Question #58698

How long will a 5MΘ star burn hydrogen as fuel, given that the Sun will do so for about 10^10
years?
1

Expert's answer

2016-03-26T09:37:04-0400

Answer on question #58698, Physics / Astronomy | Astrophysics

How long will a 5M5\mathrm{M}_{\odot} star burn hydrogen as fuel, given that the Sun will do so for about 101010^{\wedge}10 years?

Solution:

Star's lifetime equals to t=EreleasedLt = \frac{E_{\text{released}}}{L} , where EreleasedE_{\text{released}} - energy released after all hydrogen fusion into helium, LL - star luminosity.

Solar luminosity equals to 3.8281033erg/s3.828 \cdot 10^{33} \, \text{erg/s} . For the star with M=5MM = 5M_{\odot} the released amount of energy will be proportional to the mass, i.e.


5Ereleased=51010y31556926sy3.8281033ergs=51.211051erg.5 \cdot E _ {r e l e a s e d} = 5 \cdot 1 0 ^ {1 0} y \cdot 3 1 5 5 6 9 2 6 \frac {s}{y} \cdot 3. 8 2 8 \cdot 1 0 ^ {3 3} \frac {e r g}{s} = 5 \cdot 1. 2 1 \cdot 1 0 ^ {5 1} e r g.


The increasing of mass leads to the increase of luminosity which is roughly proportional to LL=(MM)α\frac{L}{L_{\odot}} = \left(\frac{M}{M_{\odot}}\right)^{\alpha} , where a3.9a \approx 3.9

LL=53.9532.\frac {L}{L _ {\odot}} = 5 ^ {3. 9} \approx 5 3 2.


Then L=532L=5323.8281033erg/sL = 532 * L_{\odot} = 532 \cdot 3.828 \cdot 10^{33} \, \text{erg/s}

The final expression as follows


t=51.211051erg5323.8281033erg/s=9.4107yt = \frac {5 \cdot 1 . 2 1 \cdot 1 0 ^ {5 1} e r g}{5 3 2 \cdot 3 . 8 2 8 \cdot 1 0 ^ {3 3} e r g / s} = 9. 4 \cdot 1 0 ^ {7} y


Answer: the star will burn its fuel in 9.4107y9.4 \cdot 10^{7} y

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