1000 Solved Problems in Modern Physics

290 4 Thermodynamics and Statistical Physics

The negative sign in (3) is omitted because asλincreasesvdecreases.

4.76 Power radiated from the sun=σ×(surface area)×Ts^4

Ps=σ 4 πRs^2 Ts^4

Power received by the earth,

PE=

πRe^2

4 πr^2

.Ps

The factorπR^2 erepresents the effective (projected) area of the earth on

which the sun’s radiation is incident at a distancerfrom the sun. The factor

4 πr^2 is the surface area of a sphere scooped with the centre on the sun. Thus

πRe^2 / 4 πr^2 is the fraction of the radiation intercepted by the earth’s surface

area.

Now power radiated by earth,

PE=σ 4 πR^2 ETE^4

For radiation equilibrium, power radiated by the earth=power received by

the earth.

σ 4 πR^2 ETE^4 =σ 4 πR^2 sTs^4.

πR^2 E

4 πr^2

orTE=Ts

(

Rs

2 r

) 1 / 2

= 5 , 800

[

7 × 108

2 × 1. 5 × 1011

] 1 / 2

=280 K= 7 ◦C

Note that the calculations are approximate in that the earth and sun are not

black bodies and that the contribution of heat from the interior of the earth has

not been taken into account.

4.77 Power radiated by the sun,Ps=σ 4 πRs^2 Ts^4
Power received by 1 m^2 of earth’s surface,

S=

σ 4 πRs^2 Ts^4

4 πr^2

=

(5. 7 × 10 −^8 )(7× 108 )^2 (5,800)^4

(1. 5 × 1011 )^2

= 1 ,400 W/m^2

4.78 P= 4 πr^2 σT^4
= 4 π(0.3)^2 (5. 67 × 10 −^8 )(10^7 )^4
= 6. 4 × 1020 W