The distance between the Sun and the Earth is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For thermal equilibrium, the power absorbed by the Earth from the Sun must equal the power radiated by the Earth.The power received by the Earth i

Vedclass · Accepted Answer

$300$

Vedclass · Answer

(B) For thermal equilibrium, the power absorbed by the Earth from the Sun must equal the power radiated by the Earth.The power received by the Earth is $P_{in} = \left( \frac{\sigma T_s^4 \cdot 4 \pi R_s^2}{4 \pi d^2} ight) \cdot \pi R_e^2$.The power radiated by the Earth is $P_{out} = e \sigma T_e^4 \cdot 4 \pi R_e^2$.Equating $P_{in} = P_{out}$:$\frac{\sigma T_s^4 R_s^2}{d^2} \cdot \pi R_e^2 = e \sigma T_e^4 \cdot 4 \pi R_e^2$Simplifying for $T_e^4$:$T_e^4 = \frac{T_s^4 R_s^2}{4 e d^2}$Given: $T_s = 6000 \ K$, $R_s = 7 	imes 10^5 \ km$, $d = 2 	imes 10^8 \ km$, $e = 0.6$.$T_e^4 = \frac{(6000)^4 \cdot (7 	imes 10^5)^2}{4 \cdot 0.6 \cdot (2 	imes 10^8)^2}$$T_e^4 = \frac{(6^4 	imes 10^{12}) \cdot (49 	imes 10^{10})}{2.4 \cdot (4 	imes 10^{16})}$$T_e^4 = \frac{1296 \cdot 49 \cdot 10^{22}}{9.6 \cdot 10^{16}} = \frac{63504}{9.6} 	imes 10^6 = 6615 	imes 10^6 = 6.615 	imes 10^9 \approx 81 	imes 10^8$$T_e = (81 	imes 10^8)^{1/4} = 3 	imes 10^2 = 300 \ K$.

The distance between the Sun and the Earth is $2 \times 10^{8} \ km$, the temperature of the Sun is $6000 \ K$, and the radius of the Sun is $7 \times 10^{5} \ km$. If the emissivity of the Earth is $0.6$, find the temperature of the Earth in thermal equilibrium (in $K$).

Similar Questions

Two spheres of the same material have radii $1 \ m$ and $4 \ m$ and temperatures $4000 \ K$ and $2000 \ K$ respectively. The energy radiated per second by the first sphere is

Two bodies $X$ and $Y$ at temperatures $T_1 \ K$ and $T_2 \ K$ respectively have the same dimensions. If their emissive powers are the same,the relation between their temperatures is:

Two spheres of the same material have radii $1 \; m$ and $4 \; m$ and temperatures $4000 \; K$ and $2000 \; K$ respectively. The ratio of the energy radiated per second by the first sphere to that by the second is

If the temperature of a black body increases from $7^{\circ}C$ to $287^{\circ}C$,then what is the ratio of the rate of energy emission?

If the temperature of the body is increased by $10\%$,the percentage increase in the emitted radiation will be ....... $\%$

Vedclass Test Series

Exam Paper Generator

Online Exam Module