Phil-X-
Active member
Its worth a shot, but for some reason I cant think straight at all right now and cant figure this out
We’ve come across the 1/r2 law once already – in Newton’s Law of Gravity. This "inverse square law" means that the quantity decreases with the square of the distance. With gravity, it is the force that gets smaller with distance. With the brightness of stars, it's the flux or the "energy received" that gets smaller. Flux is defined to be the amount of energy received from a star per area (for example Watts received per meter squared here on Earth). Luminosity (of a star or of any light) is proportional to 1/r2 or L = 1/r2 .
The symbol "r" stands for "radial distance" or just plain "distance", where the "distance"
is the length that separates the bright object from the one who is measuring the luminosity.
Recall from our discussion in class that m = apparent magnitude and M = absolute magnitude and these two are related in the following way:
m – M + 5 = 5 log D where D = distance given in parsecs
Note that 1 parsec = 3.26 light years.
[By the way, on most calculators there will be a button for obtaining the log. Calculators don't always say so, but generally they calculate what is called "log to the base 10". For example, if you want to calculate "log to the base 10 of the number 77", this can be written log10 77 or simply log 77. In this example, "log to the base 10 of 77" = 1.886.]
These are your "tools". You also need to know the absolute magnitude of the Sun. For these exercises, we set the Sun's absolute magnitude to be Mo = +4.8.
4. Calculate the apparent magnitude for a star that is identical to the Sun if we see it from a distance of 16.5 parsecs (approximately 54 light years away).
a. -1.02 b. +1.02 c. -5.88 d. +5.88
5. How far away would this same star (identical to the Sun) have to be if its apparent magnitude m = 15.35 ?
a. 3.75 pc b. 21.66 pc c. 795 pc d. 1288 pc
6. What is the relative apparent brightness of the stars described in questions 4 & 5? In other words, how much more light (or flux) is received from the star in question 4 when compared to the star in question 5?
a. 6093 x b. 78.1 x c. 35.0 x d. 12.7 x
7. If the flux from a star is 1.40 x 10-7 Watts/square meter when viewed from a distance of 16 parsecs, determine the flux from this star if viewed from a distance of 8 parsecs?
All of the following answers are given in units of Watts/square meter
a. 5.80 x 10-7 b. 9.60 x 10-7 c. 6.00 x 10-8 d. 4.76 x 10-8
kthx
We’ve come across the 1/r2 law once already – in Newton’s Law of Gravity. This "inverse square law" means that the quantity decreases with the square of the distance. With gravity, it is the force that gets smaller with distance. With the brightness of stars, it's the flux or the "energy received" that gets smaller. Flux is defined to be the amount of energy received from a star per area (for example Watts received per meter squared here on Earth). Luminosity (of a star or of any light) is proportional to 1/r2 or L = 1/r2 .
The symbol "r" stands for "radial distance" or just plain "distance", where the "distance"
is the length that separates the bright object from the one who is measuring the luminosity.
Recall from our discussion in class that m = apparent magnitude and M = absolute magnitude and these two are related in the following way:
m – M + 5 = 5 log D where D = distance given in parsecs
Note that 1 parsec = 3.26 light years.
[By the way, on most calculators there will be a button for obtaining the log. Calculators don't always say so, but generally they calculate what is called "log to the base 10". For example, if you want to calculate "log to the base 10 of the number 77", this can be written log10 77 or simply log 77. In this example, "log to the base 10 of 77" = 1.886.]
These are your "tools". You also need to know the absolute magnitude of the Sun. For these exercises, we set the Sun's absolute magnitude to be Mo = +4.8.
4. Calculate the apparent magnitude for a star that is identical to the Sun if we see it from a distance of 16.5 parsecs (approximately 54 light years away).
a. -1.02 b. +1.02 c. -5.88 d. +5.88
5. How far away would this same star (identical to the Sun) have to be if its apparent magnitude m = 15.35 ?
a. 3.75 pc b. 21.66 pc c. 795 pc d. 1288 pc
6. What is the relative apparent brightness of the stars described in questions 4 & 5? In other words, how much more light (or flux) is received from the star in question 4 when compared to the star in question 5?
a. 6093 x b. 78.1 x c. 35.0 x d. 12.7 x
7. If the flux from a star is 1.40 x 10-7 Watts/square meter when viewed from a distance of 16 parsecs, determine the flux from this star if viewed from a distance of 8 parsecs?
All of the following answers are given in units of Watts/square meter
a. 5.80 x 10-7 b. 9.60 x 10-7 c. 6.00 x 10-8 d. 4.76 x 10-8
kthx
