When launching rockets into space, they start straight but begin to lean, I assume due to the Earth's rotation.

For something on a much smaller scale, such as a golf tee, is there a measurable trajectory alteration, or is it something so incredibly minuscule that there's virtually no effect?

  • $\begingroup$ This reminds me of Modern Warfare, where they claimed that snipers had to compensate for the Coriollis effect. In that case and especially this one, it doesn't matter. $\endgroup$
    – Wutaz
    Sep 16, 2013 at 20:16
  • $\begingroup$ During a space shuttle launch, the shuttle tilts backwards to allow the pilot to see the ground and keep it on a relatively vertical course. $\endgroup$
    – Sheathey
    Sep 17, 2013 at 2:18
  • $\begingroup$ Well, maybe not snipers, but certainly artillery needed to understand Coriolis. $\endgroup$
    – DWin
    Jul 17, 2014 at 23:53

4 Answers 4


The answer: the ball appears to be deflected ~10 cm.

The calculation: For simplicity, say we tee off at the north pole. The effects are a bit weaker at more typical locations, you multiply by sin(latitude) = 0.64 for a 40 degree (central california or washington DC) latitude.

The Coriolis effect exists because the Earth rotates while the ball is in flight (we ignore air friction). We can picture the hole attached to the edge of a circle which rotates horizontally while our ball is in flight. This will mean that the hole is deflected sideways by the time the ball gets there. However, since we are also rotating, we think that the ball was deflected.

A realistic golf swing puts the ball (not the tee!) in flight 10 seconds and is 200m long. Our circle rotates around once per day, or 10*360/(60*60*24) = 0.04 degrees during the flight. For a 200m radius, 0.04 degrees corresponds to a sideways motion of (circumference)*(# of revolutions) = (200m*2*pi)*(0.04/360) = 0.15m = 15cm. At a 40 degree latitude, you get about 10 cm.

This is enough to make you miss your hole in one, but I don't think even the pro golfers can achieve a 10cm accuracy on a 200m swing. For putting, the deflection is much less since distance is far smaller. So don't worry about aiming 10cm clockwise the next time your going for that hole in one.

Edit: Our calculation does not include the initial Eastward velocity due to the motion of the golfer. This doubles the effect (at least on the poles), but air and/or ground friction will reduce the the effect and it's not unreasonable to guess that they approximately cancel.

  • $\begingroup$ Great answer. Note, though, that the latitude of more typical locations like Mumbai or Sao Paulo is a bit lower. $\endgroup$ Sep 16, 2013 at 21:03
  • 2
    $\begingroup$ 35 degree launch angle @ 45 m/s = 194m travelled, 5.2 seconds flight time. 20 seconds of flight time is far too long. $\endgroup$ Sep 16, 2013 at 22:00
  • $\begingroup$ Yes, 20 is way bit long. However, golf balls spin and gain lift form the atmosphere. I set it to 10 seconds. $\endgroup$ Sep 16, 2013 at 23:10
  • $\begingroup$ About the putting: Is the ball even slightly affected by earths rotation? It never leaves the ground, basically you "roll" it towards the hole. So does the rotation of earth have any effect on it? $\endgroup$
    – padarom
    Jun 25, 2014 at 11:45
  • 1
    $\begingroup$ @Padarom: Yes it would still affect it. The ground doesn't magically prevent the "Corellolis effect" from existing. However, the ground friction would partially counteract the effect by "dragging" the ball along with Earths rotation. $\endgroup$ Jun 26, 2014 at 17:24

An alternate derivation for a due-north ball, ignoring the diminishing effect of latitude, that confirms Kevin's order of magnitude:

Acceleration due to the Coriolis effect: $a_C = -2 \, {\Omega \times v}$

$\Omega = 2 \pi/day$

$v = 45 m/s$

$a_C = -0.00654498469 m/s^2$

Horizontal displacement $d$ is given by

$d = 1/2a_C\,t^2 $

Using earlier estimate of 5.2s flight time for a ~200m drive:

$d = -17.6976386 cm$


Rockets lean as they climb on purpose, in order to obtain the high orbital velocities needed to stay in space once they get there. For a nice explanation, see Orbital Speed at xkcd what-if, but the gist is the following. Being in orbit means going so fast that the Earth begins to curve away from you as you fall down towards it. The classic image to keep in mind is

enter image description here

Source: Astronomy Education at the University of Nebraska-Lincoln

To actually orbit, you need to be going fast enough for this to happen. Rockets lean sideways to get such an acceleration; that's the 'cannon' if you will.

While there is an appreciable effect on the rocket's motion (or its apparent motion from Earth's surface) due to the Earth's rotation as it climbs into orbit, this is more than compensated by the intentional manoeuvering.

And, of course, the effect on a golf ball is negligible.

  • 3
    $\begingroup$ The Coriolis effect on a golf ball could be calculated which would make this answer complete. On a typical length drive, is the deviation microns or millimeters? $\endgroup$
    – tpg2114
    Sep 16, 2013 at 20:15
  • 4
    $\begingroup$ This is correct but really only addresses the misconception in the question text, the answer is small, but calculable and is only negligible compared the the repeatability of the golf swing, not compared to the size of the ball. $\endgroup$ Sep 16, 2013 at 20:38

Golf swing
As has been pointed out in other answers: at best golfers can put a 200m swing in something like a 10m circle. So there will never be a measurable deviation; the aim of a golf swing is limited in accuracy.

Long range rifle shooting
There won't be any measurable rotation-of-Earth effect in the case long range rifle shooting either. A long range rifle that has been dialed in is extremely accurate, but the travel time of the bullet is too short to build up a measurable effect.

However, it may well be that high power crossbows can show the effect. A crossbow that is designed to repeat very well can be dialed to a high accuracy. At the same time the bolt is under way long enough for the effect to build up.

The procedure of dialing in incorporates all effects that influence where the projectile ends up. For firing north-south the deviation due to the Earth rotation is the same either way, so when a crossbow is dialed in firing south-to-north then firing north-to-south the aim will still be just as good.

But there will be a discernible effect, I expect, for east-west direction. After dialing in shooting west-to-east there will be a deviation shooting east-to-west. The simplest case of that is at the equator. Shooting towards the West the projectile ends up circumnavigating the Earth's center at a slower velocity than the equator itself, and shooting towards the East the projectile ends up circumnavigating the Earth's center at a faster velocity than the equator itself. In all the consequence is that the westbound projectile will drop faster than the eastbound projectile.

The amount of drop is part of wat gets incorporated in the correction when a rifle scope is dialed in. So I expect a (high power) crossbow that has been dialed in in one direction (along east-west) will miss the target by a predictable margin when shooting in the opposite direction.


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