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

`L_f = K ((r)/(100))^2 * (2T)^4`

`L_f = K (r^2)/(10000) * 16 T^4`

`L_f = (16)/(10000) k r^2 T^4 = (1)/(625) k r^2 T^4`

`L_f = (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

Step-by-step 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 ((r)/(100))^2 * (2T)^4`

`L_f = K (r^2)/(10000) * 16 T^4`

`L_f = (16)/(10000) k r^2 T^4 = (1)/(625) k r^2 T^4`

`L_f = (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