Assuming the Sun to be a spherical body of radius R at a temperature of TK , evaluate the total radiant powered incident of Earth at a distance r from the Sun Where r0 is the radius of the Earth and \sigma is Stefan's constant.

\pi r_0^2R^2\sigma T^4 \over r^2

4\pi r_0^2R^2\sigma T^4 \over r^2

r_0^2R^2\sigma T^4 \over 4\pi r^2

Heat and Thermodynamics PYQ — JEE Main 2006 · Shift 0 · Q108 with Solution

View written solutionFree

Correct answer: B

Power radiated by the Sun

Since the Sun is approximated as a spherical black body of radius $R$ and temperature $T$ , its total power output is given by Stefan–Boltzmann law:

P_{\text{Sun}} = 4\pi R^2 \sigma T^4

Intensity at the location of Earth

At a distance $r$ from the Sun, this power is spread uniformly over a sphere of radius $r$ . So the intensity there is:

I = \frac{P_{\text{Sun}}}{4\pi r^2} = \frac{4\pi R^2 \sigma T^4}{4\pi r^2} = \frac{R^2 \sigma T^4}{r^2}

Power incident on Earth

The Earth does not intercept radiation over its full surface area $4\pi r_0^2$ . It intercepts sunlight over its projected area, which is a circular disc of radius $r_0$ :

A_{\text{proj}} = \pi r_0^2

Hence, the total radiant power incident on Earth is:

P_{\text{incident}} = I \cdot A_{\text{proj}} = \left(\frac{R^2 \sigma T^4}{r^2}\right)(\pi r_0^2)

P_{\text{incident}} = \pi r_0^2 R^2 \sigma \frac{T^4}{r^2}

Match with the options

This corresponds to:

\boxed{\pi r_0^2 R^2 \sigma \frac{T^4}{r^2}}

So the correct option is B.

More from Heat and Thermodynamics

View written solutionFree

Correct answer: B

Power radiated by the Sun

Since the Sun is approximated as a spherical black body of radius $R$ and temperature $T$ , its total power output is given by Stefan–Boltzmann law:

P_{\text{Sun}} = 4\pi R^2 \sigma T^4

Intensity at the location of Earth

At a distance $r$ from the Sun, this power is spread uniformly over a sphere of radius $r$ . So the intensity there is:

I = \frac{P_{\text{Sun}}}{4\pi r^2} = \frac{4\pi R^2 \sigma T^4}{4\pi r^2} = \frac{R^2 \sigma T^4}{r^2}

Power incident on Earth

The Earth does not intercept radiation over its full surface area $4\pi r_0^2$ . It intercepts sunlight over its projected area, which is a circular disc of radius $r_0$ :

A_{\text{proj}} = \pi r_0^2

Hence, the total radiant power incident on Earth is:

P_{\text{incident}} = I \cdot A_{\text{proj}} = \left(\frac{R^2 \sigma T^4}{r^2}\right)(\pi r_0^2)

P_{\text{incident}} = \pi r_0^2 R^2 \sigma \frac{T^4}{r^2}

Match with the options

This corresponds to:

\boxed{\pi r_0^2 R^2 \sigma \frac{T^4}{r^2}}

So the correct option is B.

Heat and Thermodynamics question

Heat and Thermodynamics question

More from Heat and Thermodynamics

Heat and Thermodynamics question

Heat and Thermodynamics question

More from Heat and Thermodynamics