by henk21cm » Thu May 01, 2008 7:19 pm
Sputnick wrote:Would a planet in a relatively dense part of these clusters have no nightime? (the night sky being as bright as the day sky?)
I don't think so. Two simple estimates.
1) uniform distribution of stars. The diameter of the globular cluster is 150 ly. The volume is (4*π*R^3)/3, so 14 million ly^3. There are 10 million stars, so 0.7 star per qubic ly. That is comparable to our Sol.
2) distribution inversely proportional to the distance to the center. The distribution of stars, ρ is therefore ρ = α/r We do not yet know the value of α , however the number of stars, N, is know and it is equal to:
N = ∫ dr ∫ d φ ∫ d θ ρ cos(θ ) r^2
while integrating over r=0,R, φ =0,2π , θ = -π/2, π/2
Evaluating these integrals leads to:
N= 2παR^2, so α = 1E7/(2π*150^2) ≅ 70 /ly^3
When we locate our planet at 1 ly from the center, we count 70 stars per qubic ly. For simplicity sake we assume 64 stars per ly^3. That is easy, so the distance between the stars is then 0.25 ly. When we see our sun at a distance of 0.25 ly, i.e. 16 kAU, his brightness will be 256 million times less than at earth. Since a magnitude is a factor 2.5, the star will be 21 magnitudes less bright. Our sun at earth has a magnitude of -27, so the brightness of the closest star is -6, roughly the same as Venus, when she is brightest.
There are 4 of these bright Venusses. Then there are 4 lesser bright stars, at 1.4 times the distance, so twice less bright. And so on. As a result i do not see daylight during night. Not even as much as during full moon.
Note that this sort of approximation implicitly uses extinction by dusty clouds, since i stopped calculating at the next nearest neighbours. Otherwise it might lead to something similar to
Olbers Paradox.
[quote="Sputnick"]Would a planet in a relatively dense part of these clusters have no nightime? (the night sky being as bright as the day sky?) [/quote]
I don't think so. Two simple estimates.
1) uniform distribution of stars. The diameter of the globular cluster is 150 ly. The volume is (4*π*R^3)/3, so 14 million ly^3. There are 10 million stars, so 0.7 star per qubic ly. That is comparable to our Sol.
2) distribution inversely proportional to the distance to the center. The distribution of stars, ρ is therefore ρ = α/r We do not yet know the value of α , however the number of stars, N, is know and it is equal to:
N = ∫ dr ∫ d φ ∫ d θ ρ cos(θ ) r^2
while integrating over r=0,R, φ =0,2π , θ = -π/2, π/2
Evaluating these integrals leads to:
N= 2παR^2, so α = 1E7/(2π*150^2) ≅ 70 /ly^3
When we locate our planet at 1 ly from the center, we count 70 stars per qubic ly. For simplicity sake we assume 64 stars per ly^3. That is easy, so the distance between the stars is then 0.25 ly. When we see our sun at a distance of 0.25 ly, i.e. 16 kAU, his brightness will be 256 million times less than at earth. Since a magnitude is a factor 2.5, the star will be 21 magnitudes less bright. Our sun at earth has a magnitude of -27, so the brightness of the closest star is -6, roughly the same as Venus, when she is brightest.
There are 4 of these bright Venusses. Then there are 4 lesser bright stars, at 1.4 times the distance, so twice less bright. And so on. As a result i do not see daylight during night. Not even as much as during full moon.
Note that this sort of approximation implicitly uses extinction by dusty clouds, since i stopped calculating at the next nearest neighbours. Otherwise it might lead to something similar to [url=http://en.wikipedia.org/wiki/Olbers"_paradox]Olbers Paradox[/url].