How much does the Earth's rotation affect a golf ball?

4,377

Solution 1

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.

Solution 2

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:

$d = -17.6976386 cm$

Solution 3

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.

Share:
4,377

Related videos on Youtube

Qmechanic
Author by

Qmechanic

Updated on August 01, 2022

Comments

  • Qmechanic
    Qmechanic over 1 year

    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?

    • Wutaz
      Wutaz about 10 years
      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.
    • Sheathey
      Sheathey about 10 years
      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.
    • DWin
      DWin over 9 years
      Well, maybe not snipers, but certainly artillery needed to understand Coriolis.
  • tpg2114
    tpg2114 about 10 years
    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?
  • dmckee --- ex-moderator kitten
    dmckee --- ex-moderator kitten about 10 years
    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.
  • Emilio Pisanty
    Emilio Pisanty about 10 years
    Great answer. Note, though, that the latitude of more typical locations like Mumbai or Sao Paulo is a bit lower.
  • Ryan Cavanaugh
    Ryan Cavanaugh about 10 years
    35 degree launch angle @ 45 m/s = 194m travelled, 5.2 seconds flight time. 20 seconds of flight time is far too long.
  • Kevin Kostlan
    Kevin Kostlan about 10 years
    Yes, 20 is way bit long. However, golf balls spin and gain lift form the atmosphere. I set it to 10 seconds.
  • padarom
    padarom over 9 years
    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?
  • Kevin Kostlan
    Kevin Kostlan over 9 years
    @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.