Physics of Light Sail Spacecraft Question) LightSail-1 is a spacecraft that will
ID: 1486866 • Letter: P
Question
Physics of Light Sail Spacecraft
Question)
LightSail-1 is a spacecraft that will be launched in late 2010. It is designed to fly far above the Earth (to avoid atmospheric drag), and will test the practicality of using sunlight for propulsion. The spacecraft has a mass of 4.5 kg, and the sail has an area of 32.5 m^2. Assuming that all sunlight strikes the sail at normal incidence (i.e., perpendicular to the plane of the sail) and is perfectly reflected. What will be the spacecraft’s speed one hour after the sail opens? Assume that the energy intensity of sunlight is 1400 W/m^2. Ignore all gravitational effects.
My Attempt)
I know the answer is 0.24m/s but I have absolutely no idea how to get to that answer. My line of thinking is to multiply energy intensity 1400W/m^2 by the area of the solar sail 32.5m^2 to get 45500W. Since a watt is 1J/s I thought to multiply 45500J/s by 3600s in an hour. Then to set that number equal to the kinetic energy and I got a velocity of 8532m/s, which is wrong. Can someone please tell me where my reasoning went wrong and how to get to the answer?
Explanation / Answer
here the energy is not obsorbed by the spacecraft. The total energy is reflected back. but you assumed the spacecraft is obsorbing total energy.
given, I = 1400 W/m^2
we know, light speed c = 3*10^8 m/s
Area of wing, A = 32.5 m^2
m = 4.5 kg
radiation pressure on perfectly reflecting surface,
P = 2*I/c
= 2*1400/(3*10^8)
= 9.333*10^-6 pa
force on wings due to this radiation pressure, F = P*A
= 9.333*10^-6*32.5
= 3.033*10^-4 N
From Newton's second law,
acceleration of spacecraft, a = F/m
= 3.033*10^-4/4.5
= 6.74*10^-5 m/s^2
at time t = 1 hour = 60*60 s
use, v = u + a*t
= 0 + 6.74*10^-5*60*60
= 0.24 m/s <<<<<<<------------------Answer