
Concept explainers
The factor by which the radius of the star and luminosity change, if in an outburst its surface temperature doubles, the average density decreases by a factor of 8, and its mass remains the same.

Answer to Problem 50Q
Solution:
The star’s radius becomes
Explanation of Solution
Given data:
The temperature of the star’s surface doubles.
The average density of the star decreases to
The mass of the star will remain the same.
Formula used:
The expression for density is:
Here,
The expression for the volume of a spherical body is:
Here,
The expression for the luminosity from the Stefan-Boltzmann’s law is:
Here,
Explanation:
Consider the star’s surface temperature before outburst to be
According to the problem, after the outburst, the surface temperature of the star doubles. Thus,
Consider the average density before outburst to be
According to the question, the average density of the star decreased by a factor of 8 after the outburst. Thus,
Consider the star’s radius before the outburst to be
According to the question, the mass of the star remains the same after the outburst. Thus,
Recall the expression for density and write it for before the outburst
Here,
Similarly, write the expression for the density of the star after the outburst.
Here,
Divide equations (1) and (2).
Simplify the above expression.
Rearrange the above expression for
Substitute
Solving the above expression further,
So, the radius becomes 2 times the initial value.
Recall the expression for luminosity and write it for before the outburst as.
Similarly, write the expression for the luminosity of the star after the outburst.
Divide equation (3) and (4).
Substitute
Solve the above expression further, for
So, after the outburst, the luminosity becomes 64 times the initial value.
Conclusion:
So, the luminosity increases by 64 times and the radius increases by 2 times.
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