Understanding Our Universe
Understanding Our Universe
3rd Edition
ISBN: 9780393614428
Author: PALEN, Stacy, Kay, Laura, Blumenthal, George (george Ray)
Publisher: W.w. Norton & Company,
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Chapter 5, Problem 45QAP

(a)

To determine

The flux from the each square meter of the planet’s surface if its temperature is 400K.

(a)

Expert Solution
Check Mark

Answer to Problem 45QAP

The flux per square meter of the planet is 2.72×1013K/m2.

Explanation of Solution

The radius of the Earth is 6371km whereas the radius of the planet is 1.7times the radius of the Earth i.e. 10830.7km.

Write the expression for the surface area of the planet.

A=4πR2        (I)

Here, A is the total surface area of the planet and R is the radius of the planet.

Write the expression for the flux from a square meter of the planet’s surface.

ϕ=TA        (II)

Here, ϕ is the flux from the surface and T is the temperature of the star.

Conclusion:

Substitute 10830.7km for R in equation (I).

A=4π(10830.7km(1000m1km))2=12.566(1.17×1014)m21.47×1015m2

Substitute 400K for T and 1.47×1015m2 for A in equation (II).

ϕ=400K1.47×1015m2=2.72×1013K/m2

Thus, the flux per square meter of the planet is 2.72×1013K/m2.

(b)

To determine

The luminosity of the planet that has the temperature of 400K.

(b)

Expert Solution
Check Mark

Answer to Problem 45QAP

The luminosity of the planet will be 2.133×1018W.

Explanation of Solution

Write the expression for the Luminosity of the planet.

L=σAT4        (III)

Here, L is the luminosity of the planet and σ is the Stefan-Boltzmann constant.

Conclusion:

Substitute 5.67×108W/m2K4 for σ, 1.47×1015m2 for A and 400K for T in equation (III).

L=(5.67×108W/m2K4)(1.47×1015m2)(400K)4=2.133×1018W

Thus, the luminosity of the planet will be 2.133×1018W.

(c)

To determine

The peak wavelength of the radiations emitted from the planet.

(c)

Expert Solution
Check Mark

Answer to Problem 45QAP

The peak wavelength of emission of the planet is 7242.5nm.

Explanation of Solution

Write the expression for the peak wavelength of emission from the planet.

λpeak=2.897×103mKT        (IV)

Here, λpeak is the peak wavelength of emission.

Conclusion:

Substitute 400K for T in equation (IV).

λpeak=2.897×103mK400K=7.24×106m(109nm1m)7242.5nm

Thus, the peak wavelength of emission of the planet is 7242.5nm.

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