As new-forming stars grow by gravitationally attracting more material, they become brighter. If a star becomes too bright, the radiation pressure from the star prevents gravity from pulling in more material. What power output PP would a star with a mass 35.135.1 times that of our Sun need to have in order to be unable to pull in a 3.96 μg3.96 μg dust particle of radius rp=0.935 mm,rp=0.935 mm, assuming the particle absorbs all the light incident on its πr2pπrp2 cross‑section? The mass ????⊙M⊙ of the Sun is 1.99×1030 kg,1.99×1030 kg, and the gravitational constant GG equals 6.67×10−11 m3/(kg·s2).

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Answer:

The power output of the star is 2.215x10²⁸W

Explanation:

Given data:

ms = mass of the star = 35.135 times msun

rp = radius = 0.935 mm = 9.35x10⁻⁴m

p = 3.46 μg = 3.46x10⁻⁹g

msun = mass of the Sun = 1.99x10³⁰kg

G = gravitational constant = 6.67x10⁻¹¹m³/kg s²

c = speed of light = 3x10⁸m/s

Question: What power ouput would a star, P = ?

First, calculate the mass of the star:

[tex]m_{s} =35.135*1.99x10^{30} =6.992x10^{31}kg[/tex]

The power output of the star:

[tex]P=\frac{Gm_{s}*p*4c }{r^{2} } =\frac{6.67x10^{-11}*6.992x10^{31}*3.46x10^{-9}*4*3x10^{8}}{(9.35x10^{-4})^{2} } =2.215x10^{28}W[/tex]

The power output of a star with the given mass is [tex]2.53 \times 10^{28} \ W[/tex].

The given parameters;

  • mass of the sun, m = 1.99 x 10³⁰ kg
  • gravitational constant, G = 6.67 x 10⁻¹¹ kg
  • radius of the dust particle, rp = 0.935 mm

The power output of a star is calculated as follows;

[tex]P = \frac{GMp \times (4c)}{r^2}[/tex]

where;

M is the mass of the star

c is speed of light

The mass of the star is calculated as follows;

[tex]M = 35.135 \times 1.99 \times 10^{30}\\\\M = 6.99 \times 10^{31} \ kg[/tex]

[tex]P = \frac{6.67 \times 10^{-11} \times 6.99 \times 10^{31} \times 3.96 \times 10^{-9} \times 4 \times 3\times 10^8}{(0.935 \times 10^{-3})^2} \\\\P = 2.53 \times 10^{28} \ W[/tex]

Thus, the power output of a star with the given mass is [tex]2.53 \times 10^{28} \ W[/tex].

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