In 1974, Mike Austin set the World Record for the longest drive in golf at 515 yards. To put this in perspective, the ball flew 65 yards past the fifth hole, a par 4, at the Winterwood Golf Course in Las Vegas, Nevada. If you’re still not convinced of this feat, the average distance of the top 115 drivers in 2017 was 290 yards. This means Mike Austin’s drive is 1.77 times that distance.
He did what?!
In 1971, astronaut Alan Shepard performed the infamous Apollo 14 ‘golf stunt’. If you didn’t know, Shepard snuck a golf club and two golf balls onto the Apollo 14 mission. He hid the balls in his sock, the head of the club in his helmet and the shaft to a piece of equipment used for taking soil samples. Considering NASA was in no mood to mess around after the events of Apollo 13, if Shepard had been caught, he would have been in a lot of trouble.
According to Shepard he shanked the first ball into a crater but managed to hit the second flush and send it flying. Of course, the restriction of the suit on mobility meant that the distance covered was probably not that far, Shepard reckoned about 200 yards.
What if we could take a proper golf swing like we do on Earth, with no space suit restrictions, yet conduct it in the low-gravity, low-resistance environment of the moon?
Well, there are lots of calculations and parameters to consider, but the end result turns out to be just short of 2.5 miles (3938.4 metres). Wow. that’s quite some distance – imagine an 18-hole golf course accommodating those figures. The moon buggy would definitely be handy between greens!
Here are the calculations – if you really want to know
On Earth, you would expect to drive the ball at a relatively low angle, minimise wind resistance and rely on aerodynamics in order to maximise the distance the ball would travel. On the moon however it is a very different story, as air resistance is minimal, we would expect to maximise results by hitting the ball at a 45-degree angle. This is because you would be giving equal force to both the horizontal acceleration and vertical acceleration.
We start by taking the ‘triangle’ of forces that the ball will undergo, as we are assuming with no air resistance the ball will not change its speed until it hits the ground at which point, I will assume, it will stop. When we hit a ball, we can think of it as having two speeds, one vertical speed, one horizontal speed, so if we hit the ball at 45 degrees, two forces go into the ball and as such the horizontal and vertical forces will be equal:
So, using the Sins rule, we know that:
hence there is an upward vertical velocity of 56.57m/s and a horizontal velocity of 56.57m/s in the direction the ball is hit.
We can check this using Pythagoras’s theorem:
Where a and b are the two shorter sides of the triangle and is the longest, known as the hypotenuse. So, we get:
As this is valid (with some rounding) we know we are on the right track.
Now we must find the flight time of the ball, we are using 1.625m/s2 as the moon’s gravity, an approximation but valid nonetheless. We will use the flight time of the ball to calculate the time the ball spends in the air.
To find the flight time we need to use a SUVAT equation, in this case it is:
s represents the (vertical) displacement of the ball, as the ball is going up then back down, the displacement at the end will equal zero as displacement is a vector quantity. This does assume that the ball will be landing on a surface at the same level as the one we struck from.
u is the initial velocity in the direction we calculated using the triangle above.
a is the acceleration of the object.
t is the time, in this case it is the time it takes for the ball to go up and then back down.
So, plugging this in we get:
Which after a few lines of rearranging yields the answer:
The reason the 1.625m/s2 is negative in this case, is because the gravity is acting against the ball’s upward velocity.
So, with a flight time of 69.62 seconds we can calculate the distance covered, as stated before I am assuming there is no air resistance to the ball hence it does not decelerate in the air.
The equation we use is:
As the ball is staying at a constant speed, the ‘a‘ in this equation is equal to zero hence it simplifies to:
Filling in our flight time and horizontal speed:
So, our final figure for distance covered is 3938.4m (2.447 miles), which is over 4,000 yards.
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Author: William Stallibrass (edit Neil Fozzard).