What is the effect of distance on the force created by a laser in outer space?

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SUMMARY

The discussion focuses on calculating the time it takes for one astronaut to reach the Space Shuttle when propelled by a 123.0-W laser. The force exerted by the laser is determined using the formula Force = Power / speed of light, resulting in a force of 4.1E-7 N. The astronaut's mass is 129.0 kg, leading to an acceleration of 3.178E-9 m/s². The participant questions the impact of distance on the calculations and suggests that a totally reflecting surface should double the force, which is confirmed by another user.

PREREQUISITES
  • Understanding of basic physics concepts such as force, mass, and acceleration.
  • Familiarity with the equation Force = Power / speed of light.
  • Knowledge of kinematic equations, particularly Df = Do + VoT + 1/2aT².
  • Concept of reflection and its effects on force application.
NEXT STEPS
  • Study the principles of laser propulsion in space environments.
  • Learn about the effects of distance on force in physics.
  • Explore advanced kinematic equations for varying acceleration scenarios.
  • Investigate the physics of totally reflecting surfaces and their impact on force dynamics.
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Students studying physics, aerospace engineers, and anyone interested in the applications of laser technology in space propulsion.

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Homework Statement


Two astronauts are at rest in outer space, one 15.3 m from the Space Shuttle and the other 30.6 m from the Shuttle. Using a 123.0-W laser, the astronaut located 30.6 m away from the Shuttle decides to propel the other astronaut toward the Space Shuttle. He focuses the laser on a piece of totally reflecting fabric on her space suit. If her total mass with equipment is 129.0 kg, how long will it take her to reach the Space Shuttle?


Homework Equations



Force=Power/c



The Attempt at a Solution



Since Force=Power/speed of light

F=(123W)/(3x10^8)
F=4.1E-7 since F=ma (4.1E-7)/129kg = accel= 3.178E-9 m/s^2

the laser causes constant acceleration so...
Df=Do + volt+ 1/2aT^2
and I solved for T... with Df = 15.3 meters, Do and Vo both equal 0.

but this didn't work and I have no idea where to go from here. Does the distance of the second astronaut (the one shooting the laser) affect the calculation? Any help?
Thanks!
 
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This is just a suggestion:
If it's a totally reflecting surface shouldn't the force (F) of the beam be doubled?

Just like if water hits the back of a spoon as compared to the front of a spoon. The water hitting the front is deflected and therefore puts a greater force on the spoon.
 
I just tried your hunch and you were correct. Thanks a bunch!
 

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