Question

The outer surface of a spacecraft in space has an emissivity of \(0.8\) and a solar absorptivity of \(0.3\). If solar radiation is incident on the spacecraft at a rate of \(950 \mathrm{~W} / \mathrm{m}^{2}\), determine the surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed.

Short answer

Answer: The surface temperature of the spacecraft is approximately 291.52 K.

Step-by-step solution

Calculate the absorbed solar energy

To calculate the absorbed solar energy, we need to use the solar absorptivity (A) and the incident solar radiation (S). The absorbed solar energy (Q_absorbed) can be calculated using the following equation: $$Q_\text{absorbed} = A \cdot S$$ In our case, A = 0.3 and S = 950 W/m². Plugging in these values, we get: $$Q_\text{absorbed} = 0.3 \cdot 950 = 285 \mathrm{~W} / \mathrm{m}^{2}$$

Write the energy balance equation

To write the energy balance equation, we need to consider that the total emitted radiation by the spacecraft surface (Q_emitted) equals the absorbed solar energy. Thus: $$Q_\text{emitted} = Q_\text{absorbed}$$ We also know that the emissivity (ε) and the surface temperature (T) are related by the Stefan-Boltzmann law: $$Q_\text{emitted} = ε \cdot σ \cdot T^4$$ Where σ is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2}\mathrm{K}^{-4}\)). Combining these equations, we get the energy balance equation: $$ε \cdot σ \cdot T^4 = Q_\text{absorbed}$$

Solve for the surface temperature

Now, we can plug in the known values to solve for the surface temperature (T): $$0.8 \cdot 5.67 \times 10^{-8} \cdot T^4 = 285$$ Divide both sides by 0.8 and \(5.67 \times 10^{-8}\): $$T^4 = \frac{285}{0.8 \cdot 5.67 \times 10^{-8}}$$ Now, we can take the fourth root of both sides to find T: $$T = \sqrt[4]{\frac{285}{0.8 \cdot 5.67 \times 10^{-8}}} \approx 291.52 \mathrm{K}$$ Therefore, the surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed is approximately 291.52 K.

Key concepts

Emissivity

Emissivity is a crucial concept in understanding how objects emit thermal radiation. It is a measure of an object's ability to emit energy as thermal radiation and is denoted by the Greek letter \( \varepsilon \). Emissivity ranges from 0 to 1, where:

• 0 indicates that an object is a perfect reflector that emits no radiation,
• 1 indicates a perfect blackbody that emits the maximum possible radiation at a given temperature.

Most real-life objects have emissivity values between these two extremes. In the context of our spacecraft problem, the emissivity of the surface is 0.8, meaning it is quite efficient at radiating energy away as thermal radiation. This high emissivity is useful for the spacecraft to maintain its temperature stability in space, where thermal management is crucial.Knowing the emissivity helps predict how much energy will be radiated away, especially important in balancing the energy absorbed from sources like the Sun.

Solar Radiation

Solar radiation refers to the radiant energy emitted by the Sun, which reaches Earth and other bodies in space. This energy is primarily in the form of visible light but also includes infrared and ultraviolet light. The intensity of solar radiation incident on an object depends on its distance from the Sun and the angle of incidence. In the exercise, the spacecraft is exposed to solar radiation at a rate of \(950 \text{ W/m}^2\). The spacecraft's surface absorbs this energy based on its solar absorptivity property, which is 0.3 in this case. This means 30% of the incident solar energy is absorbed by the spacecraft. Understanding solar radiation is critical in designing spacecraft and other solar-powered devices, as it influences how much energy the object can capture and use. By managing the absorption and reflection of solar radiation, it is possible to regulate the temperature and energy balance of space vehicles.

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is a fundamental physical law used to describe how much thermal radiation an object emits based on its temperature. Specifically, it quantifies the power radiated from a blackbody in terms of its temperature. The law is expressed as:\[ Q_{\text{emitted}} = \varepsilon \cdot \sigma \cdot T^4 \]where:

• \(Q_{\text{emitted}}\) is the radiant energy emitted per unit area,
• \(\varepsilon\) is the emissivity of the object's surface,
• \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \text{ W/m}^2\text{K}^4\)),
• \(T\) is the absolute temperature of the object in Kelvin.

In our scenario, this law helps calculate the balance point where the energy absorbed from solar radiation equals the energy emitted. By rearranging and solving the equation, we can determine the surface temperature of the spacecraft. The surface temperature is critical because it ensures the spacecraft's systems function within their optimal thermal range, avoiding overheating or freezing in the harsh environment of space.