I just saw a post claiming that a professional golfer's t-shot would travel 4,170 feet if executed on the surface of the moon. Is it possible to be that accurate in accounting for gravity and atmospheric friction? If so, can you also account for those variables on the swing itself, as well as the flight of the ball?
Posted Apr 6, 2022, 6:13 AM UTC
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This is the calculation for an imaginary ideal scenario:
The top speed of the fastest shots is around 350 km/h (IIRC world record is
360 km/h).247.5 km/hWhile this is not at all the angle pro golfers on Earth use, the ideal angle for maximum distance in an environment with no air resistance (like the Moon) is 45°.
That makes the vertical and horizontal velocities identical: v_x = v_y.
Pythagoras can then be used to calculate that without having to deal with actual trigonometry functions: (350 km/h)^2 = v_x^2 + v_y^2
(350 km/h)^2 = v_x^2 + v_x^2
(350 km/h)^2 = 2 v_x^2 1/2 * (350 km/h)^2 = v_x^2 v_x = 1/sqrt(2) * 350 km/h
v_x =
Gravitational acceleration on the Moon is around g = 1.62 m/s^(2).
The fomula to calculate fall speed is v_x = v_x0 + g * t. v_x being the final vertical speed, v_x0 the inital vertical speed, g the gravitational acceleration, and t the time until the final measurement.
We can use this to figure out how long it takes for the ball to reach the apex of its flight path where the vertical velocity is 0 by setting v_x = 0:
0 = -247.5 km/h + 1.62 m/s^(2) * t
247.5 km/h = 1.62 m/s^(2) * t
68.75 m/s = 1.62 m/s^(2) * t
t =~ 42.44 s
So, it will take the ball around 42.44 seconds to reach its apex. Since there is no air resistance, the upward motion and downward motion are symmetric in path and speed, so we know it just takes twice that long for the full distance:
t_full = 2 * t = 2 * 42.44 s = 84.88 s
Since there is no air resistance, the horizontal speed remains constant throughout the flight, so all we need is to multiply the initial speed by the flight duration to get the total distance:
d = v_y * t_full = 247.5 km/h * 84.88 s = 68.75 m/s * 84.88 s =~ 5,835.5 m =~ 19,145 ft
That is quite a bit more than the 4,170 feet you listed. Why?
I found this article that goes into a bit more detail on that number.
They went with a much lower launch angle (10.52°) and a lower speed (274.2 km/h). The launch angle especially changes things by a lot: While a higher percentage of the speed is vertical, thereby increasing the distance, the low angle means a much shorter flight time.
The end of the article has another scenario with 45° and 306.93 km/h where they end up with ~4580 m which is much closer to our values; the discrepancy being due to the still notably lower speed.
So, the number you posted is a decent average prediction if you just dropped a pro golfer on the moon and let them take a single shot with their regular technique.
The number I calculated is (possibly) a decent prediction on a 'world' record after many attempts and adjusting technique for the Moon environment.
All assuming that there is no impediment from the space suit and whatnot.
Would the 10.8° launch angle be due to the face of a golf club driver's loft?
In part, yes. The other part is swing technique and where exactly the ball is hit.
But the driver is designed this way because this is the ideal launch angle (or close enough) for a golf ball on Earth.
As I wrote, the lower angle transfers more of the force put into the swing to vertical movement than horizontal movement. In a place with an atmosphere (and therefore air resistance), such a lower launch angle is just better than 45°. The reason being that air resistance slows down the ball throughout its flight, meaning the longer flight duration of the higher angle won't be as beneficial as it is without air resistance. So you'd rather get the most horizontal momentum right away since vertical momentum only extends your flight duration. And if you're not getting much more horizontal movement out of the tail end of that flight because air resistance reduced the speed, there's little point in wanting as much vertical speed.
Great explanation. Thanks
That's a brilliant response and very interesting. Then you very much.