Question

At our distance from the Sun, the intensity of solar radiation is 1370 W/m2 . As discussed in class and in the textbook, the temperature of the Earth is affected by the greenhouse effect of the atmosphere. This phenomenon describes the effect of absorption of infrared radiation by the surface so as to make the surface temperature of the Earth higher than if it were airless. The average global surface temperature of the Earth is ~288 K. Calculate the surface temperature of the Earth if it did not have an atmosphere. You may assume an emissivity of 1. The radius of the Earth is 6370 km and its distance from the Sun is 1.496 × 10^11 m.

Answer 1

Answer 2

When gases are present, the radiation emitted by earth is absorbed back by the atmosphere so Pout<Pin. But in case there's no atmosphere, Pout= Pin. For this case, solution is provided below..

Answer 3

Given:

• Solar intensity ($S$) = 1370 W/m²

• Earth's radius ($R_E$) = 6370 km = $6.37\times 10^6\text{m}$

• Distance from Sun ($d$) = $1.496\times 10^{11}\text{m}$

• Emissivity ($ϵ$) = 1 (perfect blackbody)

• No atmosphere → Earth behaves like a blackbody in radiative equilibrium.

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Step 1: Calculate Incoming Solar Power (Absorbed by Earth)

Earth absorbs sunlight on its cross-sectional area ($\pi R_E^2$):

$$P_{\text{absorbed}}=S\times \pi R_E^2=1370\times \pi (6.37\times 10^6{)}^2$$$$P_{\text{absorbed}}\approx 1.75\times 10^{17}\text{W}$$

Step 2: Earth Radiates Energy Back to Space (Stefan-Boltzmann Law)

At equilibrium, absorbed power = emitted power:

$$P_{\text{emitted}}=\sigma T^4\times (4\pi R_E^2)$$

(where $\sigma =5.67\times 10^{-8}{\text{W/m}}^2{\text{K}}^4$, and Earth emits over its full surface area $4\pi R_E^2$)

Set $P_{\text{absorbed}}=P_{\text{emitted}}$:

$$1370\times \pi R_E^2=\sigma T^4\times 4\pi R_E^2$$

Cancel $\pi R_E^2$:

$$1370=4\sigma T^4$$

Solve for $T$:

$$T^4=\frac{1370}{4\times 5.67\times 10^{-8}}$$$$T^4\approx 6.04\times 10^9⟹T\approx 255\text{K}$$

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Final Answer:

The Earth's surface temperature without an atmosphere would be ~255 K (−18°C).