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- anonymous

There's a problem in Halliday I - chapter 1
In my book it's question 19 (but it's a brasillian version). I'll try to translate it from portuguese :P
"Suppose you are laid at the beach, near the equator, looking the sun going down in a calm sea, and turns on a cronometer in the (exactly) moment that the sun dissapears. Then, you stand up, shifting (?) your eyes up a distance H=1.7m, and turns of the cronometers in the moment that the sun dissapears again. If the time appointed by the cronometer is t=11,1s, what is the radius of the Earth?"
Can someone help me, plese?

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- anonymous

- katieb

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- anonymous

*turns off

- anonymous

Sorry, bad drawing. Better one coming:

- anonymous

|dw:1343440772302:dw|

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- anonymous

There ya go. That angle theta is the angle through which the planet rotates in the time between you raise your hed 1.7 m and the time the sun sets again. Note that
\[ \cos(\theta) = \frac{R}{R+\delta r} = \frac{R}{R+\delta r} \cdot \frac{R -\delta r}{R - \delta r} \approx 1 - \frac{\delta r}{R} \]
if
\[ \delta r <<< R\]
Now, the angular velocity of the planet is
\[ \omega = \frac{\theta}{\delta t} = \frac{2\pi}{T_{day}}\]
where delta t is our 11 seconds or so and Tday is the number of seconds in a day (86,400).
This yields the equation
\[ \cos(\theta) = \cos\left(2\pi \frac{\delta t}{T_{day}}\right) = 1 - \frac{\delta r}{R}\]
Therefore, solving for R, we find
\[ R = \frac{ \delta r}{1 - \cos\left(2\pi \frac{\delta t}{T_{day}}\right)} \]
You can plug your values in here. Alternatively, you can go a step further and note that for small x,
\[ \cos(x) \approx 1 - \frac{x^2}{2} \]
which yields
\[R \approx \frac{\delta r}{2 \pi^2 \left(\frac{\delta t}{T_{day}}\right)^2 } = \delta r \cdot \left(\frac{T^2_{day}}{2\pi^2 \delta t^2}\right) \]
These should yield the same answer. Plugging in the given values, I get
\[R \approx 5.2 \cdot 10^6 m\]
Which is fairly close to the equatorial average of
\[ 6.4 \cdot 10^6 m\]

- anonymous

wow, thanks Jemurray3. The answer is indeed 5.2, haha
I own you one ;)

- anonymous

Sure, just make sure you understand how it works. Drawing a picture is often rather useful.

- anonymous

yes, sir. I'm reading your solution again :)

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