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Maths Question.

vardas0antras
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12/11/2010 3:00:47 AM
Posted: 5 years ago
How many golf balls does one need to go around the equator while placing one golf ball besides another to make a long line and ultimately a circle? Lets assume that these golf balls are unmovable and they can float. This is just am estimation. Personally I have solved only half of this... Yeah I bet some of you find this ridiculously easy.
"When he awoke in a tomb three days later he would actually have believed that he rose from the dead" FREEDO about the resurrection of Jesus Christ
Mirza
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12/11/2010 4:55:01 AM
Posted: 5 years ago
I looked up, and the diameter of a golf ball is 42.67mm.

42.67mm to km = 0.00004267km

The length of equator: 40,075km

You divide 40,075 with 0.00004267 =

approximately 939,184,439 golf balls
vardas0antras
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12/11/2010 10:11:31 AM
Posted: 5 years ago
At 12/11/2010 7:06:56 AM, badger wrote:
how can something be unmoveable and still able to float?

When I said unmovable I meant that once the golf ball is placed you can't move it, whatever on ground or water.
"When he awoke in a tomb three days later he would actually have believed that he rose from the dead" FREEDO about the resurrection of Jesus Christ
Grape
Posts: 989
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1/13/2011 5:36:24 PM
Posted: 5 years ago
At 12/11/2010 4:55:01 AM, Mirza wrote:
I looked up, and the diameter of a golf ball is 42.67mm.

42.67mm to km = 0.00004267km

The length of equator: 40,075km

You divide 40,075 with 0.00004267 =

approximately 939,184,439 golf balls

I agree with Mirza's calculation (I found the same values for the golf ball and Earth's equatorial circumference and I assume his division was correct). If the significance of the digits matters to you than you should change that to 939,200,000 golf balls. The remaining digits cannot be known precisely because the size of the golf ball and the planet are not given exactly.
J.Kenyon
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1/13/2011 6:33:55 PM
Posted: 5 years ago
At 1/13/2011 5:36:24 PM, Grape wrote:
At 12/11/2010 4:55:01 AM, Mirza wrote:
I looked up, and the diameter of a golf ball is 42.67mm.

42.67mm to km = 0.00004267km

The length of equator: 40,075km

You divide 40,075 with 0.00004267 =

approximately 939,184,439 golf balls

I agree with Mirza's calculation (I found the same values for the golf ball and Earth's equatorial circumference and I assume his division was correct). If the significance of the digits matters to you than you should change that to 939,200,000 golf balls. The remaining digits cannot be known precisely because the size of the golf ball and the planet are not given exactly.

Incorrect. You're assuming that the diameter of each golf ball will be laid end to end. If you could imagine the line circumscribed around the earth at the equator laid out flat on a plane, Mirza's calculation would be correct, however, because the earth is spherical, it'll be different. Try to visualize 1 ball with 3 balls surrounding it. The balls would have to be huge, right? The sum of their diameters would massively exceed the the circumference of the ball in the middle.

Where 'r' is radius of the earth and 'g' is the diameter of a golf ball:

180/Tan[g/(r+g)]

At least I THINK that's what it is. Someone good at math will probably have to correct me.
MikeLoviN
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1/13/2011 7:07:48 PM
Posted: 5 years ago
Assuming the earth is spherical (which it actually isn't, but its close enough for our purposes), imagine looking straight down on the earth from directly above it. You would essentially see a 2-dimensional circle with a circumference equal to that of the equator.

Mirza was on the right track, but as J.Kenyon pointed out, his answer is valid only for the case were the equator is laid out in a straight line. Since this isn't actually the case, you have to account for the extra radius added on to the equator by each golf ball, calculate the new circumference and then do the division.

The equatorial radius of the earth is 6378.1370 km http://en.wikipedia.org...

The radius of a golf ball is 21.335 mm = 0.000021335 km

So the new radius of the earth at the equator with the golf balls laid across it is 6378.137021335 km

Now, the circumference of the circle through the center of each golf ball is equal to 2*PI*R (where R is the new radius we calculated): C = 40075.01681963024468 km

Therefore the # of golf balls is C/(diameter of each ball) = 939,184,832
lovelife
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1/13/2011 7:32:32 PM
Posted: 5 years ago
At 1/13/2011 7:07:48 PM, MikeLoviN wrote:
Assuming the earth is spherical (which it actually isn't, but its close enough for our purposes), imagine looking straight down on the earth from directly above it. You would essentially see a 2-dimensional circle with a circumference equal to that of the equator.

Mirza was on the right track, but as J.Kenyon pointed out, his answer is valid only for the case were the equator is laid out in a straight line. Since this isn't actually the case, you have to account for the extra radius added on to the equator by each golf ball, calculate the new circumference and then do the division.

The equatorial radius of the earth is 6378.1370 km http://en.wikipedia.org...

The radius of a golf ball is 21.335 mm = 0.000021335 km

So the new radius of the earth at the equator with the golf balls laid across it is 6378.137021335 km

Now, the circumference of the circle through the center of each golf ball is equal to 2*PI*R (where R is the new radius we calculated): C = 40075.01681963024468 km

Therefore the # of golf balls is C/(diameter of each ball) = 939,184,832

So it would be about 400 more balls?
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SuperRobotWars
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1/13/2011 8:29:10 PM
Posted: 5 years ago
How many golf balls would it take to equal the mass of the moon?
Minister Of Trolling
: At 12/6/2011 2:21:41 PM, badger wrote:
: ugly people should beat beautiful people ugly. simple! you'd be killing two birds with the one stone... women like violent men and you're making yourself more attractive, relatively. i met a blonde dude who was prettier than me not so long ago. he's not so pretty now! ha!
:
: ...and well, he wasn't really prettier than me. he just had nice hair.
belle
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1/13/2011 9:29:41 PM
Posted: 5 years ago
At 1/13/2011 7:07:48 PM, MikeLoviN wrote:
Assuming the earth is spherical (which it actually isn't, but its close enough for our purposes), imagine looking straight down on the earth from directly above it. You would essentially see a 2-dimensional circle with a circumference equal to that of the equator.

Mirza was on the right track, but as J.Kenyon pointed out, his answer is valid only for the case were the equator is laid out in a straight line. Since this isn't actually the case, you have to account for the extra radius added on to the equator by each golf ball, calculate the new circumference and then do the division.

The equatorial radius of the earth is 6378.1370 km http://en.wikipedia.org...

The radius of a golf ball is 21.335 mm = 0.000021335 km

So the new radius of the earth at the equator with the golf balls laid across it is 6378.137021335 km

Now, the circumference of the circle through the center of each golf ball is equal to 2*PI*R (where R is the new radius we calculated): C = 40075.01681963024468 km

Therefore the # of golf balls is C/(diameter of each ball) = 939,184,832

which really makes absolutely no difference given the magnitudes involved. if this were a physics class you'd have to cut off at 939,200,000 b/c none of those other numbers are significant. measure the radii of the earth and golf ball more carefully and then come back to this...
evidently i only come to ddo to avoid doing homework...
SuperRobotWars
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1/13/2011 9:42:35 PM
Posted: 5 years ago
At 1/13/2011 8:29:10 PM, SuperRobotWars wrote:
How many golf balls would it take to equal the mass of the moon?

A golf balls mass is around 45.93 grams and the moons mass is around 7.36 × 1022 kilograms so . . . it is about 163 769.214 golf balls . . . [?]
Minister Of Trolling
: At 12/6/2011 2:21:41 PM, badger wrote:
: ugly people should beat beautiful people ugly. simple! you'd be killing two birds with the one stone... women like violent men and you're making yourself more attractive, relatively. i met a blonde dude who was prettier than me not so long ago. he's not so pretty now! ha!
:
: ...and well, he wasn't really prettier than me. he just had nice hair.
MikeLoviN
Posts: 746
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1/13/2011 10:05:13 PM
Posted: 5 years ago
At 1/13/2011 9:29:41 PM, belle wrote:
At 1/13/2011 7:07:48 PM, MikeLoviN wrote:
Assuming the earth is spherical (which it actually isn't, but its close enough for our purposes), imagine looking straight down on the earth from directly above it. You would essentially see a 2-dimensional circle with a circumference equal to that of the equator.

Mirza was on the right track, but as J.Kenyon pointed out, his answer is valid only for the case were the equator is laid out in a straight line. Since this isn't actually the case, you have to account for the extra radius added on to the equator by each golf ball, calculate the new circumference and then do the division.

The equatorial radius of the earth is 6378.1370 km http://en.wikipedia.org...

The radius of a golf ball is 21.335 mm = 0.000021335 km

So the new radius of the earth at the equator with the golf balls laid across it is 6378.137021335 km

Now, the circumference of the circle through the center of each golf ball is equal to 2*PI*R (where R is the new radius we calculated): C = 40075.01681963024468 km

Therefore the # of golf balls is C/(diameter of each ball) = 939,184,832

which really makes absolutely no difference given the magnitudes involved. if this were a physics class you'd have to cut off at 939,200,000 b/c none of those other numbers are significant. measure the radii of the earth and golf ball more carefully and then come back to this...

I was assuming it was the process and not the final answer that matters. If the scale difference wasn't so drastic, this would have a very noticeable affect. There's a reason you'll never hear a competent engineer or mathematician say "well that looks about right, lets go with it"...
J.Kenyon
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1/13/2011 10:14:34 PM
Posted: 5 years ago
At 1/13/2011 10:05:13 PM, MikeLoviN wrote:
At 1/13/2011 9:29:41 PM, belle wrote:
which really makes absolutely no difference given the magnitudes involved. if this were a physics class you'd have to cut off at 939,200,000 b/c none of those other numbers are significant. measure the radii of the earth and golf ball more carefully and then come back to this...

I was assuming it was the process and not the final answer that matters. If the scale difference wasn't so drastic, this would have a very noticeable affect. There's a reason you'll never hear a competent engineer or mathematician say "well that looks about right, lets go with it"...

No, it's not that the figures don't matter, they're not significant in the sense that they are meaningless; they fall within the margin of error given the imprecise measurements. http://en.wikipedia.org... That's why I only posted an equation: 180/[tan/(r+g)] instead of a solution.

Btw, belle, was my equation correct?
belle
Posts: 4,113
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1/13/2011 10:20:46 PM
Posted: 5 years ago
not sure where you got the tangent from... i think you're thinking about it too hard.

it would just be PI(g+e)/g where g is the radius of the golf ball and e is the radius of the earth
evidently i only come to ddo to avoid doing homework...
darkkermit
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1/13/2011 10:50:50 PM
Posted: 5 years ago
Oh boy math. We need more mathematical questions to add to the geekiness of DDO. Now J.Kenyon is correct, that you have to account for the curvature, if you want to do this calculation. However the curvature would be so small it would be insignificant since the earth's radius is so large compared to the golf balls. If we were to look at the world outside, it looks roughly flat. The circumfrence is so large, that the curvature is very small for thousands of meter.

The fact that that the earth is not a perfect circle but has ridges (ex. mountains, holes) would have a greater effect then the curvature of earth.
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MikeLoviN
Posts: 746
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1/14/2011 12:22:00 AM
Posted: 5 years ago
At 1/13/2011 10:14:34 PM, J.Kenyon wrote:
At 1/13/2011 10:05:13 PM, MikeLoviN wrote:
At 1/13/2011 9:29:41 PM, belle wrote:
which really makes absolutely no difference given the magnitudes involved. if this were a physics class you'd have to cut off at 939,200,000 b/c none of those other numbers are significant. measure the radii of the earth and golf ball more carefully and then come back to this...

I was assuming it was the process and not the final answer that matters. If the scale difference wasn't so drastic, this would have a very noticeable affect. There's a reason you'll never hear a competent engineer or mathematician say "well that looks about right, lets go with it"...

No, it's not that the figures don't matter, they're not significant in the sense that they are meaningless; they fall within the margin of error given the imprecise measurements. http://en.wikipedia.org... That's why I only posted an equation: 180/[tan/(r+g)] instead of a solution.

Btw, belle, was my equation correct?

Fair enough, but the OP asked for a solution, so I put numbers into my equation. Belle is right, its [PI*(g+e)]/g, which is what I had ... don't know where you got the tangent from. I'm guessing you're somehow thinking in terms of radians (which would explain the 180).

BTW, I understand the concept of sig figs. I purposely tossed the extra numbers in there to give as much a difference as possible in the final answer in an attempt to demonstrate that accounting for the curvature (normally) matters. It just doesn't seem that way here because of the scale difference. If we were talking about more practical scenarios, you wouldn't be able to get away with that straight line estimation, at least not outside of high school.
Ragnar_Rahl
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1/14/2011 12:35:28 AM
Posted: 5 years ago
How can there be a plural of math?
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J.Kenyon
Posts: 4,194
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1/14/2011 12:38:10 AM
Posted: 5 years ago
At 1/14/2011 12:22:00 AM, MikeLoviN wrote:
At 1/13/2011 10:14:34 PM, J.Kenyon wrote:
At 1/13/2011 10:05:13 PM, MikeLoviN wrote:
At 1/13/2011 9:29:41 PM, belle wrote:
which really makes absolutely no difference given the magnitudes involved. if this were a physics class you'd have to cut off at 939,200,000 b/c none of those other numbers are significant. measure the radii of the earth and golf ball more carefully and then come back to this...

I was assuming it was the process and not the final answer that matters. If the scale difference wasn't so drastic, this would have a very noticeable affect. There's a reason you'll never hear a competent engineer or mathematician say "well that looks about right, lets go with it"...

No, it's not that the figures don't matter, they're not significant in the sense that they are meaningless; they fall within the margin of error given the imprecise measurements. http://en.wikipedia.org... That's why I only posted an equation: 180/[tan/(r+g)] instead of a solution.

Btw, belle, was my equation correct?

Fair enough, but the OP asked for a solution, so I put numbers into my equation. Belle is right, its [PI*(g+e)]/g, which is what I had ... don't know where you got the tangent from. I'm guessing you're somehow thinking in terms of radians (which would explain the 180).

I used trigonometry; my equation gave me roughly the same answer yours did. It's kind of difficult to explain without a diagram, but it works. Fairly ingenious if I do say so myself :P
J.Kenyon
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1/14/2011 12:46:22 AM
Posted: 5 years ago
At 1/14/2011 12:38:10 AM, J.Kenyon wrote:
At 1/14/2011 12:22:00 AM, MikeLoviN wrote:
Fair enough, but the OP asked for a solution, so I put numbers into my equation. Belle is right, its [PI*(g+e)]/g, which is what I had ... don't know where you got the tangent from. I'm guessing you're somehow thinking in terms of radians (which would explain the 180).

I used trigonometry; my equation gave me roughly the same answer yours did. It's kind of difficult to explain without a diagram, but it works. Fairly ingenious if I do say so myself :P

Basically, imagine a triangle circumscribed around a circle, then a square, then a pentagon, then a hexagon, then an octagon, etc.

Hint: what's the interior angle of an octagon?
Cerebral_Narcissist
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1/14/2011 2:08:48 AM
Posted: 5 years ago
At 12/11/2010 3:00:47 AM, vardas0antras wrote:
How many golf balls does one need to go around the equator while placing one golf ball besides another to make a long line and ultimately a circle? Lets assume that these golf balls are unmovable and they can float. This is just am estimation. Personally I have solved only half of this... Yeah I bet some of you find this ridiculously easy.

At what height are these golf balls floating above the earth, are they all at uniform height?

Otherwise it's very simple to work out.
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Korashk
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1/14/2011 2:16:35 AM
Posted: 5 years ago
Are you supposed to factor in the geography of the land at the equator? I doubt it but its something to consider.
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MikeLoviN
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1/14/2011 10:56:44 AM
Posted: 5 years ago
At 1/14/2011 12:46:22 AM, J.Kenyon wrote:
At 1/14/2011 12:38:10 AM, J.Kenyon wrote:
At 1/14/2011 12:22:00 AM, MikeLoviN wrote:
Fair enough, but the OP asked for a solution, so I put numbers into my equation. Belle is right, its [PI*(g+e)]/g, which is what I had ... don't know where you got the tangent from. I'm guessing you're somehow thinking in terms of radians (which would explain the 180).

I used trigonometry; my equation gave me roughly the same answer yours did. It's kind of difficult to explain without a diagram, but it works. Fairly ingenious if I do say so myself :P

Basically, imagine a triangle circumscribed around a circle, then a square, then a pentagon, then a hexagon, then an octagon, etc.

So basically you approximated the circumference as an infinite sum of straight lines / tangents?

Hint: what's the interior angle of an octagon?

135 degrees

I do understand you're method though. Interesting way of doing it... though you are assuming that the distance from the center of one ball to the point of contact with an adjacent ball forms a right angle with the (r+g) radius (ie. it's tangent to the circle). Since the balls are so small comparatively, it works perfectly fine here and you do infact get the right answer. But this is still technically an approximation and would not be that accurate if the size difference wasn't so huge (that right triangle you used wouldn't actually be a right triangle).

BTW, You're equation is actually 180/[arctan{g/(r+g)}] ...correct?

Good job though, I'll give credit where credit is due... It would have taken me a while before I even considered using trig to do this, but that's because I've been conditioned to always find the easy way out.
J.Kenyon
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1/14/2011 11:51:05 AM
Posted: 5 years ago
At 1/14/2011 10:56:44 AM, MikeLoviN wrote:
At 1/14/2011 12:46:22 AM, J.Kenyon wrote:
At 1/14/2011 12:38:10 AM, J.Kenyon wrote:
At 1/14/2011 12:22:00 AM, MikeLoviN wrote:
Fair enough, but the OP asked for a solution, so I put numbers into my equation. Belle is right, its [PI*(g+e)]/g, which is what I had ... don't know where you got the tangent from. I'm guessing you're somehow thinking in terms of radians (which would explain the 180).

I used trigonometry; my equation gave me roughly the same answer yours did. It's kind of difficult to explain without a diagram, but it works. Fairly ingenious if I do say so myself :P

Basically, imagine a triangle circumscribed around a circle, then a square, then a pentagon, then a hexagon, then an octagon, etc.

So basically you approximated the circumference as an infinite sum of straight lines / tangents?

Hint: what's the interior angle of an octagon?

135 degrees

I do understand you're method though. Interesting way of doing it... though you are assuming that the distance from the center of one ball to the point of contact with an adjacent ball forms a right angle with the (r+g) radius (ie. it's tangent to the circle). Since the balls are so small comparatively, it works perfectly fine here and you do infact get the right answer. But this is still technically an approximation and would not be that accurate if the size difference wasn't so huge (that right triangle you used wouldn't actually be a right triangle).

Ah, my bad. I should have asked what's half the internal angle of an octagon. That will give you a right angle formed between the opposing side and the adjacent side.

BTW, You're equation is actually 180/[arctan{g/(r+g)}] ...correct?

Good job though, I'll give credit where credit is due... It would have taken me a while before I even considered using trig to do this, but that's because I've been conditioned to always find the easy way out.

I haven't taken a math class in two years, so I've forgotten a lot of stuff. I remembered my trigonometry, though,so it was just a matter of making do with the tools available.
Ren
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1/14/2011 12:31:32 PM
Posted: 5 years ago
At 12/11/2010 3:00:47 AM, vardas0antras wrote:
How many golf balls does one need to go around the equator while placing one golf ball besides another to make a long line and ultimately a circle? Lets assume that these golf balls are unmovable and they can float. This is just am estimation. Personally I have solved only half of this... Yeah I bet some of you find this ridiculously easy.

Out of curiosity, I'm just gonna try to solve this without reading the thread.

Let's start with the diameter of the earth. I think I'll use the metric system; I like it more.

...Google says 12,756.32 km.

Alright, so that will be the diameter of the circle that the golfballs make. Therefore, each golf ball will be 6,378.16 km away from the center.

The circumference of a golf ball is 13.405866 cm. The diameter is 4.2672 cm.

So, using the diameter, we can get a rough estimate by dividing it out. 12,756.32 * 1000 = 12,756,320 * 100 = 1,275,632,000/4.2672 = 298,938,882.6397 golf balls.

Now, let's find out specifically and see how they align.

Each golf ball will have the same distance from each midpoint, which is literally the diameter of each golf ball. Therefore, from center to center, there will be 4.2672 cm. If each endpoint were measured to the center of the planet (center of the circle that the golf balls would make), it would measure the same--6,378.16 km--forming an isosceles. That means that two of those angles have the same measurement. So, it would be fitting to find the tangent angle value. That would be O/A, or 4.2672 /1,275,632,000 = 0.0000000033425 degree angle at the center, meaning that each respective angle coming from the center of each golf ball is 89.99. That sufficiently accepts that each golf ball is nearly adjacent by 90 degrees, meaning that I can conceivably treat the distance that each golf ball would accommodate as that which would be on a flat plane. Accordingly, the estimate that I found was sufficiently accurate.

298,938,882.6397

Now, let's check out the thread. :D
Ren
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1/14/2011 12:57:15 PM
Posted: 5 years ago
At 1/14/2011 12:31:32 PM, Ren wrote:

Ah-HA! I found something.
6,378.16 km
That would be O/A, or 4.2672 /1,275,632,000 = 0.0000000033425 degree angle at the center,

That was based on the diameter, not the circumference, of the circle the golf balls would make. So, let's try again.

6,378.16 km * 1,000 = 6,378,160 * 100 = 63,781,600.

4.2672/63,781,600 = 0.000000069033.

It still pretty much remains the same. A simple way to explain it, is that tan values below .0175 will not render an angle with any significant degree measurement.

...meaning that each respective angle coming from the center of each golf ball is 89.99. That sufficiently accepts that each golf ball is nearly adjacent by 90 degrees, meaning that I can conceivably treat the distance that each golf ball would accommodate as that which would be on a flat plane. Accordingly, the estimate that I found was sufficiently accurate.

298,938,882.6397

Now, let's check out the thread. :D
Ren
Posts: 7,102
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1/14/2011 1:10:23 PM
Posted: 5 years ago
And I used the diameter rather than the circumference. Wtf. :S

This is what I get for trying to do Math stoned.

40,075.16 km, which is like, a billion cm, versus the 4.1 cm for the diameter of a golf ball, which means that it would take a little less than a billion golf balls, which is what everyone else said.

So, yeah. Congrats, everyone.
mattrodstrom
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1/14/2011 2:15:44 PM
Posted: 5 years ago
you guys are all nuts.

that stuff's not fun...
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