# Gap-size puzzle

• 2010-09-16, 16:15
Imagine that you have perfect sphere the size of the Earth. Around its equator is a piece of string, tight to the sphere so there is no gap and joined it such away that it is perfectly smooth (no knots, bumps etc). You cut the piece of string and then insert another piece just 3ft long which is again joined so the joint is smooth. The resultant new circle of string is positioned so that it shares the same centre as the sphere i.e. so that there is a uniform gap between string and sphere all the way round (looking from above you would see two concentric circles. What order of magnitude would you think the gap between sphere and new circle of string would be? Is it, for instance, going to be so small that you can't see it easily, or what?

Most people will be able to do the maths on this puzzle and I think you will be surprised at the answer!

Maths is posted below:-
• 2010-10-07, 07:17
MikeMercury
it must be some kind of trick. my guess for the trick is that it would float pi (3.14159...) feet above the earth. No reason, just a guess.
• 2010-10-07, 11:08
TerryB
Hang on, this must be unit dependant.
The gap will be a lot more than 6 inches.....
• 2010-10-07, 11:49
elito
The sphere's perimeter without adding the 3ft will be P1 = 2 x pi x R1, where R1 is the radius
After adding the 3ft, the spheres perimeter P2 = 2 x pi x R2
What we want to know is the value of R2 - R1 and we will call that D
D = R2 - R1
D = ( P2 / 2 x pi ) - ( P1 / 2 x pi)
D = ((P1 + 3) / 2 x pi) - (P1 / 2 x pi)
D = (P1 - P1 + 3) / 2 x pi
D = 3 / 2 x 3.14
D = 0.16 ft
• 2010-10-07, 14:22
TerryB
Or to generalise, where A is the additional length then the gap will be
Gap = A x 1/( 2 x Pi)
Gap = A x 0.159
• 2010-10-08, 12:19
jwboyes
The derivation provided by Elias is essentially correct but there is a math error going from step 1 to step 2 in that PI has moved from the denominator to the numerator. Try this (using his setup):

D = R2 - R1
D = ( P2 / (2 x pi) ) - ( P1 / (2 x pi))
D = ((P1 + 3) / (2 x pi)) - (P1 / (2 x pi))
D = (P1 - P1 + 3) / (2 x pi)
D = 3 / (2 x 3.14)
D = 0.477 ft or about 6 inches

Note that coming to step 5, P1 drops out of the equation. This means that the three feet of added rope will change the radius of the circle of the string by about 6 inches without regard to the original circumference. Therefore if the original sphere was a point sphere (zero radius) the new diameter would be about a foot which is about what you would expect if the problem were phrased that way.
• 2010-10-08, 13:29
rdcook01
Much greater than I would have expected. I expected it to be very small. The change in the circumference is minuscule: 3/(24,000x5,280) about 1/(8000x5000), ~1/40,000,000. Surprisingly it gets it all back when you multiply by the radius of the earth, to get the change in radius i.e. height. The radius is also a non trivial number when expressed in feet: 4000x5000, ~20,000,000
• 2013-04-14, 10:09
Maudibe
I agree with rdcook. Looking at that logically doesn't make much sense to me...adding 3 feet (something so small) to the circumference of the earth (which is so large) would have such an impact on the readius of the earth, but it is true!

Let's use actual numbers assuming that the earth is a perfect sphere. Googling the circumference of the earth, general concensus is that it is 24,901 miles or 131,477,280 feet or 1,577,727,360 inches (in billions).

HTML Code:

```[pre]         Circumference of the Earth  Circumference of the Earth + 36 inches (3feet)                                                We know:                      We know:                Circumference=2*Pi*r              Circumference=2*Pi*r         Rearranging:                      Rearranging:                C/(2*Pi)=r                      C/(2*Pi)        =r         Substituting:                      Substituting:        inches        1577727360/(2*Pi)=r              1577727396/(2*Pi)=r         Calculating:                      Calculating:        inches        251002080.00=r                      251002085.73=r                                 inches        251002085.73 - 251002080.00 = 5.73        [/pre]```
It is not what you would expect. But if you realize that the ratio between circumference and radius is approx 6.3 to 1 (6.3:1). So for every 6.3 inches of circumference, the radius increases by 1 inch (36 inches:5.75 inches). 5.75 inches is also an insignificant amount compared to the earth's radius.
• 2013-04-15, 12:53
RussB
Now calculate the square mile area of the space between the two 'strings' and compare that to the area of just the smaller string. You may be surprised there, to make it easy assume that the earth is 25,000 miles in circumference.
• 2013-04-15, 16:39
Maudibe
Could that be right? ~ 2.25 sq miles? I used 24901 because I built it in the formulas from my previous post
• 2013-04-15, 21:13
Maudibe
Yep, that is correct. If you stacked 2.25 square miles next to each other (1 mile by 2.25 miles), you could rip off 24,880 strips of 5.73" wide x 1 mile long. If you line all the strips end to end, you have the circumference of the earth...whew!!
• 2013-05-06, 00:51
BigMac56
One thing that seems to have been forgotten is that the original question about the size of the gap.
The difference in size of the two circles may be 5.75", but the gap would only be half of that, or 2.875".
• 2013-05-07, 07:25
Maudibe
BigMac,

The radius is increased by 5.75 inches which is the gap at any point around the circumference. If it were the diameter that increased by 5.75" then you would be correct.