Distance Traveled by a Laser Pointer in Outer Space

  • #1

Homework Statement


A 50 g, 420 mw laser pointer is floating in outer space (don’t ask how it got there) at rest with respect to an observer. The laser pointer is turned on and let go. If the battery runs continuously for 250 hours before dying, what is the final speed achieved by the laser pointer with respect to the observer?

Homework Equations


P=W/t (Power)
Ek=pc
W=Ek (conservation of energy)
n=v/c
p=mv(momentum)

The Attempt at a Solution




Since it is in space, I can say that n=1 so v=c(it is essentially in vacuum). From this I can derive that
W=Ek
from the conservation of energy law. Knowing this I can replace
P=Ek/t.

This is where I assume the question is non-relativistic which I am not sure if I can. Let's say I do make the assumption. I can then get

Pt= Ek = pc
Pt = pc = mvc
v = Pt/mc
v = (420 * 10-3 W) * (900,000 s)/(0.05 kg)(3.00 * 108 m/s)
= 2.52 m/s

I am pretty sure I can not say Pt = pc since pc is the energy for light and Pt is referring to the laser pointer itself but I am not sure what else to do
 
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Answers and Replies

  • #2
Orodruin
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2.5 m/s seems pretty non-relativistic to me.

You have only treated non-relativistic objects non-relativistically. The light, as you say, must have E/c as its momentum and the momentum of the pointer must be of equal magnitude and opposite direction.
 
  • #3
Ah okay so if use momentum to find the velocity then I would get

p=mv = E/c

but I'm not sure where to go from here since I don't know E or v
So I'm going to try using conservation of energy

P*t=W = Ek = 0.5mv2
v=[2P*t/m]1/2
v = 3888 m/s

This answer is very different than 2.5 m/s however I think this is more correct than 2.5 m/s
 
  • #4
Orodruin
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So I'm going to try using conservation of energy
You are not using it correctly. In fact, only a minor part of the energy will be carried by the laser pointer. Most of the energy is carried away by the emitted light.

This answer is very different than 2.5 m/s however I think this is more correct than 2.5 m/s
Why would you think so?
 
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  • #5
You are not using it correctly. In fact, only a minor part of the energy will be carried by the laser pointer. Most of the energy is carried away by the emitted light.

If I want to find the energy carried away by the emitted light would I have to find the number of photons that were emitted over 250 hours?

Why would you think so?
Because I am certain that in my original post my statement

Pt= Ek = pc

makes no sense as I am relating the energy of a photon to the total energy of the pointer.
 
  • #6
Orodruin
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Because I am certain that in my original post my statement

Pt= Ek = pc

makes no sense as I am relating the energy of a photon to the total energy of the pointer.
This is not true. You are relating the light momentum to the light energy and then relating it to the light momentum. Nowhere did you involve the pointer kinetic energy.
 
  • #7
I see. Then would it be correct in saying that

P:t/c = p <--photon
mv=p <--- pointer
therefore
P*t/c = mv
v = P*t/m*c
 
  • #8
Orodruin
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Yes, as long as ##v \ll c##, this is a good approximation.
 
  • #9
I see. Thanks!
 

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