Question

Assuming the Sun to be a perfectly spherical black body of radius \(R\) and is at a temperature \(T\ \text{K}\), then the total radiant power incident on the Earth at a distance \(r\) from the sun is ( consider \(R_{o}\) is the radius of the Earth and \(\sigma\) is Stefan's constant).

• A. \(\frac{\pi R^{2}R_{o}^{2}\sigma T^{4}}{r^{2}}\)
• B. \(\frac{\pi R^{2}\sigma T^{4}}{R_{o}^{2}r^{2}}\)
• C. \(\frac{\pi R_{o}^{2}\sigma T^{4}}{R^{2}r^{2}}\)
• D. \(\frac{R_{o}^{2}\sigma T^{4}}{\pi R^{2}r^{2}}\)

Hint:

Find the power emitted by the Sun using Stefan's law, spread it over a sphere of radius r, and multiply the intensity by the area of the Earth's disc.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The Sun is a black body, so the power it emits is \(P=\sigma T^{4}\times4\pi R^{2}\).

Step 2: Intensity at the Earth
At distance \(r\), the power spreads over area \(4\pi r^{2}\), so the intensity is \(I=\dfrac{4\pi R^{2}\sigma T^{4}}{4\pi r^{2}}=\dfrac{R^{2}\sigma T^{4}}{r^{2}}\).

Step 3: Area intercepted
The Earth presents a disc of area \(\pi R_{o}^{2}\) to the Sun.

Step 4: Result
Power received \(=I\times\pi R_{o}^{2}=\dfrac{\pi R^{2}R_{o}^{2}\sigma T^{4}}{r^{2}}\).

Final Answer:
The power received is pi R squared Ro squared sigma T to the fourth, divided by r squared. \[ \boxed{\text{(A) }\dfrac{\pi R^{2}R_{o}^{2}\sigma T^{4}}{r^{2}}} \]