Solving the energy crisis, one giant construction project at a time

The Stefan-Boltzmann Law states that if $s$ is the Stefan-Boltzmann constant, which is $5.67 \times 10^{-8} W m^{-2} K^{-4}$ , $A$ is the area of the body in $m^2$ , and $T$ is the temperature of the body in $K$ , the power radiated in watts is $sAT^4$ . Since the radius of the sun is $7 \times 10^8$ meters, the surface area, which is $4\pi r^2$ is $196 \times 10^{16} \pi = 6.1575 \times 10^{18} m^2$ . $T^4 = 6000^4 = 1296000000000 = 1.296 \times 10^{15}$ . Hence, the energy emitted is: $sAT^4 = (5.67 \times 10^{-8} W m^{-2} K^{-4}) \times (6.1544 \times 10^{18} m^2) \times (1.296 \times 10^{15} K^4) = 4.5225 \times 10^{26} W$ . Since the total needs is $15$ Terawatts, which is $1.5 \times 10^{13}$ , the fraction of energy emitted that needs to be captured is $\frac{1.5 \times 10^{13}}{4.5225 \times 10^{26}} = \boxed{3.317 \times 10^{-14}}$ .

The Stefan-Boltzmann Law states that the irradiance $E$ (energy per unit area per unit time) of a black body is jointly proportional to the fourth power of the temperature $T$ and the emissivity $\epsilon$ : $E=\sigma \epsilon T^4$ . In this equation, $\sigma$ is the Stefan-Boltzmann constant, which is approximately $5.6704\times 10^{-8}\,\, \text{J}\cdot \text{s}^{-1}\cdot \text{m}^{-2}\cdot \text{K}^{-4}$ We can then use this to find the total power of the black body by multiplying by the surface area, which is $4\pi r^2$ , to get that the total power of the Sun: $P=4\pi r^2 E=4\pi r^2\sigma\epsilon T^4$ . Since we are assuming that the Sun is a perfect black body, $\epsilon=1$ . Now, all that is left to do to find the total power is to plug in the given numbers: $P=4\pi(7\times10^8 \,\text{m})^2(5.6704\times 10^{-8}\, \text{J}\cdot \text{s}^{-1}\cdot \text{m}^{-2}\cdot \text{K}^{-4})(6000 \,\text{K})^4 \\ =4\pi(49\times 10^{16})(5.6704\times 10^{-8})(1296\times 10^{12})\, \text{W} \\ \approx 4.525\times 10^{26} \, \text{W}$ To find the fraction of the total energy the sphere needs to capture, we divide the energy need by the total Solar energy: $\dfrac{15\times 10^{12}\, \text{W}}{4.525\times 10^{26} \,\text{W}} \approx \boxed{3.3149\times 10^{-14}}$ .

Stefan-Boltzmann Law stated that power emitted by a black body per surface area is given the formula, P=\sigma T^4 . The \sigma is the Stefan-Boltzmann constant, which is equal 5.67 \times 10^{-8} W m^{-2} K^{-4} .

Therefore, P =5.67 \times 10^{-8} \times 6000^4 =73483200 Watts

Then the total surface area of the sun =4 \times \pi \times (7 \times 10^8)^2 =6.157521601 \times 10^{18}

After that, take Power \times total surface area of sun The value will be the total power emitted, 4.525063119 \times 10^{26} Now take 15Terawatts and divide it by the total power and the answer is 3.314871 \times 10^{-14}

The Stefan-Boltzmann Law relates luminosity (flux, or energy output) with the stars radius and temperature. It is of the form $L=4\pi R^2\sigma T^4$ where $\sigma=5.67\times10^{-8}$ is the Stefan-Boltzmann constant, T is absolute temperature, and L is in Watts. Plugging in our numbers, we get $\begin{aligned}L&=4\pi (7\times10^8)^2\cdot5.67\times10^{-8}\cdot6000^4\\&=4.52\times10^{26}\end{aligned}.$

We want to find the fraction of that number that is represented by 15 terawatts, or $1.5\times10^{13}$ watts, so we divide to get $\frac{1.5\times10^{13}}{4.52\times10^{26}}=\boxed{3.32\times10^{-14}}.$

In subject matter of Radiation, it is known a formula:

P = e * σ * A * T**4

Where: P = Power in (W) e = fraction of emitter σ = Steven-Boltzmann constants = 5.67 * 10E-8 (W/m 2 K 4) A = Surface Area (m 2) = 4 π r 2 ( if it is ball) T = Temperature (K 4) Based on the question, we need to find the value of e, then: 15 * 10 12 = e * (5.67 10E-8) * (4 3.14 49 10 16) (1296 10 12)

Just finish the algebraic and finally you will find the value of e = ± 3.3*10E-14

The Power radiated from a ideal black body is given by the Stefan Boltzmann Law which states that $\\ P=A \sigma {T^4}$ where P is the power, A is the area, $\sigma$ is the Stefan boltzmann constant and T is the absolute temperature of the black body. Let $x$ fraction of the energy emitted by the sun. Clearly, $P\cdot x =15*10^12$ . Substituting, we get, $x=3.31E-14$

Since the sun is assumed to be a perfect black body, we can find it's total power output using Stefan-Boltzmann's Law:

$$P = A\sigma T^4$$ $$P = 4\pi (7 \cdot 10^8)^2 \cdot 5.67 \cdot 10^{-8} \cdot 6000^4 = 4.52 \cdot 10^{26} \text{ W}$$.

Dividing the total consumed power of $12 \text{ TW} = 12 \cdot 10^{12} \text{ W}$ by the sun power output $P$ we get the answer $\boxed{3.31 \cdot 10^{-14}}$ .

By the Stefan-Boltzmann law for a perfect black body we have

$j^{\star}$ = $\sigma T^{4}$

where $j^{\star}$ is the radiant exitance.

Now the $P$ is given by

$P$ $=$ $A \times j^{*}$

The area of sun is given by $4\pi$ $r^{2}$

So the $P_(req)$ = $4\pi$ $r^{2} \times$ $\sigma T^{4}$

And hence the fraction of energy required is:

$Fraction$ = $\frac {4 \pi r^{2}}{\sigma T^{4}}$

Finally plugging in the values we get the fraction as $3.31E-14$

Since we are modelling the Sun is a perfect blackbody, it's emissivity will be $1$ . By the Stefan Boltzmann law, the total power radiated from the sun per unit surface area will be: $E= \sigma T^4$ where $\sigma$ is the Stefann Boltzmann constant, and $T$ is the temperature. Since the total surface area of the sun is $4 \pi r^2$ (where $r$ is the radius of the Sun), the total power emitted from the sun will be $P = 4 \pi r^2 \times E = 4 \pi r^2 \sigma T^4$ If the power we need to capture is $P_{req}$ , the fraction of the Sun's energy we need to consume will be: $\text{Fraction}= \frac{ 4 \pi r^2 \sigma T^4}{P_{req}}$ Plugging in the values, we get the fraction to be $\boxed{3.3149 \times 10^{-14}}$ .

By the Stefan–Boltzmann law: $j^{\star} = \sigma T^{4},$ where $j^{\star}$ is the power emitted per unit area, $\sigma$ is the Stefan–Boltzmann constant (about $5.670373\times 10^{-8}~Wm^{-2}K^{-4}$ ), and $T$ is the temperature of the sun.

Hence to get the total power output from the sun ( $P$ ) we multiply $j^{\star}$ by the sun's surface area ( $A$ ). The sun is roughly a sphere, so $A=4\pi \times (7\times10^8)^2$ .

Therefore: $P=4\pi \times (7\times10^8)^2 \times 5.670373\times 10^{-8} \times 6000^4,$ $P\approx 4.525\times 10^{26}.$

So the fraction needed for the Dyson sphere is $\frac{15\times 10^{12}}{4.525\times 10^{26}},$ which is about $3.31\times 10^{-14}$ .