What is the distance from which our Sun appears to be just one of many stars in space rather than being obviously larger and brighter than all the others?
We need to first mention that the smallest size which the human eye can easily resolve an object is approximately one arc-minute, or one sixtieth of a degree. So, the Sun will appear as a ‘dot’ when its angular diameter equals or is less than one arc-minute. At what distance will the Sun’s angular diameter equal one arc-minute?
We’ll determine this distance by using the following formula:
α=2arctan(D/2r)
where α = is the object’s angular diameter; D = object’s diameter; r = distance from object.
one arc-minute = 1/60th of a degree = 0.0167
the sun’s true diameter = 865,370 miles
Assuming that our calculation is correct, we determine:
The distance at which the Sun’s angular diameter of one arc-minute is approximately 2.969 billion miles, or about at Pluto’s perihelion, or point of least separation distance from the Sun.*
However, at this distance, the Sun will still appear about 185 times brighter than the full moon. (One could still read a book by Sunlight at this distance.)
How far would one have to travel in order for the Sun to appear as bright as Sirius, the brightest star in the night sky?
At approximately 1.46 light years from the Sun, our parent star and Sirius will appear equally bright.**
This assumes that one is traveling toward Sirius. This equals the distance of the remotest Oort Cloud shell, at which the Sun’s gravitational influence becomes quite tenuous.
As a person’s separation distance from the Sun increases, its apparent magnitude (brightness) decreases. The Sun will actually vanish from our view at a distance of about 56 light years.**
*Pluto’s distance varies between perihelion (about 2.7 billion miles) to aphelion (4.58 billion miles)
**d⋅10M−4.855/10^(M−4.855/5) +1 = d☉
The formula for calculating the distance at which the Sun and a star with a given absolute magnitude appear equally bright.
*** m- M = 5logd – 5
m = apparent magnitude; M = absolute magnitude; d = distance in parsecs
We assume that the dimmest star’s apparent magnitude is 6.0. The Sun’s absolute magnitude equals 4.83.
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