5. The absolute magnitude of a star, M_(5) is related to its luminosity, L,by the formula M=4.72-log(L)/(L_(0)) , where L_(0) is the luminosity of the sun. The luminosity is the rate at which the star emits light, and is measured in watts. a) Determine the absolute magnitude of the sun. b) The absolute magnitude of Sirius, the star that,other than the sun.appears brightest from Earth, is 1.41 . Is Sirius more or less luminous than the sun?By what factor? c) Repeat part b)for Canopus the star that appears second brightest from Earth,which has an absolute magnitude of -4.7 d) A distant object called a quasar has a luminosity of about 10^38W The sun's luminosity is about 4times 10^26W Determine the absolute magnitude of the quasar.

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5. The absolute magnitude of a star, M_(5) is related to its luminosity, L,by the formula M=4.72-log(L)/(L_(0)) , where L_(0) is the luminosity of the sun. The luminosity is the rate at which the star emits light, and is measured in watts. a) Determine the absolute magnitude of the sun. b) The absolute magnitude of Sirius, the star that,other than the sun.appears brightest from Earth, is 1.41 . Is Sirius more or less luminous than the sun?By what factor? c) Repeat part b)for Canopus the star that appears second brightest from Earth,which has an absolute magnitude of -4.7 d) A distant object called a quasar has a luminosity of about 10^38W The sun's luminosity is about 4times 10^26W Determine the absolute magnitude of the quasar.

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**a) Determine the absolute magnitude of the sun.**The absolute magnitude of the sun is calculated by using the given formula and setting L = L₀ (since we're considering the sun itself):M = 4.72 - log(L₀/L₀)M = 4.72 - log(1)M = 4.72 - 0M = 4.72Therefore, the absolute magnitude of the sun is 4.72.**b) The absolute magnitude of Sirius is 1.41. Is Sirius more or less luminous than the sun? By what factor?**Let Mₛ be the absolute magnitude of Sirius (1.41) and Lₛ be its luminosity. We know the sun's absolute magnitude (M⊙ = 4.72) and its luminosity (L₀). We can set up two equations:1.41 = 4.72 - log(Lₛ/L₀)4.72 = 4.72 - log(L₀/L₀)Subtracting the second equation from the first:-3.31 = -log(Lₛ/L₀) + log(L₀/L₀)3.31 = log(Lₛ/L₀)Now, we can solve for the ratio Lₛ/L₀:Lₛ/L₀ = 10³·³¹ Lₛ/L₀ ≈ 2041.7Since Lₛ/L₀ > 1, Sirius is more luminous than the sun. It is approximately 2042 times more luminous.**c) Repeat part b) for Canopus, which has an absolute magnitude of -4.7.**Let Mc be the absolute magnitude of Canopus (-4.7) and Lc be its luminosity.-4.7 = 4.72 - log(Lc/L₀)-9.42 = -log(Lc/L₀)9.42 = log(Lc/L₀)Lc/L₀ = 10⁹·⁴²Lc/L₀ ≈ 2.63 x 10⁹Canopus is much more luminous than the sun, by a factor of approximately 2.63 billion.**d) A quasar has a luminosity of about 10³⁸ W. The sun's luminosity is about 4 x 10²⁶ W. Determine the absolute magnitude of the quasar.**Let Lq be the luminosity of the quasar (10³⁸ W) and Mq be its absolute magnitude.Mq = 4.72 - log(Lq/L₀)Mq = 4.72 - log(10³⁸ / 4 x 10²⁶)Mq = 4.72 - log(2.5 x 10¹¹)Mq = 4.72 - 11.40Mq = -6.68The absolute magnitude of the quasar is approximately -6.68.

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