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 2 years ago
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?
 2 years ago
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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Jemurray3
 2 years ago
Best ResponseYou've already chosen the best response.3Sorry, bad drawing. Better one coming:

Jemurray3
 2 years ago
Best ResponseYou've already chosen the best response.3dw:1343440772302:dw

Jemurray3
 2 years ago
Best ResponseYou've already chosen the best response.3There 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 \[ 6.4 \cdot 10^6 m\]

seidi.yamauti
 2 years ago
Best ResponseYou've already chosen the best response.0wow, thanks Jemurray3. The answer is indeed 5.2, haha I own you one ;)

Jemurray3
 2 years ago
Best ResponseYou've already chosen the best response.3Sure, just make sure you understand how it works. Drawing a picture is often rather useful.

seidi.yamauti
 2 years ago
Best ResponseYou've already chosen the best response.0yes, sir. I'm reading your solution again :)
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