Question #61087

A main sequence star has mass 2×1031 kg and radius 3×109
m. Obtain an estimate of
the average temperature throughout the star.

Expert's answer

Answer on Question #61087, Physics / Astronomy | Astrophysics

A main sequence star has mass 2×1031kg2 \times 10^{31} \, \mathrm{kg} and radius 3×109m3 \times 10^{9} \, \mathrm{m} . Obtain an estimate of the average temperature throughout the star.

Find: Tstar?T_{\text{star}} - ?

Given:

mstar=2×1031 kg\mathrm{m_{star}} = 2 \times 10^{31} \mathrm{~kg}

rstar=3×109 m\mathrm{r_{star}} = 3 \times 10^{9} \mathrm{~m}

msun=1,9891×1030kg\mathrm{m_{sun} = 1,9891\times 10^{30}kg}

rsun=6,96×108m\mathrm{r_{sun} = 6,96\times 10^{8}m}

Tsun=5778 K\mathrm{T}_{\mathrm{sun}} = 5778 \mathrm{~K}

Solution:

The relationship between luminosity LL , temperature TT and radius rr :

Lstar=Lsunrstarrsun2TstarTsun4\mathrm{L_{star}} = \mathrm{L_{sun}}\quad \frac{\mathrm{r_{star}}}{\mathrm{r_{sun}}}^2\quad \frac{\mathrm{T_{star}}}{\mathrm{T_{sun}}}^4 (1)

m=mstarmsun(2)\mathrm{m} = \frac{\mathrm{m_{star}}}{\mathrm{m_{sun}}} (2)

Of (2) \Rightarrow m=10.1

If m=10.1m = 10.1 (2<m<20), then l=LstarLsun=m3.5l = \frac{L_{\text{star}}}{L_{\text{sun}}} = m^{3.5} (3)

(3) in (1): (10.1)3.5=3×1096,96×1092TstarTsun4(10.1)^{3.5} = \frac{3 \times 10^{9}}{6,96 \times 10^{9}}^2 \frac{T_{\text{star}}}{T_{\text{sun}}}^4 (4)

Of (4) TstarTsun=176.23\Rightarrow \frac{T_{\text{star}}}{T_{\text{sun}}} = 176.23 (5)

Of (5) TstarTsun=3.64\Rightarrow \frac{T_{\text{star}}}{T_{\text{sun}}} = 3.64 (6)

Of (6) \Rightarrow Tstar=21032 K

Answer:

21032 K

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