[Physics] How much does the Earth’s rotation affect a golf ball

coriolis-effectearthnewtonian-gravityprojectile

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?

Best Answer

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.