# Kinetic reaction from emission of electromagnetic radiation

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?

• Great question, and thanks for your accept. I've just added a little more background below "Amazing right!?"
– uhoh
Apr 3, 2021 at 2:10

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.

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.

• I just yesterday came back to this question and thought of a different way to approach it. e = mc^2, right? So can't you solve for m given e and then multiply both sides by the velocity to get the momentum? m = e/c^2 --> mv = e/c^2 * v, v = c --> momentum = ec/c^2 = e/c --> force = p / c where p is power and c is the speed of light. This is the same formula that you mentioned, but I'm new to physics (I'm in 8th grade) and I'm not sure that all of my steps are doable generally or if they'll lead me to the wrong answer in some case. What are my mistakes, if any? Apr 16, 2021 at 13:48
• @WesleyAdams In stack Exchange we don't ask new questions in comments. I recommend that you post a new question asking about that, you can link back to your question here. I'm no expert, but I can tell you that you have to use the proper relativistic equations for energy and mass and it doesn't look like you did. Even NASA people get it wrong sometimes, see this answer.
– uhoh
Apr 16, 2021 at 14:05