anonymous
  • 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?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
*turns off
anonymous
  • anonymous
Sorry, bad drawing. Better one coming:
anonymous
  • anonymous
|dw:1343440772302:dw|

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anonymous
  • 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
  • anonymous
wow, thanks Jemurray3. The answer is indeed 5.2, haha I own you one ;)
anonymous
  • anonymous
Sure, just make sure you understand how it works. Drawing a picture is often rather useful.
anonymous
  • anonymous
yes, sir. I'm reading your solution again :)

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