Verified answer

First lets find the Apparent magnitude of the whole cluster.

L= [(n1) x (10^(1.9-0.4(m1)] + [(n2) x (10^(1.9-0.4(m2)]

m = 4.75 - 2.5 log(L)

L is the luminosity of the cluster

n1 is the number of stars with the first apparent magnitude, 80,000

m1 is the first apparent magnitude, 4

n2 is the numbers of stars with the second apparent magnitude, 160,000

m2 is the second apparent magnitude, 5

m is the integrated apparent magnitude, which is what we have to find.

L = [(80,000) x (10^(1.9-0.4(4)] + [(160,000) x (10^(1.9-0.4(5)]

L = 159,621 + 127,092.5

L = 286,713.5

m = 4.75-2.5 log(286,713.5)

m = -8.89 «------------- Apparent magnitude of the cluster

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Now lets solve for the Absolute Magnitude

M = m-5 x log(D/10)

M is the absolute magnitude

m is the apparent magnitude, -8.89

D is the distance to the cluster in parsecs, 10,720 parsecs

M = (-8.89)-5 x log(10,720/10)

M = -24.04 «---------- Absolute Magnitude of the cluster

This is a very bright cluster!

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Actual Diameter(km) = 150,000,000 x distance(parsecs) x angular diameter(arc seconds)

Actual Diameter = 150,000,000 x 10,720 x 1,090

Actual Diameter = 1.75272 × 10^15 km = 185.266 light years