Question

A protostar's radius decreases by a factor of 100 and its surface temperature increases by a factor of two before it becomes a main sequence star. Its luminosity: A. Decreases by a factor of 160,000 B. Decreases by a factor of 625 C. Decreases by a factor of 40 D. Increases by a factor of 40 E. Increases by a factor of 625 F. Increases by a factor of 160,000

Answer 1

Answer:

$L_f = K (\frac{r}{100})^2 * (2T)^4$

$L_f = K \frac{r^2}{10000} * 16 T^4$

$L_f = \frac{16}{10000} k r^2 T^4 = \frac{1}{625} k r^2 T^4$

$L_f = \frac{1}{625} L_i$

So then we see that the final luminosity decrease by a factor of 625 so then the correct answer for this case would be:

B. Decreases by a factor of 625

Explanation:

For this case we can use the formula of luminosity in terms of the radius and the temperature given by:

$L_i = K r^2 T^4$

Where L_i = initial luminosity, r= radius and T = temperature.

We know that we decrease the radius by a factor of 100 and the temperature increases by a factor of 2 so then the new luminosity would be:

$L_f = K (\frac{r}{100})^2 * (2T)^4$

$L_f = K \frac{r^2}{10000} * 16 T^4$

$L_f = \frac{16}{10000} k r^2 T^4 = \frac{1}{625} k r^2 T^4$

$L_f = \frac{1}{625} L_i$

So then we see that the final luminosity decrease by a factor of 625 so then the correct answer for this case would be:

B. Decreases by a factor of 625