Assuming the sun to be a spherical body of radius R at a temperature of T K, evaluate the total radi

Q: Why do we use the projected area $\pi r_0^2$ instead of the surface area $4\pi r_0^2$ or hemisphere area $2\pi r_0^2$?

The Earth intercepts radiation exactly like a flat circular disk of radius $r_0$ facing the Sun. The radiation travels in parallel lines (effectively) across the Earth's diameter, so the "shadow" it casts is a circle, not a sphere.

Q: Why is the emissivity $e$ taken as 1?

The problem states to assume the Sun is a spherical body at temperature $T$ K. In such idealized physics problems, stars are treated as perfect black bodies, which means they are perfect emitters with $e=1$.

Q: Does the distance $r$ include the radius of the Sun and Earth?

Yes, $r$ is the center-to-center distance. Since $r$ is vastly larger than both $R$ and $r_0$, the slight differences at the surfaces are negligible, and we use $r$ for the radius of the radiation sphere.

Q: Why doesn't the Earth's temperature appear in the formula?

The question asks for the radiant power *incident* on the Earth, not the net heat exchanged. The incident power depends only on the source (Sun) and the geometry of the setup.

Q: What happens to the rest of the Sun's energy?

It radiates out into empty space! The Earth only intercepts a tiny fraction of the total energy—specifically, the ratio of its projected area to the massive spherical area $4\pi r^2$.

Question

Assuming the sun to be a spherical body of radius $R$ at a temperature of $T$ K, evaluate the total radiant power, incident on earth, at a distance $r$ from the sun.
where, $r_0$ is the radius of the earth and $\sigma$ is Stefan's constant.

Options:

(A) $\frac{4\pi r_0^2 R^2 \sigma T^4}{r^2}$

(B) $\frac{\pi r_0^2 R^2 \sigma T^4}{r^2}$

(C) $\frac{r_0^2 R^2 \sigma T^4}{4\pi r^2}$

(D) $\frac{R^2 \sigma T^4}{r^2}$

Sigma Percentile · Accepted Answer

### The Cosmic Geometry of Starlight

Imagine standing outside on a warm, sunny day. The heat you feel on your skin has traveled millions of kilometers through the freezing vacuum of space. But how much of the Sun's total fury actually reaches our tiny blue planet? This classic problem is a beautiful intersection of thermodynamics and geometry. Let's break it down step by step.

### The Sun as a Perfect Radiator

Our journey begins at the source: the Sun. We are told to model the Sun as a perfect spherical black body of radius $R$ and surface temperature $T$.

According to **Stefan's Law**, the total power $P$ (energy per second) radiated by a black body is proportional to its surface area and the fourth power of its absolute temperature. 
$$P = \sigma A T^4$$

Since the Sun is a sphere, its surface area is $A = 4\pi R^2$. Substituting this in, we get the total power output of the Sun:
$$P = \sigma (4\pi R^2) T^4$$

This is an unimaginably large number, representing the total energy blasting out in all directions every single second.

### The Journey Through Space

As this energy travels outward, it doesn't get lost, but it does get diluted. Imagine a giant, invisible sphere expanding outward from the Sun. By the time the radiation reaches the Earth at a distance $r$, that original power $P$ is now spread thinly over the surface of a sphere of radius $r$.

To find the **intensity** $I$ (power per unit area) at the Earth's location, we divide the total power by the surface area of this massive sphere ($4\pi r^2$):
$$I = \frac{P}{4\pi r^2} = \frac{\sigma (4\pi R^2) T^4}{4\pi r^2}$$

Notice how the $4\pi$ terms cancel out? This simplifies our intensity to:
$$I = \frac{\sigma R^2 T^4}{r^2}$$

This equation perfectly captures the **Inverse Square Law**. If you double the distance $r$, the intensity drops by a factor of four!

### Earth's Cosmic Shadow

Now, we look at the Earth, a sphere of radius $r_0$. How much of this incoming radiation does it actually catch?

A common mistake is to multiply the intensity by the Earth's total surface area ($4\pi r_0^2$) or its sun-facing hemisphere ($2\pi r_0^2$). But think about how a shadow works. If you hold a ball in front of a flashlight, the shadow it casts on the wall is a flat circle, not a hemisphere.

The Earth intercepts the Sun's rays exactly like a flat circular disk of radius $r_0$. Therefore, the **effective projected area** is simply the area of this circle:
$$A_{	ext{projected}} = \pi r_0^2$$

### The Final Calculation

To find the total radiant power received by the Earth, we multiply the local intensity of the sunlight by the Earth's projected area:
$$P_{	ext{received}} = I 	imes A_{	ext{projected}}$$

Substituting our previous results:
$$P_{	ext{received}} = \left( \frac{\sigma R^2 T^4}{r^2} ight) 	imes (\pi r_0^2)$$

Rearranging the terms to match the options, we get our final, elegant result:
$$P_{	ext{received}} = \frac{\pi r_0^2 R^2 \sigma T^4}{r^2}$$

And there we have it! A beautiful formula that connects the blazing surface of the Sun to the life-giving energy received by our home planet.

Select Answer:

Visualized Solution

Visualizing the Source

Stefan's Law

Energy Propagation

Calculating Intensity

Earth's Projected Area

Total Power Received

Final Expression

The Inverse Square Law

The Sigma Insight: Heat Transfer

The Cosmic Geometry of Starlight

The Sun as a Perfect Radiator

The Journey Through Space

Earth's Cosmic Shadow

The Final Calculation

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