Physics Principle
The power ($P$) radiated by a perfect blackbody is given by the Stefan-Boltzmann Law:
$$P = \sigma A T^4$$
where:
$P$ is the total radiant power (in Watts).
$\sigma$ is the Stefan-Boltzmann constant ($\sigma \approx 5.67 \times 10^{-8}\ W\cdot m^{-2}\cdot K^{-4}$).
$A$ is the total radiating surface area (in $m^2$).
$T$ is the absolute temperature (in $K$).
For a sphere, the surface area $A$ is given by $A = 4\pi R^2$, where $R$ is the radius.
Substituting the area for a sphere into the Stefan-Boltzmann Law:
$$P = \sigma (4\pi R^2) T^4$$
$$P = 4\pi \sigma R^2 T^4$$
Given Data
The problem involves two blackbodies, body 1 and body 2, with the following properties:
| Property | Body 1 | Body 2 |
|---|---|---|
| Radius | $R_1$ | $R_2$ |
| Temperature ($T$) | $T_1 = 1000\ K$ | $T_2 = 2000\ K$ |
| Power ($P$) | $P_1 = 1\ kW$ | $P_2 = 1\ kW$ |
Since the power emitted is the same, $P_1 = P_2$.
Calculation
Set the power equations for both blackbodies equal to each other:
$$P_1 = P_2$$
$$4\pi \sigma R_1^2 T_1^4 = 4\pi \sigma R_2^2 T_2^4$$
The constants $4\pi$ and $\sigma$ cancel out:
$$R_1^2 T_1^4 = R_2^2 T_2^4$$
Rearrange the equation to find the ratio $R_1/R_2$:
$$\frac{R_1^2}{R_2^2} = \frac{T_2^4}{T_1^4}$$
Take the square root of both sides:
$$\frac{R_1}{R_2} = \sqrt{\frac{T_2^4}{T_1^4}}$$
$$\frac{R_1}{R_2} = \frac{T_2^2}{T_1^2}$$
Now, substitute the given temperature values:
$$\frac{R_1}{R_2} = \left(\frac{T_2}{T_1}\right)^2$$
$$\frac{R_1}{R_2} = \left(\frac{2000\ K}{1000\ K}\right)^2$$
$$\frac{R_1}{R_2} = (2)^2$$
$$\frac{R_1}{R_2} = 4$$
Final Answer
The ratio of the radii, $R_1/R_2$, is 4.

