The earth receives at its surface radiation from the sun at the rate of 1400 Wm^-2. The distance of

Q: Why do we treat the Sun as a black body?

While no object is a perfect black body, stars like the Sun are excellent approximations. They absorb almost all radiation that falls on them and emit a continuous spectrum of radiation that closely matches the theoretical black body curve for their temperature.

Q: What happens to the energy that doesn't hit the Earth?

The Sun radiates energy in all directions spherically. The Earth only intercepts a tiny fraction of this energy (the part that hits its cross-sectional area). The rest continues traveling through space, eventually hitting other planets, dust, or traveling indefinitely.

Q: Why does the $4\pi$ cancel out in the intensity equation?

The $4\pi$ comes from the surface area of a sphere ($4\pi r^2$). The Sun emits from its spherical surface ($4\pi R^2$), and this energy spreads over a larger spherical surface at Earth's distance ($4\pi r^2$). Since both are spherical geometries, the $4\pi$ factor is common and cancels out.

Q: Is the temperature calculated the core temperature or surface temperature?

This is the surface temperature (specifically, the effective temperature of the photosphere). The radiation we receive and measure is emitted from the Sun's outer layers. The core temperature is vastly higher, around 15 million Kelvin!

Q: How do we calculate the fourth root without a calculator?

In competitive exams like JEE, you can often simplify the expression inside the bracket to a number whose fourth root is easier to estimate, or use approximations. For instance, $10^{14}$ becomes $10^{3.5}$, and you can use logarithmic approximations or binomial expansion for the remaining significant digits.

Question

The earth receives at its surface radiation from the sun at the rate of $1400 	ext{ Wm}^{-2}$. The distance of the centre of the sun from the surface of the earth is $1.5 	imes 10^{11} 	ext{ m}$ and the radius of the sun is $7 	imes 10^8 	ext{ m}$. Treating the sun as a black body, it follows from the above data that its surface temperature is ...... K.

Sigma Percentile · Accepted Answer

The universe is a grand stage of energy transfer, and the Sun is our primary actor. Have you ever wondered how we can determine the surface temperature of a star that is millions of kilometers away? We can't exactly send a thermometer there! Instead, we rely on the beautiful principles of thermodynamics and the behavior of light.

### Analyzing the Setup

Imagine the Sun as a perfect **black body**. In physics, a black body is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. More importantly for us, it is also a perfect emitter of radiation.

The Sun, with a radius $R$, radiates energy uniformly in all directions into the vastness of space. The Earth sits at a distance $r$ from the center of the Sun, catching a tiny fraction of this energy. The energy received per unit area per unit time at the Earth's surface is called the **intensity** or the solar constant, denoted by $I$.

### The Master Equation

To find the Sun's temperature, we need to connect the power it emits to the intensity we measure on Earth. First, let's look at the total power emitted by the Sun. According to the **Stefan-Boltzmann Law**, the total power $P$ radiated by a black body is proportional to its surface area and the fourth power of its absolute temperature $T$.

$$
P = \sigma A T^4 = \sigma (4\pi R^2) T^4
$$

Here, $\sigma$ is the Stefan-Boltzmann constant. This immense power travels through space, spreading out over larger and larger spherical areas. By the time it reaches the Earth at distance $r$, this power is spread over an imaginary sphere of radius $r$. The intensity $I$ is the power per unit area of this giant sphere:

$$
I = \frac{P}{4\pi r^2}
$$

Now, we substitute the expression for $P$ into our intensity equation:

$$
I = \frac{\sigma (4\pi R^2) T^4}{4\pi r^2}
$$

Notice how the $4\pi$ terms elegantly cancel out, leaving us with a neat relationship:

$$
I = \frac{\sigma R^2 T^4}{r^2}
$$

### Final Calculation

Our goal is to find the surface temperature $T$. Let's rearrange the equation to isolate $T$:

$$
T^4 = \frac{I \cdot r^2}{\sigma R^2}
$$

Taking the fourth root of both sides gives us our master formula:

$$
T = \left( \frac{I \cdot r^2}{\sigma R^2} ight)^{\frac{1}{4}}
$$

Now, we carefully plug in the given values. The intensity $I = 1400 	ext{ W/m}^2$, the distance $r = 1.5 	imes 10^{11} 	ext{ m}$, the Sun's radius $R = 7 	imes 10^8 	ext{ m}$, and the Stefan-Boltzmann constant $\sigma = 5.67 	imes 10^{-8} 	ext{ W/m}^2	ext{K}^4$.

$$
T = \left[ \frac{1400 	imes (1.5 	imes 10^{11})^2}{5.67 	imes 10^{-8} 	imes (7 	imes 10^8)^2} ight]^{\frac{1}{4}}
$$

Squaring the terms inside the bracket and grouping the powers of ten simplifies the expression significantly.

$$
T = \left[ \frac{1400 	imes 2.25 	imes 10^{22}}{5.67 	imes 10^{-8} 	imes 49 	imes 10^{16}} ight]^{\frac{1}{4}}
$$

After evaluating the fraction, we find:

$$
T \approx 5803 	ext{ K}
$$

And there we have it! By simply measuring the sunlight falling on a square meter of Earth, we have deduced the surface temperature of our star to be approximately **$5803 	ext{ K}$**. This is the power of physics—connecting the local to the cosmic!

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Total Power Radiated by the Sun

Intensity at Earth's Distance

Combining the Equations

Rearranging for Temperature

Substituting the Values

Final Calculation

The Way Forward

The Sigma Insight: Heat Transfer

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Total Power Radiated by the Sun

Intensity at Earth's Distance

Combining the Equations

Rearranging for Temperature

Substituting the Values

Final Calculation

The Way Forward

The Sigma Insight: Heat Transfer

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

The intensity of radiation emitted by the sun has its maximum value at a wavelength of 510 nm and that emitted by the north star has the maximum value at 350 nm. If these stars behave like black bodies, then the ratio of the surface temperature of the sun and the north star is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A body with area A and temperature T and emissivity e=0.6 is kept inside a spherical black body. What will be the maximum energy radiated?

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

A spherical black body with a radius of 12 cm radiates 450 W power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be

A body with area $A$ and temperature $T$ and emissivity $e = 0.6$ is kept inside a spherical black body. What will be the maximum energy radiated?