21ST CENT.ASTRONOMY(LL)W/CODE WKBK PKG.
21ST CENT.ASTRONOMY(LL)W/CODE WKBK PKG.
6th Edition
ISBN: 9780393874921
Author: PALEN
Publisher: Norton, W. W. & Company, Inc.
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Chapter 5, Problem 43QP

(a)

To determine

The average temperature on the planet.

(a)

Expert Solution
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Answer to Problem 43QP

The average temperature on the planet is 271.7K_.

Explanation of Solution

The black body temperature of a planet is given by,

    T=[Lsun(1a)16πσ]141d        (I)

Here, T is the temperature, Lsun is the luminosity of sun, a is the fraction of sunlight that reflects from a planet, σ is the Stefan’s constant, d is the distance.

All factors except the albedo are equal so that the change in temperature compared with zero albedo is given by,

    TT0=[Lsun(1a)16πσ]141d[Lsun(1a0)16πσ]141d=(1a)14(1a0)14        (II)

Conclusion:

Substitute 279K for T0 and 0 for a0 in equation (II) to find T.

    T=2791a4        (III)

Substitute 0.1 for a in equation (III) to find T.

    T=2791(0.1)4=271.7K

Therefore, the average temperature on the planet is 271.7K_.

(b)

To determine

The average temperature on the planet if its albedo were 0.9

(b)

Expert Solution
Check Mark

Answer to Problem 43QP

The average temperature on the planet is 156.9K_.

Explanation of Solution

The black body temperature of a planet is  given by,

    T=[Lsun(1a)16πσ]141d        (I)

Here, T is the temperature, Lsun is the luminosity of sun, a is the fraction of sunlight that reflects from a planet, σ is the Stefan’s constant, d is the distance.

All factors except the albedo are equal so that the change in temperature compared with zero albedo is given by,

    TT0=[Lsun(1a)16πσ]141d[Lsun(1a0)16πσ]141d=(1a)14(1a0)14        (II)

Conclusion:

Substitute 279K for T0 and 0 for a0 in equation (II) to find T.

    T=2791a4        (III)

Substitute 0.9 for a in equation (III) to find T.

    T=2791(0.9)4=156.9K

Therefore, the average temperature on the planet is 156.9K_.

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Consider a planet where Solar constant = 1360 W /m2 and albedo =0.30.  If n=0, what is the equilibrium surface temperature of the planet? If S increases to 1370 W/m2 and n remains zero, what is the new equilibrium surface temperature? If the planet has a water vapor feedback so that the number of layers n is a function of surface temperature nT = (T – 254.5)/100, what is the equilibrium surface temperature? Considering the answers above, calculate the climate feedback factor g for this planet.
Earth's daylight surface disk absorbs about 1036 W per m2 from the Sun. Using 6400 km for the Earth's radius, how much of this radiative power is emitted by each square meter of the spherical Earth? Hint: Compare the ratio of the disk area to the spherical surface area.
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