Question

Assuming the Sun is a black body at a temperature of $T_s$ and that the Earth is also a black body. Diameter of Sun subtends a very small angle $\theta$ radians at the centre of Earth. (Neglect sources of heat from the Earth itself). If an opaque cloud of dust particles in the form of a spherical shell with a radius equal to half the radius of the Earth's orbit were interposed between Earth and Sun centred on the Sun. Mark the CORRECT statement(s): [Assume cloud of dust and earth to behave as black body]

• A. Equilibrium temperature of cloud ($T_D$) is $T_s\sqrt{\theta}$
• B. Equilibrium temperature of cloud ($T_D$) is $T_s\sqrt{\frac{\theta}{2}}$
• C. Equilibrium temperature of Earth ($T_E$) is $\frac{T_D}{2}$
• D. Equilibrium temperature of Earth ($T_E$) is $\frac{T_D}{4}$

Answer 1

Answer: (A), (C)

A, C

Answer 2

The Sun, a black body at temperature $T_s$, emits power $P_s$. The intensity of solar radiation at a distance $r$ is $I_s(r) = \frac{P_s}{4\pi r^2}$.

Equilibrium Temperature of the Dust Cloud ($T_D$): The dust cloud is a spherical shell of radius $R_D = d/2$, where $d$ is the Earth-Sun distance. It absorbs all incident solar radiation and emits radiation as a black body at temperature $T_D$. At equilibrium, the power emitted by the dust cloud equals the power absorbed from the Sun. Power absorbed by dust cloud = $I_s(R_D) \times (\text{surface area of dust shell}) = \frac{P_s}{4\pi R_D^2} \times (4\pi R_D^2) = P_s$. Power emitted by dust cloud = $\sigma A_{D, surf} T_D^4 = \sigma (4\pi R_D^2) T_D^4$. Equating absorbed and emitted power: $\sigma (4\pi R_D^2) T_D^4 = P_s = \sigma (4\pi R_s^2) T_s^4$ $R_D^2 T_D^4 = R_s^2 T_s^4$ $T_D = T_s \left(\frac{R_s}{R_D}\right)^{1/2}$ Given $R_D = d/2$, $T_D = T_s \left(\frac{R_s}{d/2}\right)^{1/2} = T_s \sqrt{\frac{2R_s}{d}}$. The Sun's diameter subtends a small angle $\theta$ at Earth, so $\theta \approx \frac{2R_s}{d}$, which implies $\frac{R_s}{d} \approx \frac{\theta}{2}$. Substituting this into the expression for $T_D$: $T_D = T_s \sqrt{2 \left(\frac{R_s}{d}\right)} \approx T_s \sqrt{2 \left(\frac{\theta}{2}\right)} = T_s \sqrt{\theta}$. This confirms Option (A) is correct.

Equilibrium Temperature of the Earth ($T_E$): The dust cloud is opaque and completely covers the Sun from Earth's perspective. Thus, Earth receives radiation solely from the dust cloud. The total power emitted by the dust cloud is $P_{emit, D} = P_s$. The intensity of radiation from the dust cloud at Earth's distance $d$ is $I_D(d) = \frac{P_{emit, D}}{4\pi d^2} = \frac{P_s}{4\pi d^2}$. Earth absorbs this radiation over its cross-sectional area $A_E = \pi R_E^2$. Power absorbed by Earth: $P_{abs, E} = I_D(d) \times A_E = \frac{P_s}{4\pi d^2} \times (\pi R_E^2) = \frac{P_s R_E^2}{4d^2}$. At equilibrium, power emitted by Earth equals power absorbed: $\sigma (4\pi R_E^2) T_E^4 = P_{abs, E} = \frac{P_s R_E^2}{4d^2}$ Substituting $P_s = \sigma (4\pi R_s^2) T_s^4$: $\sigma (4\pi R_E^2) T_E^4 = \frac{\sigma (4\pi R_s^2) T_s^4 R_E^2}{4d^2}$ $4 T_E^4 = \frac{R_s^2 T_s^4}{d^2}$ $T_E^4 = \frac{1}{4} \left(\frac{R_s}{d}\right)^2 T_s^4$ $T_E = \frac{1}{2} \left(\frac{R_s}{d}\right) T_s$.

Now, we relate $T_E$ to $T_D$. We have $T_D = T_s \sqrt{2R_s/d}$ and $T_E = \frac{1}{2} (R_s/d) T_s$. From the expression for $T_E$, we get $\frac{R_s}{d} = \frac{2T_E}{T_s}$. Substituting this into the expression for $T_D$: $T_D = T_s \sqrt{2 \left(\frac{2T_E}{T_s}\right)} = T_s \sqrt{\frac{4T_E}{T_s}} = T_s \frac{2\sqrt{T_E}}{\sqrt{T_s}} = 2\sqrt{T_s T_E}$. Squaring both sides: $T_D^2 = 4 T_s T_E$.

Alternatively, consider the power emitted by the dust cloud. $P_{emit, D} = \sigma (4\pi R_D^2) T_D^4$. The intensity of radiation from the dust cloud at Earth's distance $d$ is $I_D(d) = \frac{P_{emit, D}}{4\pi d^2} = \frac{\sigma (4\pi R_D^2) T_D^4}{4\pi d^2} = \sigma T_D^4 \frac{R_D^2}{d^2}$. Since $R_D = d/2$, $I_D(d) = \sigma T_D^4 \frac{(d/2)^2}{d^2} = \sigma T_D^4 \frac{d^2/4}{d^2} = \frac{\sigma T_D^4}{4}$. Earth's equilibrium temperature $T_E$ is related to the absorbed flux. Power absorbed by Earth is proportional to $I_D(d)$. $T_E^4 \propto I_D(d) = \frac{\sigma T_D^4}{4}$. Therefore, $T_E^4 \propto \frac{T_D^4}{4}$, which implies $T_E \propto \frac{T_D}{2}$. This confirms Option (C) is correct.