### Author Topic: measurement problem.  (Read 2596 times)

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#### starf

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##### 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?

#### 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

#### robin_astro

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##### 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

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##### Re: measurement problem.
« Reply #4 on: 17:40:54, 09 April, 2009 »
how do you answers compare to the true value?

$$\theta=\frac{360\deg}{24h*60m*60s}*11.1s=0.04625\deg$$

you can also use pythagoras thm, so

$$(r+h)^2 = r^2 + 2rh +h^2 = d^2 + r^2$$

so

$$d^2 = 2rh + h^2$$

since h^2 is so small compared to r, its negligible, so

$$d^2 = 2rh$$

$$d=r tan \theta$$

so

$$2rh = r^2tan^2\theta$$

and rearranging

$$r = \frac{2h}{tan^2\theta$$

so

$$r = \frac{2(1.7)}{tan^20.04625}=5.22*10^6m$$

here's robin's way

$$\cos\theta=\frac{r}{r+h}$$

rearranging

$$r = \frac{cos \theta h}{1-\cos\theta}$$

so

$$r=\frac{\cos 0.04625(1.7)}{1-cos 0.04625} = 5.22*10^6m$$

#### MrYannah

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##### 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

#### starf

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• Posts: 1888
##### Re: measurement problem.
« Reply #6 on: 19:09:31, 09 April, 2009 »
for right angle triangles do you remember

$$c^2 = a^2 + b^2$$

and $$SohCahToa$$

$$\sin\theta=\frac{opp}{hyp}$$

$$\cos\theta=\frac{adj}{hyp}$$

$$\tan\theta=\frac{opp}{adj}$$

#### MrYannah

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• 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

#### MrYannah

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• 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

#### srmby

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• 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 :)

#### ttz668

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##### Re: measurement problem.
« Reply #10 on: 21:50:40, 01 March, 2010 »

Then go to link to wikipedia