Electronic aiming systems use automated rangefinders that can sometimes be based on lasers. Obviously a projectile itself will be subject to the Coriolis Force (among others) and deviate from its course slightly and this can be accounted for/calculated. Was doing some work on the accuracy of these when I reached this dilemma.

The scenario I have in mind is (for the sake of argument) a gun firing a projectile at a target to the north which should get deflected slightly east at latitude of around 51 degrees north. The laser would be shone in the same direction.

So, will the laser beam be subject to the Coriolis Force in the same manner as the projectile?

My initial reaction was that the equation for the Coriolis Force:

$$F_{Coriolis} = -2m \overrightarrow{\omega} \times \overrightarrow{\dot{x}}$$

Depends on the mass of the object, and therefore if I take photons to be massless they shouldn't be affected, but this doesn't seem fulfilling. With the huge velocity of the beam I suppose that the effects will be extremely small, but it is more the principle that's bugging me.

Extra edit: In fact would the beam necessarily follow the curvature of the earth (at extremely large distances)?

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    $\begingroup$ The laser would be slightly bent downwards due to the gravitational pull, but almost nothing. If your target is 10 000 km away (a quarter of the Earth), the light will be there in 33 ms, which means that, if the gravity was uniform, it would have deflected the beam 5 mm. In reality, of course, this would be much less, as the gravity reduces with height. $\endgroup$
    – Davidmh
    Commented Mar 13, 2014 at 9:05

1 Answer 1


Coriolis force is not an actual force, but rather an effect observed in rotating frame of reference. The light path is not actually bent, so it doesn't matter that the photon has no mass, the Earth's rotation will have an affect on the photon's apparent path.

This does not contradict your calculation of $F_{Coriolis}=0$, because you have to put this force in $F=ma$, where $m$ is, again, zero. However, you can take the limit of both sides for $m\rightarrow 0$ and calculate $a$.

  • $\begingroup$ Yes, this is roughly what I suspected - and that also as that velocity of light was (in this case) far in excess of the speed of rotation of the frame that any effect would be tiny, but theoretically still there. Thanks. $\endgroup$
    – Folau
    Commented Mar 13, 2014 at 10:00

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