You have discovered a new star in the Mitky Way: Your new star is red and has 3/5 the temper:ature
of our Sun. The new star emits a total power that is 100,000 times greater than the power
emitted by our Sun.
(a) Determine the spectraltype (i.e. spectral ctassification) of the new star.
(b) How many times bigger is the radius of the new star compared to the radius of our Sun?
Answers
Answer:
To determine the spectral class of the star we need to know its temperature. We can determine it by using the temperature of the Sun:
T_{star}=\frac 3 5 T_{Sun} = \frac 3 5 * 5778K = 3466.8KTstar=53TSun=53∗5778K=3466.8K
As the temperature of this star is below 3500K and its color is red, then it belongs to the spectral type M.
We can determine the radius of the new star by using the Stefan-Boltzmann formula:
L=\sigma A T^4=4\pi R^2 \sigma T^4;L=σAT4=4πR2σT4;
Where L - the luminosity of the star (emission power), A - the total surface of the star, T - the temperature in Kelvin. As the star has a spherical shape, we can transform A into 4*pi*R2.
The total emission power of the new star is 100,000 times bigger than the power of the Sun, so we can make an equation:
\frac {L_{Star}} {L_{Sun}} =100,000;LSunLStar=100,000;\frac {4\pi R_{star}^2 \sigma T_{Star}^4;} {4\pi R_{Sun}^2 \sigma T_{Sun}^4;} =100,000;4πRSun2σTSun4;4πRstar2σTStar4;=100,000;\frac {R_{star}^2 T_{Star}^4;} {R_{Sun}^2 T_{Sun}^4;} =100,000;RSun2TSun4;Rstar2TStar4;=100,000;
Tstar is equal to 3/5 of the Sun's temperature, so we can transform the equation above:
\frac {R_{star}^2 (\frac 3 5 T_{Sun})^4;} {R_{Sun}^2 T_{Sun}^4;} =100,000;RSun2TSun4;Rstar2(53TSun)4;=100,000;\frac {81 R_{star}^2 T_{Sun}^4;} {625 R_{Sun}^2 T_{Sun}^4;} =100,000;625RSun2TSun4;81Rstar2TSun4;=100,000;R_{star} =\sqrt {\frac {625*100,000*R_{Sun}^2} {81}};Rstar=81625∗100,000∗RSun2;R_{star} =\frac {25*100*\sqrt {10}*R_{Sun}} {9};Rstar=925∗100∗10∗RSun;R_{star} =878.41 R_{Sun};Rstar=878.41R