Universe
Universe
11th Edition
ISBN: 9781319039448
Author: Robert Geller, Roger Freedman, William J. Kaufmann
Publisher: W. H. Freeman
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Chapter 17, Problem 50Q
To determine

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.

Expert Solution & Answer
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Answer to Problem 50Q

Solution:

The star’s radius becomes 2 times its initial value and luminosity becomes 64 times the initial value.

Explanation of Solution

Given data:

The temperature of the star’s surface doubles.

The average density of the star decreases to 18 of the initial value.

The mass of the star will remain the same.

Formula used:

The expression for density is:

ρ=MV

Here, ρ is the density, M is the mass, and V is the volume.

The expression for the volume of a spherical body is:

V=43πr3

Here, r is the radius of the body.

The expression for the luminosity from the Stefan-Boltzmann’s law is:

L=4πr2σT4

Here, L is the luminosity, σ is the Stefan-Boltzmann constant, and T is the temperature of the body.

Explanation:

Consider the star’s surface temperature before outburst to be T1 and after the outburst, it is T2.

According to the problem, after the outburst, the surface temperature of the star doubles. Thus,

T2=2(T1)

Consider the average density before outburst to be ρ1 and after the outburst, it is ρ2.

According to the question, the average density of the star decreased by a factor of 8 after the outburst. Thus,

ρ2=ρ18

Consider the star’s radius before the outburst to be r1 and after the outburst, it is r2; the luminosity of the star before the outburst to be L1 and after the outburst, it is L2; the mass of the star before the outburst is M1 and after the outburst, it is M2.

According to the question, the mass of the star remains the same after the outburst. Thus,

M1=M2

Recall the expression for density and write it for before the outburst

ρ1=M1V1=M143πr13 ...... (1)

Here, 43πr13 is the volume of the star before the outburst.

Similarly, write the expression for the density of the star after the outburst.

ρ2=M1V1=M243πr23 ...... (2)

Here, 43πr23 is the volume of the star after the outburst.

Divide equations (1) and (2).

ρ1ρ2=M143πr13M243πr23

Simplify the above expression.

ρ1ρ2=M1r13M2r23

Rearrange the above expression for r1r2.

(r1r2)3=ρ2M1ρ1M2

Substitute M1 for M2 and 8ρ2 for ρ1 in the above expression.

(r1r2)3=ρ28(ρ2)=18

Solving the above expression further,

(r1r2)3=(12)3r1r2=12r2=2r1

So, the radius becomes 2 times the initial value.

Recall the expression for luminosity and write it for before the outburst as.

L1=4πr12σT14 ...... (3)

Similarly, write the expression for the luminosity of the star after the outburst.

L2=4πr22σT24 ...... (4)

Divide equation (3) and (4).

L1L2=4πr12σT144πr22σT24=r12T14r22T24

Substitute 2r1 for r2 and 2T1 for T2 in the above expression.

L1L2=r12T14(2r1)2(2T1)4=14(16)=164

Solve the above expression further, for L2,

L2=64(L1)

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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Chapter 17 Solutions

Universe

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