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May 9th, 2017, 11:25 PM  #1 
Newbie Joined: May 2017 From: Israel Posts: 1 Thanks: 0  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. 
May 10th, 2017, 12:43 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,951 Thanks: 1599  Rope.PNG Ignoring the earth's curvature, consider a rightangled 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. 
May 11th, 2017, 07:31 PM  #3 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
$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. 
May 12th, 2017, 08:38 AM  #4  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,099 Thanks: 703 Math Focus: Physics, mathematical modelling, numerical and computational solutions  Quote:
 
May 12th, 2017, 10:56 AM  #5  
Member Joined: Feb 2015 From: Southwest Posts: 96 Thanks: 24  Quote:
 
May 12th, 2017, 11:15 PM  #6 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
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 
May 15th, 2017, 01:33 AM  #7  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,099 Thanks: 703 Math Focus: Physics, mathematical modelling, numerical and computational solutions  Quote:
 
May 15th, 2017, 01:38 AM  #8 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
Well , when you explain it that way it makes sense 

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