A solar sail is a giant circle (with a radius \(R=10.0 \mathrm{~km}\) ) made of a material that is perfectly reflecting on one side and totally absorbing on the other side. In deep space, away from other sources of light, the cosmic microwave background will provide the primary source of radiation incident on the sail. Assuming that this radiation is that of an ideal black body at \(T=2.725 \mathrm{~K},\) calculate the net force on the sail due to its reflection and absorption.

Short Answer

Expert verified
Answer: The net force on the solar sail is approximately 3.258 x 10^-6 N.

Step by step solution

01

Determine the incident power on the sail

The cosmic microwave background has a temperature of 2.725 K, and considering it as an ideal black body, we can use the Stefan-Boltzmann law to determine the incident power on the sail. The Stefan-Boltzmann law states: $$P = \sigma T^4$$ where \(P\) is the power per unit area, \(\sigma = 5.67 \times 10^{-8} \mathrm{W m^{-2} K^{-4}}\) is the Stefan-Boltzmann constant, and \(T\) is the temperature of the black body. Plugging the values, we get: $$P = (5.67 \times 10^{-8} \mathrm{W m^{-2} K^{-4}}) (2.725 \mathrm{K})^4 = 3.1042 \times 10^{-6} \mathrm{W m^{-2}}$$
02

Calculate the absorbed power on the sail

Next, we need to find the absorbed power on the sail. Since the incident radiation falls on both sides of the sail, we must consider both the perfectly reflecting side and the totally absorbing side. For the absorbing side, all the incident power is absorbed, so the absorbed power per unit area is the same as the incident power: $$P_\mathrm{absorbed} = P$$ For the reflecting side, no power is absorbed, so the absorbed power per unit area is zero: $$P_\mathrm{reflected} = 0$$
03

Calculate the pressure difference on the sail

The pressure difference on the sail due to the absorbed and reflected power can be calculated using the following formula: $$\Delta P = \frac{1}{c} \left(P_\mathrm{absorbed} - P_\mathrm{reflected} \right)$$ where \(c\) is the speed of light, which is approximately \(3 \times 10^8 \mathrm{m s^{-1}}\). Plugging in the values, we get: $$\Delta P = \frac{1}{3 \times 10^8 \mathrm{m s^{-1}}} (3.1042 \times 10^{-6} \mathrm{W m^{-2}}) = 1.0347 \times 10^{-14} \mathrm{Pa}$$
04

Calculate the net force on the sail

Finally, we can find the net force on the sail using the pressure difference and the sail area: $$F_\mathrm{net} = \Delta P \times A$$ The sail is a circle with a radius of \(10 \mathrm{km}\), so its area is given by: $$A = \pi R^2 = \pi (10 \times 10^3 \mathrm{m})^2 = 3.1416 \times 10^8 \mathrm{m^2}$$ Now, we can calculate the net force: $$F_\mathrm{net} = (1.0347 \times 10^{-14} \mathrm{Pa}) \times (3.1416 \times 10^8 \mathrm{m^2}) = 3.258 \times 10^{-6} \mathrm{N}$$ The net force on the solar sail due to its reflection and absorption of the cosmic microwave background radiation is approximately \(3.258 \times 10^{-6} \mathrm{N}\).

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