Question

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).

Answer 1

Answer:

Check the explanation

Explanation:

To tackle situations like the one above, the rate of gravity of that star must be equal to the rate of power output but we don’t have radius of that star. Also temp is not mentioned. And emissivity of star is also not mentioned. So the only possible way is like Einstein mass energy relationship E=mc^2=6.5380e40

power =E/Time so this energy is transferred per sec.

Answer 2

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:

$m_{s} =35.135*1.99x10^{30} =6.992x10^{31}kg$

The power output of the star:

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