Question

Scientists have proposed using the radiation pressure of sunlight for travel to other planets in the Solar System. If the intensity of the electromagnetic radiation produced by the Sun is about \(1.40 \mathrm{~kW} / \mathrm{m}^{2}\) near the Earth, what size would a sail have to be to accelerate a spaceship with a mass of 10.0 metric tons at \(1.00 \mathrm{~m} / \mathrm{s}^{2} ?\) a) Assume that the sail absorbs all the incident radiation. b) Assume that the sail perfectly reflects all the incident radiation.

Short answer

Answer: (a) 21,429 m² (b) 10,714 m²

Step-by-step solution

Calculate the force exerted by the sunlight on the sail in both scenarios

The intensity of the sunlight is given in \(kW/m^2\). First, we need to convert the intensity to watt/m^2: \(1.40 \mathrm{~kW} / \mathrm{m}^{2} = 1400 \mathrm{~W} / \mathrm{m}^{2}\). Next, we use the intensity to calculate the force exerted by the sunlight on the sail. The force due to the radiation pressure on the sail is given by \(F = \dfrac{I \cdot A}{c}\) for absorption (scenario a), and \(F = \dfrac{2 \cdot I \cdot A}{c}\) for reflection (scenario b), where \(I\) is the intensity, \(A\) is the area of the sail, and \(c\) is the speed of light in a vacuum.

Relate the force exerted on the sail to the acceleration of the spaceship

We can use Newton's second law to establish a relationship between the force on the sail and the acceleration of the spaceship: \(F = m \cdot a\), where \(F\) is the force exerted on the sail, \(m\) is the mass of the spaceship, and \(a\) is its acceleration. Now, we can write down a relationship between the sunlight's radiation pressure force and the acceleration of the spaceship for both scenarios (absorption and reflection): (a) \(\dfrac{I \cdot A}{c} = m \cdot a\) (b) \(\dfrac{2 \cdot I \cdot A}{c} = m \cdot a\)

Calculate the sail area for both scenarios

In both scenarios, we need to find the area of the sail, \(A\). We can solve for \(A\) by rearranging the equations from step 2: (a) \(A = \dfrac{m \cdot a \cdot c}{I}\) (b) \(A = \dfrac{m \cdot a \cdot c}{2 \cdot I}\) Now, we can plug in the given values: \(m = 10.0 \cdot 10^3 \mathrm{~kg}\), \(a = 1.00 \mathrm{~m/s^2}\), \(I = 1400 \mathrm{~W/m^2}\), and \(c = 3.00 \cdot 10^8 \mathrm{~m/s}\). (a) \(A_{absorption} = \dfrac{(10.0 \cdot 10^3 \mathrm{~kg})(1.00 \mathrm{~m/s^2})(3.00 \cdot 10^8 \mathrm{~m/s})}{1400 \mathrm{~W/m^2}}\) (b) \(A_{reflection} = \dfrac{(10.0 \cdot 10^3 \mathrm{~kg})(1.00 \mathrm{~m/s^2})(3.00 \cdot 10^8 \mathrm{~m/s})}{2(1400 \mathrm{~W/m^2})}\)

Compute the sail area for both scenarios

Now, we can compute the sail area required for both scenarios: (a) \(A_{absorption} = \dfrac{(10.0 \cdot 10^3 \mathrm{~kg})(1.00 \mathrm{~m/s^2})(3.00 \cdot 10^8 \mathrm{~m/s})}{1400 \mathrm{~W/m^2}} \approx 21,429 \mathrm{~m^2}\) (b) \(A_{reflection} = \dfrac{(10.0 \cdot 10^3 \mathrm{~kg})(1.00 \mathrm{~m/s^2})(3.00 \cdot 10^8 \mathrm{~m/s})}{2(1400 \mathrm{~W/m^2})} \approx 10,714 \mathrm{~m^2}\) So the sail would need to have an area of: (a) approximately 21,429 m² if it absorbs all the incident radiation, (b) approximately 10,714 m² if it perfectly reflects all the incident radiation.