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Topic: measurement problem. (Read 2724 times)
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starf
Poster God
Posts: 1888
measurement problem.
«
on:
11:35:09, 09 April, 2009 »
Suppose that while lying on a beach watching the Sun set over a calm ocean, you start a stopwatch just as the top of the Sun disappears. You then stand, elevating your eyes by a height h = 1.70m, and stop the watch when the top of the sun disappears again. If the elapsed time on the watch is t = 11.1 s. what is the radius r of Earth?
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chris.bailey
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Re: measurement problem.
«
Reply #1 on:
14:32:33, 09 April, 2009 »
4 inches! Think I may have done something wrong....
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Roger Banks
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Re: measurement problem.
«
Reply #2 on:
14:47:19, 09 April, 2009 »
Well I divided 11.1 secs into no secs in a day ( I know it's not quite spot on) = 0.0001284 x 360 deg a day gives me Theta at 0.046224 degrees. Therefore d = 11.1/Sin Theta.
Looking at the big triangle the opposite side length is 2 x d so r=2d/Tan Theta which is xxxx - not got a calculator or log tables about!!!
Is that anything like the right approach??
Rog
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robin_astro
Senior Member
Posts: 601
Re: measurement problem.
«
Reply #3 on:
17:00:23, 09 April, 2009 »
5218km? (typo corrected)
theta = (11.1*360)/(24*60*60) = 11.1/240 deg
cos(theta) = r/(r+h)
therefore
r=h/(1-cos(theta))
so
r =1.7/(1-cos(11.1/240)
Robin
EDIT:
oops! that should be r=h*cos(theta)/(1-cos(theta)) but since theta is so small it makes s*d all difference -fortunately
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starf
Poster God
Posts: 1888
Re: measurement problem.
«
Reply #4 on:
17:40:54, 09 April, 2009 »
how do you answers compare to the true value?
[tex]\theta=\frac{360\deg}{24h*60m*60s}*11.1s=0.04625\deg[/tex]
you can also use pythagoras thm, so
[tex](r+h)^2 = r^2 + 2rh +h^2 = d^2 + r^2[/tex]
so
[tex]d^2 = 2rh + h^2[/tex]
since h^2 is so small compared to r, its negligible, so
[tex]d^2 = 2rh[/tex]
using tan instead of cos
[tex]d=r tan \theta[/tex]
so
[tex]2rh = r^2tan^2\theta[/tex]
and rearranging
[tex]r = \frac{2h}{tan^2\theta[/tex]
so
[tex]r = \frac{2(1.7)}{tan^20.04625}=5.22*10^6m[/tex]
here's robin's way
[tex]\cos\theta=\frac{r}{r+h}[/tex]
rearranging
[tex]r = \frac{cos \theta h}{1-\cos\theta}[/tex]
so
[tex]r=\frac{\cos 0.04625(1.7)}{1-cos 0.04625} = 5.22*10^6m[/tex]
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MrYannah
Full Member
Posts: 186
Re: measurement problem.
«
Reply #5 on:
18:37:17, 09 April, 2009 »
hi
from what i remember from school you figure out the smaller triangle first as you know the lenght of side h and it also have a 90 degree angle. Using trigonometry the angles of this trianle can be found and so can the lenghts of the sides. Then as this triangle is in the same proportion as the large triangle the radius of the earth can be found.
I would have to first find my old maths books if i didnt burn them then revise for a few hours/days or maybe weeks.
It this correct.
Raymond
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starf
Poster God
Posts: 1888
Re: measurement problem.
«
Reply #6 on:
19:09:31, 09 April, 2009 »
for right angle triangles do you remember
[tex]c^2 = a^2 + b^2[/tex]
and [tex]SohCahToa[/tex]
[tex]\sin\theta=\frac{opp}{hyp}[/tex]
[tex]\cos\theta=\frac{adj}{hyp}[/tex]
[tex]\tan\theta=\frac{opp}{adj}[/tex]
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MrYannah
Full Member
Posts: 186
Re: measurement problem.
«
Reply #7 on:
21:23:07, 09 April, 2009 »
i certainly do. Mr Hector in 2nd year. Is it the sin of the hypotineuse is equal to the square of
a right angled triangle. please excuse the spelling
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MrYannah
Full Member
Posts: 186
Re: measurement problem.
«
Reply #8 on:
21:28:20, 09 April, 2009 »
correction, the square of the hypotinuese is equal to the sum of the square of the sides of a right angled triangle.
as you have written
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srmby
New Member
Posts: 27
Re: measurement problem.
«
Reply #9 on:
02:56:02, 01 March, 2010 »
yeah.. this is messing with my head, I'm not even gonna try and solve this. I was never good at maths :)
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ttz668
Full Member
Posts: 116
Re: measurement problem.
«
Reply #10 on:
21:50:40, 01 March, 2010 »
The answer is 6,378 km
First type "radius of the earth" into google.
Then go to link to wikipedia
Read for about 10secs
you have the answer.
Seems that's how the kids do it these days !!
David
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