Question 21.9: Clipper Ships of Space Goal Relate the intensity of light to...
Clipper Ships of Space
Goal Relate the intensity of light to its mechanical effect on matter.
Problem Aluminized mylar film is a highly reflective, lightweight material that could be used to make sails for spacecraft driven by the light of the sun. Suppose a sail with area 1.00 \mathrm{~km}^{2} is orbiting the Sun at a distance of 1.50 \times 10^{11} \mathrm{~m}. The sail has a mass of 5.00 \times 10^{3} \mathrm{~kg} and is tethered to a payload of mass 2.00 \times 10^{4} \mathrm{~kg}. (a) If the intensity of sunlight is 1.34 \times 10^{3} \mathrm{~W} and the sail is oriented perpendicular to the incident light, what radial force is exerted on the sail? (b) About how long would it take to change the radial speed of the sail by 1.00 \mathrm{~km} / \mathrm{s} ? Assume that the sail is perfectly reflecting.
Strategy Equation 21.30
p={\frac{2 U}{c}} (complete reflection) (21.30)
gives the momentum imparted when light strikes an object and is totally reflected. The change in this momentum with time is a force. For part (b), use Newton’s second law to obtain the acceleration. The velocity kinematics equation then yields the necessary time to achieve the desired change in speed.
(a) Find the force exerted on the sail.
Write Equation 21.30, and substitute U=\mathscr{P} \Delta t=I A \Delta t for the energy delivered to the sail:
\Delta p=\frac{2 U}{c}=\frac{2 \mathscr{P} \Delta t}{c}=\frac{2 I A \Delta t}{c}
Divide both sides by \Delta t, obtaining the force \Delta p / \Delta t exerted by the light on the sail:
\begin{aligned} F & =\frac{\Delta p}{\Delta t}=\frac{2 I A}{c}=\frac{2\left(1340 \mathrm{~W} / \mathrm{m}^{2}\right)\left(1.00 \times 10^{6} \mathrm{~m}^{2}\right)}{3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}} \\ & =8.93 \mathrm{~N} \end{aligned}
(b) Find the time it takes to change the radial speed by 1.00 \mathrm{~km} / \mathrm{s}.
Substitute the force into Newton’s second law and solve for the acceleration of the sail:
a=\frac{F}{m}=\frac{8.93 \mathrm{~N}}{2.50 \times 10^{4} \mathrm{~kg}}=3.57 \times 10^{-4} \mathrm{~m} / \mathrm{s}^{2}
Apply the kinematics velocity equation:
v=a t+v_{0}
Solve for t :
t=\frac{v-v_{0}}{a}=\frac{1.00 \times 10^{3} \mathrm{~m} / \mathrm{s}}{3.57 \times 10^{-4} \mathrm{~m} / \mathrm{s}^{2}}=2.80 \times 10^{6} \mathrm{~s}
Remarks The answer is a little over a month. While the acceleration is very low, there are no fuel costs, and within a few months the velocity can change sufficiently to allow the spacecraft to reach any planet in the solar system. Such spacecraft may be useful for certain purposes and are highly economical, but require a considerable amount of patience.