WhatIf 0085: "Rocket Golf"
Moderators: Moderators General, Prelates, Magistrates
 Various Varieties
 Posts: 505
 Joined: Tue Mar 04, 2008 7:24 pm UTC
Re: WhatIf 0085: "Rocket Golf"
Randall didn't completely answer the question!
The question asks: "How many golf balls would be required to reach the Moon?"
But he only replied with the size of our ball supply, completely missing out the quantity!
I for one am appalled by this glaring omission.
The question asks: "How many golf balls would be required to reach the Moon?"
But he only replied with the size of our ball supply, completely missing out the quantity!
I for one am appalled by this glaring omission.
Re: WhatIf 0085: "Rocket Golf"
Various Varieties wrote:Randall didn't completely answer the question!
The question asks: "How many golf balls would be required to reach the Moon?"
But he only replied with the size of our ball supply, completely missing out the quantity!
I for one am appalled by this glaring omission.
Volume of a sphere is: v= 4/3 * pi * r^3
Radius of the "final" bag of golf balls is 75 miles (4,752,000 inches), radius of a golf ball is ~ .85 inches.
Volume of the total of golf balls is 449487796497116448446.60456995839. Volume of an individual golf ball is 2.5724407845144423634278278236754.
Assuming all of that is right, that leaves us with about 174,732,028,508,854,060,655 golf balls.
"Yeah, I've been known to hit one hundred quintillion or two golfballs in my life."
"I'll guess the latter."
Re: WhatIf 0085: "Rocket Golf"
CasualSax wrote:Various Varieties wrote:Randall didn't completely answer the question!
The question asks: "How many golf balls would be required to reach the Moon?"
But he only replied with the size of our ball supply, completely missing out the quantity!
I for one am appalled by this glaring omission.
Volume of a sphere is: v= 4/3 * pi * r^3
Radius of the "final" bag of golf balls is 75 miles (4,752,000 inches), radius of a golf ball is ~ .85 inches.
Volume of the total of golf balls is 449487796497116448446.60456995839. Volume of an individual golf ball is 2.5724407845144423634278278236754.
Assuming all of that is right, that leaves us with about 174,732,028,508,854,060,655 golf balls.
"Yeah, I've been known to hit one hundred quintillion or two golfballs in my life."
"I'll guess the latter."
Are you taking account of the packing fraction? Golf balls don't tessellate very well...
Re: WhatIf 0085: "Rocket Golf"
rmsgrey wrote:CasualSax wrote:Various Varieties wrote:Randall didn't completely answer the question!
The question asks: "How many golf balls would be required to reach the Moon?"
But he only replied with the size of our ball supply, completely missing out the quantity!
I for one am appalled by this glaring omission.
Volume of a sphere is: v= 4/3 * pi * r^3
Radius of the "final" bag of golf balls is 75 miles (4,752,000 inches), radius of a golf ball is ~ .85 inches.
Volume of the total of golf balls is 449487796497116448446.60456995839. Volume of an individual golf ball is 2.5724407845144423634278278236754.
Assuming all of that is right, that leaves us with about 174,732,028,508,854,060,655 golf balls.
"Yeah, I've been known to hit one hundred quintillion or two golfballs in my life."
"I'll guess the latter."
Are you taking account of the packing fraction? Golf balls don't tessellate very well...
I did not, but I assumed that Mr. Monroe did not either, so the volumes would translate to the same number of golf balls (even though the volume is wrong).
Re: WhatIf 0085: "Rocket Golf"
Randall's last line
Umm, no you wouldn't... 1000000+ tries to hit a golf ball into a hole and then finally getting a shot in doesn't equal a hole in one. It sounds more like my life at golf.
over the course of this maneuver you would be statistically likely to hit a holeinone ... at every golf course in the world.
Umm, no you wouldn't... 1000000+ tries to hit a golf ball into a hole and then finally getting a shot in doesn't equal a hole in one. It sounds more like my life at golf.
 gmalivuk
 GNU Terry Pratchett
 Posts: 26546
 Joined: Wed Feb 28, 2007 6:02 pm UTC
 Location: Here and There
 Contact:
Re: WhatIf 0085: "Rocket Golf"
Yes, he did. That's what the 65% number is here:CasualSax wrote:I did not, but I assumed that Mr. Monroe did not eitherrmsgrey wrote:Are you taking account of the packing fraction? Golf balls don't tessellate very well...
Stripping that equation down to just the mass part, we get 1.616x10^{45}kg for the initial number, which is 3.519x10^{46} 1.62 oz regulationsized golf balls.
For the final number, where each ball is going 140m/s, we get 5.523x10^{18}kg or 1.203x10^{20} balls.
(The number of digits you gave, CasualSax, is incidentally pretty absurd, considering you estimated a single ball to only 2 significant figures. Those extra digits don't give any usable information, they just make it harder to see at a glance how big the numbers are.)
It's not that many tries to hit a ball, though. It's one try for each ball.NOTNOTJON wrote:Randall's last lineUmm, no you wouldn't... 1000000+ tries to hit a golf ball into a hole and then finally getting a shot in doesn't equal a hole in one.over the course of this maneuver you would be statistically likely to hit a holeinone ... at every golf course in the world.
Re: WhatIf 0085: "Rocket Golf"
gmalivuk wrote:Yes, he did. That's what the 65% number is here:CasualSax wrote:I did not, but I assumed that Mr. Monroe did not eitherrmsgrey wrote:Are you taking account of the packing fraction? Golf balls don't tessellate very well...
Stripping that equation down to just the mass part, we get 1.616x10^{45}kg for the initial number, which is 3.519x10^{46} 1.62 oz regulationsized golf balls.
For the final number, where each ball is going 140m/s, we get 5.523x10^{18}kg or 1.203x10^{20} balls.
(The number of digits you gave, CasualSax, is incidentally pretty absurd, considering you estimated a single ball to only 2 significant figures. Those extra digits don't give any usable information, they just make it harder to see at a glance how big the numbers are.)It's not that many tries to hit a ball, though. It's one try for each ball.NOTNOTJON wrote:Randall's last lineUmm, no you wouldn't... 1000000+ tries to hit a golf ball into a hole and then finally getting a shot in doesn't equal a hole in one.over the course of this maneuver you would be statistically likely to hit a holeinone ... at every golf course in the world.
When we are doing napkin math for the sheer sake of absurdity, seeing the full number outright has more comical effect than scientific notation.
 gmalivuk
 GNU Terry Pratchett
 Posts: 26546
 Joined: Wed Feb 28, 2007 6:02 pm UTC
 Location: Here and There
 Contact:
Re: WhatIf 0085: "Rocket Golf"
Ah, then I guess my mistake was doing math to answer the question asked...
Re: WhatIf 0085: "Rocket Golf"
NOTNOTJON wrote:Randall's last lineover the course of this maneuver you would be statistically likely to hit a holeinone ... at every golf course in the world.
Umm, no you wouldn't... 1000000+ tries to hit a golf ball into a hole and then finally getting a shot in doesn't equal a hole in one. It sounds more like my life at golf.
Since none of those balls were hit from the appropriate tee box (I know of no golf course with a green on earth and a tee box in orbit), none of them are holes in one. A million tries from the tee box would count if one of them went in. Though that's a truly Presidential number of rounds of golf, or a frighteningly permissive mulligan rule.
Re: WhatIf 0085: "Rocket Golf"
flying_kiwi wrote:Olaf Klischat is correct  for almost all configurations discussed, the escape velocity of the ship is greater than the launch velocity of the ball, so it will simply fall back to the surface of the ship after launch, and (assuming collisions are at least partially inelastic and it eventually comes to rest) the net change in momentum of the ship will be zero.
In the final configuration, plugging in the density of the average golf ball I get the launch velocity a little larger than the escape velocity, so the ship WILL move. However, the final velocity is still a good deal less than the launch velocity, which reduces the effectiveness of the rocket, forcing the designer to make it larger  likely to the point you wouldn't be able to launch to escape velocity, and the rocket would fail to change its momentum.
I'm afraid it's even more complicated than that because the ship is a noninertial reference frame. The launch speed doesn't need to exceed the ship's escape velocity because before the first ball drops back to the ship, it will have accelerated away from the ball. This is just based on intuition and I'd like to see someone more mathematically skilled actually write that out as an equation.
 gmalivuk
 GNU Terry Pratchett
 Posts: 26546
 Joined: Wed Feb 28, 2007 6:02 pm UTC
 Location: Here and There
 Contact:
Re: WhatIf 0085: "Rocket Golf"
Yeah, but you could still consider it one ball at a time, to get a bound on it. If the launch speed is above the ship's escape velocity, shooting off the golf ball will definitely accelerate the ship. If it's equal or less, then the only way to accelerate is to fire off golfballs fast enough that your own velocity added to the ball's will exceed escape velocity (at a particular distance) before the ball stops.
Later I may actually do the math to find the speed at which it balances out exactly (i.e. where launch speed is equal to the full ship's escape velocity).
Edit: Now that it's later, I'll do the calculation. I'll furthermore do it with the more accurate 63.4% for random close packing and 1.13 kg/L for a golf ball.
If we set the initial equation equal to the radius for a given mass and escape velocity (where the mass is that given by the rocket equation), we need to find v so that
[;
\left(\frac{3}{4\pi}\exp\left(\frac{5300\ m/s}{v}\right)200\ kg \frac{1\ m^3}{.634*1130\ kg}\right)^{\frac{1}{3}}=\frac{2G\exp\left(\frac{5300\ m/s}{v}\right)200\ kg}{v^2}
;]
Then the velocity that works is a bit over 134m/s. Meaning that the last, fastest scenario in the whatif is the only one with any possibility of working at all, even if we ignore all the other matter in the universe.
Edit2: If we use the optimal 74% efficiency packing, the resulting speed is only about a third of a meter per second faster.
Later I may actually do the math to find the speed at which it balances out exactly (i.e. where launch speed is equal to the full ship's escape velocity).
Edit: Now that it's later, I'll do the calculation. I'll furthermore do it with the more accurate 63.4% for random close packing and 1.13 kg/L for a golf ball.
If we set the initial equation equal to the radius for a given mass and escape velocity (where the mass is that given by the rocket equation), we need to find v so that
[;
\left(\frac{3}{4\pi}\exp\left(\frac{5300\ m/s}{v}\right)200\ kg \frac{1\ m^3}{.634*1130\ kg}\right)^{\frac{1}{3}}=\frac{2G\exp\left(\frac{5300\ m/s}{v}\right)200\ kg}{v^2}
;]
Then the velocity that works is a bit over 134m/s. Meaning that the last, fastest scenario in the whatif is the only one with any possibility of working at all, even if we ignore all the other matter in the universe.
Edit2: If we use the optimal 74% efficiency packing, the resulting speed is only about a third of a meter per second faster.

 Posts: 2
 Joined: Sat Mar 01, 2014 6:24 pm UTC
Re: WhatIf 0085: "Rocket Golf"
orbik wrote:flying_kiwi wrote:Olaf Klischat is correct  for almost all configurations discussed, the escape velocity of the ship is greater than the launch velocity of the ball, so it will simply fall back to the surface of the ship after launch, and (assuming collisions are at least partially inelastic and it eventually comes to rest) the net change in momentum of the ship will be zero.
In the final configuration, plugging in the density of the average golf ball I get the launch velocity a little larger than the escape velocity, so the ship WILL move. However, the final velocity is still a good deal less than the launch velocity, which reduces the effectiveness of the rocket, forcing the designer to make it larger  likely to the point you wouldn't be able to launch to escape velocity, and the rocket would fail to change its momentum.
I'm afraid it's even more complicated than that because the ship is a noninertial reference frame. The launch speed doesn't need to exceed the ship's escape velocity because before the first ball drops back to the ship, it will have accelerated away from the ball. This is just based on intuition and I'd like to see someone more mathematically skilled actually write that out as an equation.
I haven't worked it out, but it won't help much  the ball loses most of its momentum immediately after the "launch", and it is during this time that the "equal and opposite" force on the ship will partially cancel out the impulse it gained from the initial launch.
Likewise, the gravitational acceleration on the ship and the ball from any external field will be initially identical (ignoring the finite size of the "ship"), and any difference will only become significant as they separate.
Who is online
Users browsing this forum: No registered users and 5 guests