Kinetic reaction from emission of electromagnetic radiation

Question

If an object is emitting 1 watt of electromagnetic radiation directionally and has negligible black body radiation, how much thrust is produced? How can you calculate this generally?

Answer 1

A quick way to remember this is thrust is force. Assuming that the thermal blackbody radiation is isotropic it averages to zero so we can ignore that.

Then for a spacecraft emitting EM radiation in one direction we can say that the thrust is just:

$$T_{EM} = \frac{dp}{dt}$$

where $dp/dt$ is the amount of momentum per unit time produced in one direction as photons and the other direction as the spacecraft's momentum. Total momentum is unchanged since they are in opposite directions.

The momentum of a photon is $h/\lambda$ and the energy of a photon is $hc/\lambda$.

The number of photons per second is the energy per second or power $P$ (1 Watt in this case) divided by the energy of one photon:

$$\frac{dn}{dt} = P\frac{\lambda}{hc}$$

The electromagnetic thrust $T_{EM} = dp/dt$ is just the momentum per photon times the number of photons per second $dn/dt$:

$$T_{EM} = \frac{dp}{dt} = \frac{dn}{dt} \frac{h}{\lambda} =P\frac{\lambda}{hc} \frac{h}{\lambda} = \frac{P}{c}$$

Amazing right!?

It doesn't matter what the wavelength is; it could be a laser or a broadband incandescent light or even just hot RTG! The resulting thrust is just the power of the EM radiation in a given direction divided by $c$ the speed of light. If your source is anisotropic but has a spread in directions, you have to divide power per unit solid angle by $c$ to get thrust per unit solid angle, then integrate to get net thrust.

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So 1 watt is 1 kg m^2/s^3 and $c$ is 3×10⁸ m/s so the force is 3.33×10^-9 Newtons.

If you have an intensity $I$ (e.g 1361 W/m^2 from the Sun) then the pressure is $I/c$ if you absorb it. If you perfectly reflect it straight back it's double.