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May 9th, 2017, 11:25 PM   #1
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Around the earth in the equator we attach a rope

Around the earth in the equator we attach a rope (we treat the earth as a geometric ball).

We add to the scope of the rope 1 meter.
The length of the rope that been created is now equals to the scope of the earth + 1 meter.
At some point, we pull the rope out.

The rope attached the earth along the arc ABC
when AM and AC work as tangents

The question: is it possible that point M is high enough above the earth that an elephant can pass threw it? In other words, we need to find the altitude M above the earth.

Last edited by skipjack; May 10th, 2017 at 12:04 AM.
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May 10th, 2017, 12:43 AM   #2
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Rope.PNG
Ignoring the earth's curvature, consider a right-angled triangle with hypotenuse AM = 10 m,
then X = √(10² - 9.5²) m = 3.1225 m approximately, which would allow an elephant to pass.
Try using AM = 100 m instead.
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May 11th, 2017, 07:31 PM   #3
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$C_E = 2 \pi r_E $

$C_E + 1m = 2 \pi r_f$

Now solve for $r_f$

$2 \pi r_E + 1m = 2 \pi r_f$

$ \frac{2 \pi r_E + 1m }{2 \pi } = r_f$


Googling the Earths radius in meters and substituting ...


$ \frac {2 \pi (6,371,000m) + 1m }{2 \pi } = r_f$

$6,371,000.16m = r_f$

So the height of the rope above the Earth is 16 centimeters.

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May 12th, 2017, 08:38 AM   #4
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Quote:
Originally Posted by agentredlum View Post
$C_E = 2 \pi r_E $

$C_E + 1m = 2 \pi r_f$

Now solve for $r_f$

$2 \pi r_E + 1m = 2 \pi r_f$

$ \frac{2 \pi r_E + 1m }{2 \pi } = r_f$


Googling the Earths radius in meters and substituting ...


$ \frac {2 \pi (6,371,000m) + 1m }{2 \pi } = r_f$

$6,371,000.16m = r_f$

So the height of the rope above the Earth is 16 centimeters.

So the answer is yes, but it has to be a tiny elephant

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May 12th, 2017, 10:56 AM   #5
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Quote:
Originally Posted by agentredlum View Post
$C_E = 2 \pi r_E $

$C_E + 1m = 2 \pi r_f$

Now solve for $r_f$

$2 \pi r_E + 1m = 2 \pi r_f$

$ \frac{2 \pi r_E + 1m }{2 \pi } = r_f$


Googling the Earths radius in meters and substituting ...


$ \frac {2 \pi (6,371,000m) + 1m }{2 \pi } = r_f$

$6,371,000.16m = r_f$

So the height of the rope above the Earth is 16 centimeters.

is this method effective to for the problem? Doesn't it assume the radius is equal all around? rather than if you took that extra meter and lifted it up, using it all at a certain point? While I am unsure of how to solve this, I think skipjack is on the right path, but unsure if an extra meter of rope can generate 3.1 meters of height. It could be possible because of straight lines.
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May 12th, 2017, 11:15 PM   #6
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I don't think I made a mistake in the algebra but I may have misinterpreted the question. It's still pretty amazing to me that adding 1 meter to the circumference of the Earth increases the Earth's radius by a comparativly whopping 16 cm , which is actually

$ \frac{100 cm}{2 \pi} $

Counterintuitive until you do the math

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May 15th, 2017, 01:33 AM   #7
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Quote:
Originally Posted by agentredlum View Post
I don't think I made a mistake in the algebra but I may have misinterpreted the question. It's still pretty amazing to me that adding 1 meter to the circumference of the Earth increases the Earth's radius by a comparativly whopping 16 cm , which is actually

$ \frac{100 cm}{2 \pi} $

Counterintuitive until you do the math

The assumption is that adding a tiny length to the rope would have a negligible impact on the increase in radius, giving an answer like 1 mm or something. However... 16 cm is a negligible impact on the radius: the Earth's radius is 6371 km, so the increase in circumference by 1 m corresponds to an increase by 0.0000025%. Similarly the radius increases by 0.0000025%, giving 16 cm.
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May 15th, 2017, 01:38 AM   #8
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Well , when you explain it that way it makes sense
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