How to calculate propulsion force on a laser pointer in reaction to the emitted light power in watts?

Should I just divide power by speed of light? Since power is energy per unit of time, hence dividing by speed of light gives momentum per unit of time, which is force.

Suppose I have pointer of 20 mW.

Is it's propulsion force $\frac{0.02}{299792458} = 6.67128\times10^{-11}$ Newtons?

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    $\begingroup$ Yep. With the caveats that the energy that actually goes into the light probably is less than the quoted power spec, and the divergence of the beam reduces (slightly) the net reaction force on the laser pointer. $\endgroup$ – Brionius Feb 10 '15 at 19:23

The Poynting vector $\vec{N}$ is the power per unit area of your beam.

If the beam is perfectly absorbed, then the force is given by $$ F = \frac{1}{c} \int \vec{N} \cdot d\vec{A}$$

So, providing you have the beam incident normally upon something, the force on it will just be the power of the laser divided by the speed of light.

Of course, if the light is reflected, then you have more force. In the limit of a perfect, specular reflection, the force is doubled.

If what you mean is, suppose I have a laser pointer suspended in space and I turn it on; What is the apparent accelerating force on the laser pointer? The answer is exactly the same as above. Power divided by the speed of light.

  • $\begingroup$ How the force on laser source can depend on what happening in the final point of beam? What if laser beam falls on stellar body in millions parsecs away? $\endgroup$ – Dims Feb 11 '15 at 5:28
  • $\begingroup$ @Dims You've misunderstood his meaning. The measurement is usually made of the force due to an arriving beam (because it lets us use a very light and finely balanced target, while the source would necessarily be heavier). $\endgroup$ – dmckee Feb 11 '15 at 5:29
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    $\begingroup$ Force of arriving beam is "light pressure", while I was interested in "light propulsion". I agree that these are related phenomena, but I am disputing with someone who don't believe in this relation. $\endgroup$ – Dims Feb 11 '15 at 10:50
  • $\begingroup$ @Dims You will have to clarify your question then. In no circumstances is a pressure a force. And when you ask for the "propulsion force", naturally the response you get is of what force the laser beam applies to an object. $\endgroup$ – Rob Jeffries Feb 11 '15 at 17:23
  • $\begingroup$ Sorry I am not native English. How do you call a force, which is applied to a rocket from exhaust? I was thinking that it is "propulsion" since "jet propulsion laboratory" :) $\endgroup$ – Dims Feb 11 '15 at 18:10

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