^{6}km) and d represents the distance from the Earth to the Sun (149,598,261 km). The radius of the Sun (r) will be half of D (r = D/2). Delta (d) is the angular diameter of the Sun from the distance of the Earth and alpha (a) is half of this (a = d/2).

Since d and r form two legs of a right triangle, simple trigonometry will tell us that the sine of the angle a is r / d.

sin(a) = (r/d)

Substituting (r = D/2) and (a = d/2) gives us:

sin(d/2) = [(D/2)/d]

or

d = 2 sin

^{-1}[(D/2)/d]
Therefore, for the Sun, we can calculate its angular diameter as:

d = 2 sin

^{-1}[(1.392 x 10^{6}km/2)/149,598,261 km] = 0.533° (half a degree)
OK, we know that. But we can use this same procedure to calculate the angular diameter of the Sun from each of the other 7 major planets in our solar system. All we need to do is look up their average distance from the Sun which we can do easily enough.

Mercury d = 2 sin

^{-1}[(1.392 x 10^{6}km/2)/57,910,000 km] = 1.4°
Venus d = 2 sin

^{-1}[(1.392 x 10^{6}km/2)/108,200,000 km] = 0.74°
Mars d = 2 sin

^{-1}[(1.392 x 10^{6}km/2)/227,940,000 km] = 0.35°
Jupiter d = 2 sin

^{-1}[(1.392 x 10^{6}km/2)/778,330,000 km] = 0.10°
Saturn d = 2 sin

^{-1}[(1.392 x 10^{6}km/2)/1,429,400,000 km] = 0.06°
Uranus d = 2 sin

^{-1}[(1.392 x 10^{6}km/2)/2,870,990,000 km] = 0.03°
Neptune d = 2 sin

^{-1}[(1.392 x 10^{6}km/2)/4,504,000,000 km] = 0.02°
How does this look diagrammatically?

The Sun would look a bit smaller in the Martian sky and a bit larger in the Venusian sky. The Sun from Mercury would be frighteningly large (and fry you with temperatures around 800° F). The Sun is much smaller from the outer gas giant planets since they're so much further from the Sun than the inner rocky planets. It would be much brighter than any of the other stars in the sky, but not much larger.

Another interesting tidbit. The Moon has a diameter of 3476 km and is 384,400 km from Earth. Plugging these values into the above equation gives:

Moon d = 2 sin

^{-1}[( 3476 km/2)/384,400 km] = 0.52°
So, even though the Moon is much, much smaller than the Sun, it's also much, much closer and looks about the same size in our sky (half a degree). That's why solar eclipses are possible - the Moon can block our view of the Sun given a favorable orbital geometry.

Trigonometry is cool.

Trigonometry is cool indeed. Thanks for the calculations! The graphical comparison of the Sun's angular size is really impressive.

ReplyDeleteDue to my calculations, the angular size of Sun from the nearest star (Proxima Centauri) is ca 0.007 seconds of arc. Not much, to say the least.

Interesting. What would the Sun look like from 700,000 miles? (How close Comet ISON came as it disintegrated while whipping by the Sun?

ReplyDelete700,000 mi is about 1.13 x 10^6 km. Plugging that in above gives an angular diameter of 63 degrees. That would be as if the Sun stretched from the horizon 2/3 of the way up to the zenith.

ReplyDeleteAmazing. Now I've heard that, somehow, the comet survived it's close shave.

DeleteThanks for the interesting discussion. One thing that dramatically simplifies calculations like this is the small angle approximation. With a small angle around half of a degree, there's virtually no difference between the result you get by doing 2*arctan((a/2)/b) and arctan(a/b). In fact, these angles are so small that you still get a very accurate answer by simply approximating the angle in radians as a/b and converting to degrees. This gives the same answer as the original calculation to 4 decimal places (which is how many significant digits are present in the measurements used above). An intuition for why this is correct is to realize that an angle in radians is simply the ratio of the arc length of a segment of a circle to the circle's radius. Drawing a picture of an extremely narrow arc as well as of a right triangle with the same angle, whose hypotenuse is the same as the arc's radius, will show that these two ratios approach each other the smaller the angle gets.

ReplyDelete