# Thread: Hard Physics (i think)

1. ## Hard Physics (i think)

Suppose that, while lying on a beach near the equator 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 again disappears. If the elapsed time is $t = 11.1s$, what is the radius $r$ of the earth?

I drew a diagram but I can't get anywhere with it! No extra information is given to aid the question.

2. If I understood it correctly, you can't solve this without knowing the rotation speed of the earth. Because for any r value, you can always find a rotation speed which fits the conditions.

Edit: On a second thought, I think we can find something using angular speed. It takes 24 hours for earth to complete a turn..

3. The angular velocity ω of the Earth’s rotation can be found from $\omega=\frac{2\pi}{T}$, where T = 86&#160;400&#160;s is the rotational period.

4. we haven't learnt angular velocity yet... is there no way at all to solve the problem without extra information? This problem comes very early in the book, so I would assume mostly basic math can be used. thanks for your help though

5. The 1st pic is the moment you started your stopwatch. The second pic is the moment the sun disappears (while you're stood up) and you stop. 11.1 seconds passed between two pics.

Let's see how it looks when you see the sun stood up.

As you see, the earth rotated $\theta$ degrees in 11.1 secs. We can calculate $\theta$. If the earth turns 2 $\pi$ degrees in 86400 seconds (24 hours), how much does it turn in 11.1 s? This gives $\theta \approx 0.000257\pi$.

Aha! There's a right triangle.
$Cos(\theta) = \frac {r}{r + 1.7}$
$Cos(0.000257\pi) \approx \frac {r}{r + 1.7}$
$\frac {r}{r + 1.7} \approx 0.9999996740612671$

Thus,

$r \approx 5215702 m$

$r \approx 5215 km$

6. Amazing! Thanks for that, the answers give $5.2 \times 10^6m$, so you are even more accurate!

I had not taken into account the rotation of the earth, lol

the diagrams are great too

7. Hmm, I didn't take into account the rotation, but you didn't take into account the orbit! What gives?

How does the earth orbit the sun anyway??? And what is its axis of rotation?

8. Originally Posted by DivideBy0
Hmm, I didn't take into account the rotation, but you didn't take into account the orbit! What gives?

How does the earth orbit the sun anyway??? And what is its axis of rotation?
You shouldn't have to worry about the rotational motion of the Earth about the Sun as the correction is too small to matter. (The angular speed is about $2 \times 10^{-7}$ rad/s, meaning the Earth is moving at about 19 m/s. Compare this with the angular speed of the rotation of the Earth mentioned below.)

For the record the Earth doesn't quite revolve about the Sun. Both the Sun and the Earth revolve about a common point called the "barycenter." (This point is, I believe, actually inside the Sun but is not at its center.) This happens because there is nothing "nailing" the Sun down to a specific position so the axis (the location of the barycenter) is located at the center of mass of the Earth-Sun system. But since the radius of the Sun's orbit about the barycenter is so small compared to the Earth-Sun distance we can say the Earth rotates about the center of the Sun to a very good approximation.

Then, of course, the other objects in our solar system (planets, moons, asteroids, the Oort cloud, etc.) all affect the barycenter, not to mention exerting their own gravitational forces on the Earth itself. The Solar system can be a very complicated place to do detailed calculations.

-Dan

9. Originally Posted by wingless
If I understood it correctly, you can't solve this without knowing the rotation speed of the earth. Because for any r value, you can always find a rotation speed which fits the conditions.

Edit: On a second thought, I think we can find something using angular speed. It takes 24 hours for earth to complete a turn..
You also need the condition given, that this takes place near the equator,
infact to derive the change in elevation of the suns upper edge you need to
assume that the sum sets at right angles to the horizon.

RonL

10. Originally Posted by topsquark
You shouldn't have to worry about the rotational motion of the Earth about the Sun as the correction is too small to matter. (The angular speed is about $2 \times 10^{-7}$ rad/s, meaning the Earth is moving at about 19 m/s. Compare this with the angular speed of the rotation of the Earth mentioned below.)
Since we are working with a mean solar day of 24 hours the correction to
a siderial day has already been incorporated, and we don't need to worry about it anyway.

RonL

11. Originally Posted by CaptainBlack
Since we are working with a mean solar day of 24 hours the correction to
a siderial day has already been incorporated, and we don't need to worry about it anyway.

RonL

-Dan

12. ## Re: Hard Physics (i think)

Hello Everyone. This is a quite old thread, still hoping that someone will reply back.
In the reply of wingless, i am not able to figure out how the hypotenuse of the Right Angled Triangle would be (r + 1.7). I am attaching the snapshot of my solution, can someone please guide me here, where my procedure is wrong.

13. ## Re: Hard Physics (i think)

Classic problem ... H = 1.7 m, $\cos{\theta} = \dfrac{adjacent}{hypotenuse} = \dfrac{r}{r+1.7}$

14. ## Re: Hard Physics (i think)

Thanks a lot for sharing the diagram.