Astronomy –odd numbered Homework
Chapter 8
1.
if a star has a parallax of .050 second of arc :
ð = 20 * 3.26 ly = 65.2 ly
ð = 65.2 * 63,241.1 = 4,123,320 AU
3.
if a star has a parallax of .016 second of arc, and an apparent magnitude of 6
ð distance
= 1/ .016 = 62.5 parsecs = (62.5 * 3.26) ly = 203.75 ly
 with this answer, the absolute magnitude is 2
5.
d1 = .5 km = 500 m,
The absolute magnitude of an object at a distance of 10 pc = 3.086e17 m.
d2 = 3.086 e17 m
F = (d2/d1 )^2 = [( 3.086 e17)/(500)]^2 = 3.81 e29
we have M1 = 6
According to this formula,
the absolute magnitude of a firefly M2 is
M2 = M1 + 2.5 * log (F)
= 6 + 2.5 * log (3.81e29) ~ 80
7.
we know that,
the temperature of the sun T_sun is 5,780 k à Half of that T_half is 2,890
K
if the star is ten times the radius of the sun and half as hot, the luminosity is :
= radius^2 (T_half / T_sun) ^4 = 10^2(2,890/5780)^4 = 6.25 Solar luminosity's
So, the star will be 6.25 times more luminous than the sun.
9.
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 Fall '11
 K.Ostrowski
 Astronomy, absolute magnitude, 0.5 m, 5 km, Ð

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