math riddle

[VJ]

Ok, let's have a math riddle...

Suppose you have a rope long enough to go around the earth (at the equator) exactly once. Now, you extend this rope 1 meter.
If you now raise the rope the same amount at all locations, how much can you raise it at most?


The calculations are very simple, but let's see if you would venture a guess without calculating (be honest).



Jörg

[Fat Tone]

Not sure how that happened, I meant to click 10-20cm but it registered on 5-10cm. Can someone please fix it?

As you say VJ, easy maths, but hard to do in your head!

[VJ]

Oooh, we are getting some very different guesses... Currently (after 10 voters - I haven't voted), the option "< 1cm" has the most votes with 30 %...



Jörg

[Gurm]

Lessee... the circumference of the Earth at the equator is... roughly... 40,076 km or 40,076,000 m, or 4,007,600,000 cm.

That makes the RADIUS of the earth ~637,829,888.7 cm. Now the question is how much can we increase the RADIUS if we increase the circumference by 1 meter (100cm).

If the circumference were increased to 4,007,600,100 then the radius would end up being ~637,829,904.6 cm, a difference of ... believe it or not ... a whopping 15+ cm!

Pretty counterintuitive, huh?

[Gurm]

And it's not even really a math riddle as much as a "misestimation of scale" joke.

[Nowhere]

VJ, does your moderator status mean that you see what each of us choose?

[VJ]

Nowhere: well, I'm not a moderator here, so I can't see the results. As there never has been a poll in the DI forum, I wouldn't know. But Sasq can...


Gurm: This counter-intuitivity was why I wanted people to guess first... Even more surprising, if you calculate it symbolically:

circumference: 2*pi*r
so we are looking for x such that
2*pi*r + 1 meter = 2*pi*(r+x)

or to put it in otherwords:
2*pi*r +1 meter = 2*pi*r + 2*pi*x

or
x = 1 meter / (2*pi)

Note that x is independent of the circumference of the earth, so you would get the same on Jupiter, or the same around the coke-can on your desk.
(deeper reason: there is a lineair connection between r and the circumference)

Pretty odd, eh?
(my colleague didn't believe it, so we took some UTP cable (only 'rope' we could find) and tried it around a coke-can and around a garbage can)


Edit: Last score (before submitting this result):
<1cm: 5 voters (38.46 %)
10-20cm: 4 voters (31.87%) (I added FT's vote here)
between 20-50 cm: 1 vote (7.69%)
over 50 cm: 1 vote (7.69%)
wow... a rope that long.... : 2 votes (15.38%)



Jörg

[Gurm]

Yes, but adding a meter to a rope around a coke can is a ridiculous amount for the scale of the coke can, which is why people can't grasp it.

[KvHagedorn]

Interesting.. but it's funny that none of you said anything about the elasticity of the rope.

Suppose you have a rope long enough to go around the earth (at the equator) exactly once. Now, you extend this rope 1 meter.
If you now raise the rope the same amount at all locations, how much can you raise it at most?

[cjolley]

Of course, another way of looking at it is that 1 - 50 cm at the scale of the earth is all pretty much the same thing.

Question:
How much would the mass of the earth increase if the radius were increased 15 cm?
(for simplicity's sake assume uniform density, or the density of sea water)

PS. I got it wrong, and it's a pretty safe bet I gave that as part of a homework assignment when I was teaching.

[Gurm]

Of course, another way of looking at it is that 1 - 50 cm at the scale of the earth is all pretty much the same thing.

Question:
How much would the mass of the earth increase if the radius were increased 15 cm?
(for simplicity's sake assume uniform density, or the density of sea water)

PS. I got it wrong, and it's a pretty safe bet I gave that as part of a homework assignment when I was teaching.

Hmm... uhh... well given that I don't know the density offhand, let's see:

Current volume of the earth:

2*pi*r^3, where r = radius of earth in centimeters... making this number milliliters.

New volume:

2*pi*(r+15)^3 = 2*pi*(r^3+ 30r^2 + 550r + 3375)

Now to keep these numbers from just getting OUT OF HAND, we want to find the difference between those two numbers.

The difference, if you haven't already guessed, is:

2*pi*(30r^2 + 550r + 3375). Trust me, this will help.

Now, the radius of the earth is 637,829,888 centimeters. (Result from previous discussion.)

So r^2 = 406,826,966,026,092,544 square centimeters (yeah these numbers are getting big!)

So let's add all of this up. 2 * pi * ((30*406,826,966,026,092,544) + (550*637,829,888) + 3375)... or...

2 * pi * (12,204,808,980,782,776,320 + 350,806,438,400 + 3375)

or 2 * pi * (12,204,809,331,589,218,095)

or 76,685,013,896,054,743,350 cubic centimeters... aka milliliters.

Now there are 100^3 milliliters in a cubic meter. That's a million, for those of you who don't do powers of 10 very well. That means that the EXTRA volume of the earth due to this radius increase is:

76,685,013,896,055 cubic meters. That's seventy-six TRILLIAN cubic meters. Wow. Yikes! Ok, let's keep our heads on straight here. We're almost done. The average density of the earth is:

5,515kg/m^3

Ok? So... that gives us a mass of...

422,917,851,636,743,325 kilos, or...

422,917,851,636,743.325 metric tonnes. OUCH!

Ok, did I do it right? Chuck? Almost 423 Trillion metric tonnes?

[Gurm]

Followup: The weight of the earth, as it exists right now, is estimated at approximately 6 sextillion metric tonnes. Ouch!

[cjolley]

...
422,917,851,636,743.325 metric tonnes. OUCH!

Ok, did I do it right? Chuck? Almost 423 Trillion metric tonnes?

Sorry I can't look in detail.
Today I'm designing tables & code for a database to hold business rules that were apparently written by SATAN.

But assuming you are correct, it is a good example of scaling.
The linear dimension goes up a little and the mass goes up an amazing amount.

Though, what % increase does that really represent?

[Gurm]

432 Trillion, out of 6 sextillion? Uh... not very much. LOL.

[cjolley]

Ok, the mass of the moon is 7.36 × 10^22kg

What % is it of that?