114. Is that right? Which would be arcseconds, not in radians, so I shouldn’t need to convert? What am I doing wrong? Unless, I just need to multiply it by 2?
It's just the factor of conversion from radians to arcseconds. An arcsecond is equal to the angle subtended by earth-sun distance (AU) at a distance of parsec, so another way to write this is:Vast said:Thanks SpaceTiger. I got 39.429 seconds of arc, but I’m not sure where you got the number 206265 from. Can you explain that?