Losing the Sun
Classical Mechanics
Level
pending
The Sun's absolute magnitude (i.e. its magnitude at a distance of 10 parsecs or 32.6 light years) is 4.8, which is about three times brighter than the faintest stars visible to the unaided eye. How far from us would the Sun have to be for it to be as dim as the faintest naked eye stars?
100 light years
57 light years
45 light years
72 light years
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
The apparent magnitude (m) of the dimmest naked eye stars is 6.0.
The Sun's absolute magnitude (M) = 4.8
We use the distance modulus equation
m - M = 5log(d) - 5
We need to calculate the distance (d) at which the Sun's apparent magnitude would equal 6
Remember, in this equation, d is expressed in parsecs.
m - M => 6 - 4.8 = 1.2
1.2 = 5log(d) - 5
6.2 = 5log(d)
6.2/5 = log(d)
1.24 = log(d)
d = anti-log (1.24) or 10^1.24 = 17.4 parsecs
1 parsec = 3.26 light years
17.4 x 3.26 = 56.7 rounded up to 57 light years