21st Century Astronomy
21st Century Astronomy
6th Edition
ISBN: 9780393428063
Author: Kay
Publisher: NORTON
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Chapter 16, Problem 44QP

(a)

To determine

The luminosity at 108K.

(a)

Expert Solution
Check Mark

Answer to Problem 44QP

The luminosity at 108K is 1.8×1013LO.

Explanation of Solution

In terms of solar units, the luminosity law is given by,

    L(LO)=R(RO)2T(TO)4

Here, L(LO) is the luminosity in solar units, R(RO) is the radius in solar units and T(TO) is the temperature in solar units.

Radius in solar units is,

    R(RO)=(107m)(RO7×108m)=0.014RO

Temperature in solar units is,

    T(TO)=T5780K

Using the expressions for R(RO) and T(TO) in the equation for L(LO),

    L(LO)=(0.014)2(T5780K)4=1.8×1019T4

Conclusion:

Substitute 108K for T to get L in solar units.

    L(LO)=(1.8×1019)(108K)4LO=1.8×1013LO

Therefore, luminosity at 108K is 1.8×1013LO.

(b)

To determine

The luminosity at 106K.

(b)

Expert Solution
Check Mark

Answer to Problem 44QP

The luminosity at 106K is 1.8×105LO.

Explanation of Solution

From (a), the equation for L(LO) is,

    L(LO)=1.8×1019T4

Conclusion:

Substitute 106K for T to get L in solar units.

    L(LO)=(1.8×1019)(106K)4LO=1.8×105LO

Therefore, luminosity at 106K is 1.8×105LO.

(c)

To determine

The luminosity at 104K.

(c)

Expert Solution
Check Mark

Answer to Problem 44QP

The luminosity at 104K is 1.8×103LO.

Explanation of Solution

From (a), the equation for L(LO) is,

    L(LO)=1.8×1019T4

Conclusion:

Substitute 104K for T to get L in solar units.

    L(LO)=(1.8×1019)(104K)4LO=1.8×103LO

Therefore, luminosity at 104K is 1.8×103LO.

(d)

To determine

The luminosity at 102K.

(d)

Expert Solution
Check Mark

Answer to Problem 44QP

The luminosity at 102K is 1.8×1011LO.

Explanation of Solution

From (a), the equation for L(LO) is,

    L(LO)=1.8×1019T4

Conclusion:

Substitute 102K for T to get L in solar units.

    L(LO)=(1.8×1019)(102K)4LO=1.8×1011LO

Therefore, luminosity at 102K is 1.8×1011LO.

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