The Spectral class for the brown dwarf CoD-33 ° 7795 B star whose surface temperature is 2550 K .

Question

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Answer 1

Question

Chapter 17, Problem 54Q

(a)

To determine

The Spectral class for the brown dwarf $CoD-33°7795 B$ star whose surface temperature is $2550 K$ .

(a)

Expert Solution

Answer to Problem 54Q

Solution:

L

Explanation of Solution

Given data:

The temperature of the brown dwarf star is $2550 K$ .

Introduction:

On the basis of temperature of the surface of stars, they are divided into different spectral classes.

Explanation:

Refer the table 17-2. The spectral sequence, the temperature and the brown dwarf lies to the Spectral class L.

Conclusion:

The temperature and the brown dwarf lie to the spectral class of L.

(b)

To determine

The radius (in kilometers) of $CoD-33°7795 B$ in terms of radius of the Sun. Also, compare it to the radius of Jupiter. In addition to that explain whether the name Dwarf is justified or not, if the brown dwarf has a luminosity of $0.0025L⊙$ and its surface temperature is $2550 K$ .

(b)

Expert Solution

Answer to Problem 54Q

Solution:

$0.26R⊙$ , $1.8096×105 km$ and the name Dwarf is not justified as the radius is not much smaller than the Sun.

Explanation of Solution

Given data:

Brown dwarf has a luminosity of $0.0025L⊙$ and its surface temperature is $2550 K$ .

Formula used:

The relation between luminosity of different stars is:

$LL⊙=(RR⊙)2(TT⊙)4$

Here, $L$ is the luminosity of the star, $L⊙$ is the luminosity of the Sun, $R$ is the radius of the star, $R⊙$ the radius of the Sun, $T$ is the temperature of the star and $b⊙$ is the temperature of the Sun.

Explanation:

The Sun’s surface temperature is, $T⊙=5800 K$ .

The relation between luminosity of different stars is:

$LL⊙=(RR⊙)2(TT⊙)4$

Substitute $2550 K$ for $T$ , $5800 K$ for $T⊙$ and $0.0025L⊙$ for $L$ .

$0.0025L⊙L⊙=(RR⊙)2(2550 K5800 K)4(RR⊙)2=(5800 K2550 K)4(0.0025L⊙L⊙)RR⊙=(5800 K2550 K)4(0.0025)RR⊙=0.26$

Further solve,

$R=0.26R⊙$

Refer to Appendix 6, the value of Solar radius is $1R⊙=6.9599×108 m$ .

Therefore,

$R=0.26R⊙=0.26(6.9599×108 m)=0.26(6.9599×108 m)(1 km103 m)=1.8096×105 km$

Therefore, the value of the radius in kilometer is $1.8096×105 km$ .

Now, refer to appendix 2, the value of the diameter of Jupiter is $142984 km$ .

Therefore, the radius of Jupiter is:

Expression for radius is:

$r=d2$

Use this expression to determine the radius of Jupiter.

$RJupiter=142984 km2=71492 km$

The radius of $CoD-33°7795 B$ is smaller than the radius of the Sun, but it is more than twice the radius of Jupiter. Therefore, the name Dwarf is not justified.

Conclusion:

The name Dwarf is not justified as the radius is not much smaller, and also the value of the radius is $0.26R⊙$ in term of solar radius and $1.8096×105 km$ in terms of kilometer.

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Answer 2

Question

Chapter 17, Problem 54Q

(a)

To determine

The Spectral class for the brown dwarf $CoD-33°7795 B$ star whose surface temperature is $2550 K$ .

(a)

Expert Solution

Answer to Problem 54Q

Solution:

L

Explanation of Solution

Given data:

The temperature of the brown dwarf star is $2550 K$ .

Introduction:

On the basis of temperature of the surface of stars, they are divided into different spectral classes.

Explanation:

Refer the table 17-2. The spectral sequence, the temperature and the brown dwarf lies to the Spectral class L.

Conclusion:

The temperature and the brown dwarf lie to the spectral class of L.

(b)

To determine

The radius (in kilometers) of $CoD-33°7795 B$ in terms of radius of the Sun. Also, compare it to the radius of Jupiter. In addition to that explain whether the name Dwarf is justified or not, if the brown dwarf has a luminosity of $0.0025L⊙$ and its surface temperature is $2550 K$ .

(b)

Expert Solution

Answer to Problem 54Q

Solution:

$0.26R⊙$ , $1.8096×105 km$ and the name Dwarf is not justified as the radius is not much smaller than the Sun.

Explanation of Solution

Given data:

Brown dwarf has a luminosity of $0.0025L⊙$ and its surface temperature is $2550 K$ .

Formula used:

The relation between luminosity of different stars is:

$LL⊙=(RR⊙)2(TT⊙)4$

Here, $L$ is the luminosity of the star, $L⊙$ is the luminosity of the Sun, $R$ is the radius of the star, $R⊙$ the radius of the Sun, $T$ is the temperature of the star and $b⊙$ is the temperature of the Sun.

Explanation:

The Sun’s surface temperature is, $T⊙=5800 K$ .

The relation between luminosity of different stars is:

$LL⊙=(RR⊙)2(TT⊙)4$

Substitute $2550 K$ for $T$ , $5800 K$ for $T⊙$ and $0.0025L⊙$ for $L$ .

$0.0025L⊙L⊙=(RR⊙)2(2550 K5800 K)4(RR⊙)2=(5800 K2550 K)4(0.0025L⊙L⊙)RR⊙=(5800 K2550 K)4(0.0025)RR⊙=0.26$

Further solve,

$R=0.26R⊙$

Refer to Appendix 6, the value of Solar radius is $1R⊙=6.9599×108 m$ .

Therefore,

$R=0.26R⊙=0.26(6.9599×108 m)=0.26(6.9599×108 m)(1 km103 m)=1.8096×105 km$

Therefore, the value of the radius in kilometer is $1.8096×105 km$ .

Now, refer to appendix 2, the value of the diameter of Jupiter is $142984 km$ .

Therefore, the radius of Jupiter is:

Expression for radius is:

$r=d2$

Use this expression to determine the radius of Jupiter.

$RJupiter=142984 km2=71492 km$

The radius of $CoD-33°7795 B$ is smaller than the radius of the Sun, but it is more than twice the radius of Jupiter. Therefore, the name Dwarf is not justified.

Conclusion:

The name Dwarf is not justified as the radius is not much smaller, and also the value of the radius is $0.26R⊙$ in term of solar radius and $1.8096×105 km$ in terms of kilometer.

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Answer 3

Question

Chapter 17, Problem 54Q

(a)

To determine

The Spectral class for the brown dwarf $CoD-33°7795 B$ star whose surface temperature is $2550 K$ .

(a)

Expert Solution

Answer to Problem 54Q

Solution:

L

Explanation of Solution

Given data:

The temperature of the brown dwarf star is $2550 K$ .

Introduction:

On the basis of temperature of the surface of stars, they are divided into different spectral classes.

Explanation:

Refer the table 17-2. The spectral sequence, the temperature and the brown dwarf lies to the Spectral class L.

Conclusion:

The temperature and the brown dwarf lie to the spectral class of L.

(b)

To determine

The radius (in kilometers) of $CoD-33°7795 B$ in terms of radius of the Sun. Also, compare it to the radius of Jupiter. In addition to that explain whether the name Dwarf is justified or not, if the brown dwarf has a luminosity of $0.0025L⊙$ and its surface temperature is $2550 K$ .

(b)

Expert Solution

Answer to Problem 54Q

Solution:

$0.26R⊙$ , $1.8096×105 km$ and the name Dwarf is not justified as the radius is not much smaller than the Sun.

Explanation of Solution

Given data:

Brown dwarf has a luminosity of $0.0025L⊙$ and its surface temperature is $2550 K$ .

Formula used:

The relation between luminosity of different stars is:

$LL⊙=(RR⊙)2(TT⊙)4$

Here, $L$ is the luminosity of the star, $L⊙$ is the luminosity of the Sun, $R$ is the radius of the star, $R⊙$ the radius of the Sun, $T$ is the temperature of the star and $b⊙$ is the temperature of the Sun.

Explanation:

The Sun’s surface temperature is, $T⊙=5800 K$ .

The relation between luminosity of different stars is:

$LL⊙=(RR⊙)2(TT⊙)4$

Substitute $2550 K$ for $T$ , $5800 K$ for $T⊙$ and $0.0025L⊙$ for $L$ .

$0.0025L⊙L⊙=(RR⊙)2(2550 K5800 K)4(RR⊙)2=(5800 K2550 K)4(0.0025L⊙L⊙)RR⊙=(5800 K2550 K)4(0.0025)RR⊙=0.26$

Further solve,

$R=0.26R⊙$

Refer to Appendix 6, the value of Solar radius is $1R⊙=6.9599×108 m$ .

Therefore,

$R=0.26R⊙=0.26(6.9599×108 m)=0.26(6.9599×108 m)(1 km103 m)=1.8096×105 km$

Therefore, the value of the radius in kilometer is $1.8096×105 km$ .

Now, refer to appendix 2, the value of the diameter of Jupiter is $142984 km$ .

Therefore, the radius of Jupiter is:

Expression for radius is:

$r=d2$

Use this expression to determine the radius of Jupiter.

$RJupiter=142984 km2=71492 km$

The radius of $CoD-33°7795 B$ is smaller than the radius of the Sun, but it is more than twice the radius of Jupiter. Therefore, the name Dwarf is not justified.

Conclusion:

The name Dwarf is not justified as the radius is not much smaller, and also the value of the radius is $0.26R⊙$ in term of solar radius and $1.8096×105 km$ in terms of kilometer.

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Answer 4

Question

Chapter 17, Problem 54Q

(a)

To determine

The Spectral class for the brown dwarf $CoD-33°7795 B$ star whose surface temperature is $2550 K$ .

(a)

Expert Solution

Answer to Problem 54Q

Solution:

L

Explanation of Solution

Given data:

The temperature of the brown dwarf star is $2550 K$ .

Introduction:

On the basis of temperature of the surface of stars, they are divided into different spectral classes.

Explanation:

Refer the table 17-2. The spectral sequence, the temperature and the brown dwarf lies to the Spectral class L.

Conclusion:

The temperature and the brown dwarf lie to the spectral class of L.

(b)

To determine

The radius (in kilometers) of $CoD-33°7795 B$ in terms of radius of the Sun. Also, compare it to the radius of Jupiter. In addition to that explain whether the name Dwarf is justified or not, if the brown dwarf has a luminosity of $0.0025L⊙$ and its surface temperature is $2550 K$ .

(b)

Expert Solution

Answer to Problem 54Q

Solution:

$0.26R⊙$ , $1.8096×105 km$ and the name Dwarf is not justified as the radius is not much smaller than the Sun.

Explanation of Solution

Given data:

Brown dwarf has a luminosity of $0.0025L⊙$ and its surface temperature is $2550 K$ .

Formula used:

The relation between luminosity of different stars is:

$LL⊙=(RR⊙)2(TT⊙)4$

Here, $L$ is the luminosity of the star, $L⊙$ is the luminosity of the Sun, $R$ is the radius of the star, $R⊙$ the radius of the Sun, $T$ is the temperature of the star and $b⊙$ is the temperature of the Sun.

Explanation:

The Sun’s surface temperature is, $T⊙=5800 K$ .

The relation between luminosity of different stars is:

$LL⊙=(RR⊙)2(TT⊙)4$

Substitute $2550 K$ for $T$ , $5800 K$ for $T⊙$ and $0.0025L⊙$ for $L$ .

$0.0025L⊙L⊙=(RR⊙)2(2550 K5800 K)4(RR⊙)2=(5800 K2550 K)4(0.0025L⊙L⊙)RR⊙=(5800 K2550 K)4(0.0025)RR⊙=0.26$

Further solve,

$R=0.26R⊙$

Refer to Appendix 6, the value of Solar radius is $1R⊙=6.9599×108 m$ .

Therefore,

$R=0.26R⊙=0.26(6.9599×108 m)=0.26(6.9599×108 m)(1 km103 m)=1.8096×105 km$

Therefore, the value of the radius in kilometer is $1.8096×105 km$ .

Now, refer to appendix 2, the value of the diameter of Jupiter is $142984 km$ .

Therefore, the radius of Jupiter is:

Expression for radius is:

$r=d2$

Use this expression to determine the radius of Jupiter.

$RJupiter=142984 km2=71492 km$

The radius of $CoD-33°7795 B$ is smaller than the radius of the Sun, but it is more than twice the radius of Jupiter. Therefore, the name Dwarf is not justified.

Conclusion:

The name Dwarf is not justified as the radius is not much smaller, and also the value of the radius is $0.26R⊙$ in term of solar radius and $1.8096×105 km$ in terms of kilometer.

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Answer 5

Chapter 17, Problem 54Q

(a)

To determine

The Spectral class for the brown dwarf $CoD-33°7795 B$ star whose surface temperature is $2550 K$ .

(a)

Expert Solution

Answer to Problem 54Q

Solution:

L

Explanation of Solution

Given data:

The temperature of the brown dwarf star is $2550 K$ .

Introduction:

On the basis of temperature of the surface of stars, they are divided into different spectral classes.

Explanation:

Refer the table 17-2. The spectral sequence, the temperature and the brown dwarf lies to the Spectral class L.

Conclusion:

The temperature and the brown dwarf lie to the spectral class of L.

(b)

To determine

The radius (in kilometers) of $CoD-33°7795 B$ in terms of radius of the Sun. Also, compare it to the radius of Jupiter. In addition to that explain whether the name Dwarf is justified or not, if the brown dwarf has a luminosity of $0.0025L⊙$ and its surface temperature is $2550 K$ .

(b)

Expert Solution

Answer to Problem 54Q

Solution:

$0.26R⊙$ , $1.8096×105 km$ and the name Dwarf is not justified as the radius is not much smaller than the Sun.

Explanation of Solution

Given data:

Brown dwarf has a luminosity of $0.0025L⊙$ and its surface temperature is $2550 K$ .

Formula used:

The relation between luminosity of different stars is:

$LL⊙=(RR⊙)2(TT⊙)4$

Here, $L$ is the luminosity of the star, $L⊙$ is the luminosity of the Sun, $R$ is the radius of the star, $R⊙$ the radius of the Sun, $T$ is the temperature of the star and $b⊙$ is the temperature of the Sun.

Explanation:

The Sun’s surface temperature is, $T⊙=5800 K$ .

The relation between luminosity of different stars is:

$LL⊙=(RR⊙)2(TT⊙)4$

Substitute $2550 K$ for $T$ , $5800 K$ for $T⊙$ and $0.0025L⊙$ for $L$ .

$0.0025L⊙L⊙=(RR⊙)2(2550 K5800 K)4(RR⊙)2=(5800 K2550 K)4(0.0025L⊙L⊙)RR⊙=(5800 K2550 K)4(0.0025)RR⊙=0.26$

Further solve,

$R=0.26R⊙$

Refer to Appendix 6, the value of Solar radius is $1R⊙=6.9599×108 m$ .

Therefore,

$R=0.26R⊙=0.26(6.9599×108 m)=0.26(6.9599×108 m)(1 km103 m)=1.8096×105 km$

Therefore, the value of the radius in kilometer is $1.8096×105 km$ .

Now, refer to appendix 2, the value of the diameter of Jupiter is $142984 km$ .

Therefore, the radius of Jupiter is:

Expression for radius is:

$r=d2$

Use this expression to determine the radius of Jupiter.

$RJupiter=142984 km2=71492 km$

The radius of $CoD-33°7795 B$ is smaller than the radius of the Sun, but it is more than twice the radius of Jupiter. Therefore, the name Dwarf is not justified.

Conclusion:

The name Dwarf is not justified as the radius is not much smaller, and also the value of the radius is $0.26R⊙$ in term of solar radius and $1.8096×105 km$ in terms of kilometer.

Answer 6

Chapter 17, Problem 54Q

(a)

To determine

The Spectral class for the brown dwarf $CoD-33°7795 B$ star whose surface temperature is $2550 K$ .

(a)

Expert Solution

Answer to Problem 54Q

Solution:

L

Explanation of Solution

Given data:

The temperature of the brown dwarf star is $2550 K$ .

Introduction:

On the basis of temperature of the surface of stars, they are divided into different spectral classes.

Explanation:

Refer the table 17-2. The spectral sequence, the temperature and the brown dwarf lies to the Spectral class L.

Conclusion:

The temperature and the brown dwarf lie to the spectral class of L.

Answer 7

Chapter 17, Problem 54Q

(a)

To determine

The Spectral class for the brown dwarf $CoD-33°7795 B$ star whose surface temperature is $2550 K$ .

Answer 8

(a)

Expert Solution

Answer to Problem 54Q

Solution:

L

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given data:

The temperature of the brown dwarf star is $2550 K$ .

Introduction:

On the basis of temperature of the surface of stars, they are divided into different spectral classes.

Explanation:

Refer the table 17-2. The spectral sequence, the temperature and the brown dwarf lies to the Spectral class L.

Conclusion:

The temperature and the brown dwarf lie to the spectral class of L.

Answer 13

(b)

To determine

The radius (in kilometers) of $CoD-33°7795 B$ in terms of radius of the Sun. Also, compare it to the radius of Jupiter. In addition to that explain whether the name Dwarf is justified or not, if the brown dwarf has a luminosity of $0.0025L⊙$ and its surface temperature is $2550 K$ .

(b)

Expert Solution

Answer to Problem 54Q

Solution:

$0.26R⊙$ , $1.8096×105 km$ and the name Dwarf is not justified as the radius is not much smaller than the Sun.

Explanation of Solution

Given data:

Brown dwarf has a luminosity of $0.0025L⊙$ and its surface temperature is $2550 K$ .

Formula used:

The relation between luminosity of different stars is:

$LL⊙=(RR⊙)2(TT⊙)4$

Here, $L$ is the luminosity of the star, $L⊙$ is the luminosity of the Sun, $R$ is the radius of the star, $R⊙$ the radius of the Sun, $T$ is the temperature of the star and $b⊙$ is the temperature of the Sun.

Explanation:

The Sun’s surface temperature is, $T⊙=5800 K$ .

The relation between luminosity of different stars is:

$LL⊙=(RR⊙)2(TT⊙)4$

Substitute $2550 K$ for $T$ , $5800 K$ for $T⊙$ and $0.0025L⊙$ for $L$ .

$0.0025L⊙L⊙=(RR⊙)2(2550 K5800 K)4(RR⊙)2=(5800 K2550 K)4(0.0025L⊙L⊙)RR⊙=(5800 K2550 K)4(0.0025)RR⊙=0.26$

Further solve,

$R=0.26R⊙$

Refer to Appendix 6, the value of Solar radius is $1R⊙=6.9599×108 m$ .

Therefore,

$R=0.26R⊙=0.26(6.9599×108 m)=0.26(6.9599×108 m)(1 km103 m)=1.8096×105 km$

Therefore, the value of the radius in kilometer is $1.8096×105 km$ .

Now, refer to appendix 2, the value of the diameter of Jupiter is $142984 km$ .

Therefore, the radius of Jupiter is:

Expression for radius is:

$r=d2$

Use this expression to determine the radius of Jupiter.

$RJupiter=142984 km2=71492 km$

The radius of $CoD-33°7795 B$ is smaller than the radius of the Sun, but it is more than twice the radius of Jupiter. Therefore, the name Dwarf is not justified.

Conclusion:

The name Dwarf is not justified as the radius is not much smaller, and also the value of the radius is $0.26R⊙$ in term of solar radius and $1.8096×105 km$ in terms of kilometer.

Answer 14

(b)

To determine

The radius (in kilometers) of $CoD-33°7795 B$ in terms of radius of the Sun. Also, compare it to the radius of Jupiter. In addition to that explain whether the name Dwarf is justified or not, if the brown dwarf has a luminosity of $0.0025L⊙$ and its surface temperature is $2550 K$ .

Answer 15

(b)

Expert Solution

Answer to Problem 54Q

Solution:

$0.26R⊙$ , $1.8096×105 km$ and the name Dwarf is not justified as the radius is not much smaller than the Sun.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given data:

Brown dwarf has a luminosity of $0.0025L⊙$ and its surface temperature is $2550 K$ .

Formula used:

The relation between luminosity of different stars is:

$LL⊙=(RR⊙)2(TT⊙)4$

Here, $L$ is the luminosity of the star, $L⊙$ is the luminosity of the Sun, $R$ is the radius of the star, $R⊙$ the radius of the Sun, $T$ is the temperature of the star and $b⊙$ is the temperature of the Sun.

Explanation:

The Sun’s surface temperature is, $T⊙=5800 K$ .

The relation between luminosity of different stars is:

$LL⊙=(RR⊙)2(TT⊙)4$

Substitute $2550 K$ for $T$ , $5800 K$ for $T⊙$ and $0.0025L⊙$ for $L$ .

$0.0025L⊙L⊙=(RR⊙)2(2550 K5800 K)4(RR⊙)2=(5800 K2550 K)4(0.0025L⊙L⊙)RR⊙=(5800 K2550 K)4(0.0025)RR⊙=0.26$

Further solve,

$R=0.26R⊙$

Refer to Appendix 6, the value of Solar radius is $1R⊙=6.9599×108 m$ .

Therefore,

$R=0.26R⊙=0.26(6.9599×108 m)=0.26(6.9599×108 m)(1 km103 m)=1.8096×105 km$

Therefore, the value of the radius in kilometer is $1.8096×105 km$ .

Now, refer to appendix 2, the value of the diameter of Jupiter is $142984 km$ .

Therefore, the radius of Jupiter is:

Expression for radius is:

$r=d2$

Use this expression to determine the radius of Jupiter.

$RJupiter=142984 km2=71492 km$

The radius of $CoD-33°7795 B$ is smaller than the radius of the Sun, but it is more than twice the radius of Jupiter. Therefore, the name Dwarf is not justified.

Conclusion:

The name Dwarf is not justified as the radius is not much smaller, and also the value of the radius is $0.26R⊙$ in term of solar radius and $1.8096×105 km$ in terms of kilometer.

Answer 20

Ch. 17 - Prob. 1CC Ch. 17 - Prob. 2CC Ch. 17 - Prob. 3CC Ch. 17 - Prob. 4CC Ch. 17 - Prob. 5CC Ch. 17 - Prob. 6CC Ch. 17 - Prob. 7CC Ch. 17 - Prob. 8CC Ch. 17 - Prob. 9CC Ch. 17 - Prob. 10CC

The Spectral class for the brown dwarf CoD-33 ° 7795 B star whose surface temperature is 2550 K .

Concept explainers

The Spectral class for the brown dwarf $CoD-33°7795 B$ star whose surface temperature is $2550 K$ .

Answer to Problem 54Q

Explanation of Solution

The radius (in kilometers) of $CoD-33°7795 B$ in terms of radius of the Sun. Also, compare it to the radius of Jupiter. In addition to that explain whether the name Dwarf is justified or not, if the brown dwarf has a luminosity of $0.0025L⊙$ and its surface temperature is $2550 K$ .

Answer to Problem 54Q

Explanation of Solution

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Chapter 17 Solutions

The Spectral class for the brown dwarf CoD-33 ° 7795 B star whose surface temperature is 2550 K .

Concept explainers

The Spectral class for the brown dwarf CoD-33°<m:math><m:mrow><m:mtext>CoD-33</m:mtext><m:mo>°</m:mo><m:mtext>7795 B</m:mtext></m:mrow></m:math> star whose surface temperature is 2550 K<m:math><m:mrow><m:mn>2550</m:mn><m:mtext> K</m:mtext></m:mrow></m:math>.

Answer to Problem 54Q

Explanation of Solution

Answer to Problem 54Q

Explanation of Solution

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Chapter 17 Solutions

The Spectral class for the brown dwarf $CoD-33°7795 B$ star whose surface temperature is $2550 K$ .