Ship Disappearance Distance and Asymptotic Geometry

[member 428835]

Homework Statement

An observer from sea level watches a ship sail away from shore. If the ships height is $\epsilon \ll 1$ after scaling with the radius of the earth, show that the distance the ship disappears will be about $\sqrt{2\epsilon}$.

Homework Equations

Nothing comes to mind.

The Attempt at a Solution

Not quite sure where to start here. I'm thinking of a sort of right triangle made with the observer, the ship base, and the ship height, but I just don't know the distance between the ship and observer.

I imagine the square root appears from the pythagorean theorem (the three points being the observer, ship base and ship height) though I'm unsure. Anyone give me a nudge in the right direction?

 

[Chestermiller]

You have to take into account the curvature of the earth.

 

[member 428835]

You have to take into account the curvature of the earth.

So the distance $d$ between the observer and the ship's base would be $d=r\theta$, $r$ being the Earth's radius, but I don't know what $\theta$ is.

 

[Ray Vickson]

Homework Statement

An observer from sea level watches a ship sail away from shore. If the ships height is $\epsilon \ll 1$ after scaling with the radius of the earth, show that the distance the ship disappears will be about $\sqrt{2\epsilon}$.

Homework Equations

Nothing comes to mind.

The Attempt at a Solution

Not quite sure where to start here. I'm thinking of a sort of right triangle made with the observer, the ship base, and the ship height, but I just don't know the distance between the ship and observer.

I imagine the square root appears from the pythagorean theorem (the three points being the observer, ship base and ship height) though I'm unsure. Anyone give me a nudge in the right direction?

Start by making a sketch of the situation, so you can see what is going on. Then you will be better able to translate that into equations and formulas.

 

[member 428835]

Start by making a sketch of the situation, so you can see what is going on. Then you will be better able to translate that into equations and formulas.

Yea, I did this right away, but all I know is the Earth's radius and the ship height. It just feels open ended.

 

[Ray Vickson]

Homework Statement

An observer from sea level watches a ship sail away from shore. If the ships height is $\epsilon \ll 1$ after scaling with the radius of the earth, show that the distance the ship disappears will be about $\sqrt{2\epsilon}$.

Homework Equations

Nothing comes to mind.

The Attempt at a Solution

Not quite sure where to start here. I'm thinking of a sort of right triangle made with the observer, the ship base, and the ship height, but I just don't know the distance between the ship and observer.

I imagine the square root appears from the pythagorean theorem (the three points being the observer, ship base and ship height) though I'm unsure. Anyone give me a nudge in the right direction?

Actually, the three relevant points are the observer, the top of the ship and the center of the earth. If w is the angle between the observer and the ship (as measured from the center of the earth) you can get the observer-ship distance in terms of the Earth's radius and the angle w.

 

[member 428835]

Actually, the three relevant points are the observer, the top of the ship and the center of the earth. If w is the angle between the observer and the ship (as measured from the center of the earth) you can get the observer-ship distance in terms of the Earth's radius and the angle w.

I think I did this in Post 3. In this case the distance would be $rw$. I could find the distance from the observer and the top of the ship too if you think this is relevant?

 

[Ray Vickson]

I think I did this in Post 3. In this case the distance would be $rw$. I could find the distance from the observer and the top of the ship too if you think this is relevant?

If $\epsilon$ is much smaller than the Earth's radius, do you think there would be any noticeable difference between the distance to the bottom of the ship vs. the top of the ship? Of course, the answer is YES if you want exact results, but you being asked about "lowest-order" approximate answers. What is your answer in that case?

 

[member 428835]

If $\epsilon$ is much smaller than the Earth's radius, do you think there would be any noticeable difference between the distance to the bottom of the ship vs. the top of the ship? Of course, the answer is YES if you want exact results, but you being asked about "lowest-order" approximate answers. What is your answer in that case?

The walking distance along the Earth would still be $d = rw$. The distance from one point to another would be $\sqrt{2r^2(1-\cos w)}$. However, the distance from the viewer to the top of the boat is $$\sqrt{r^2+(r+h)^2-2 r(r+h)\cos w}=\sqrt{2r^2+2rh+h^2-2 r^2\cos w-rh\cos w}\\
= r\sqrt{2+2\epsilon+\epsilon^2-2\cos w - \epsilon\cos w}.$$
If $w \approx 0$ (not very far along the total circumference) then $\cos w \approx 1 \implies d = r\sqrt{2\epsilon}$ (ignoring $\epsilon^2$).

 

[Ray Vickson]

The walking distance along the Earth would still be $d = rw$. The distance from one point to another would be $\sqrt{2r^2(1-\cos w)}$.

Yes, but you are supposed to be working with a low-order approximation for small $w$. What then?

 

[member 428835]

Yes, but you are supposed to be working with a low-order approximation for small $w$. What then?

Shoot I think I was editing and posted the response you were looking for? See post 9.

 

[Chestermiller]

I did this much differently. I had: $$\cos{\theta}=\cos{(s/r)}=\frac{r}{h+r}$$
where s is the arc length along the great circle. Using the small angle approximation for cosine, I get:
$$\cos{\theta}\approx 1-\frac{\theta^2}{2}\approx 1-\frac{1}{2}\left(\frac{s}{r}\right)^2$$Also, assuming h<<r, $$\frac{r}{h+r}\approx1-\frac{h}{r}$$

 

[Ray Vickson]

Shoot I think I was editing and posted the response you were looking for? See post 9.

OK now.

 

[I like Serena]

Alternatively...
We have a right triangle with sides $r$, $r+\epsilon$, and the distance $d$.
We can neglect any differences between the arc length and the side of the triangle corresponding to the distance.
Now apply Pythagoras:
$$d=\sqrt{(r+\epsilon)^2 - r^2}=...$$

 

[member 428835]

Wow lots of different ways people are solving this! I'm going to mark it solved. Thanks for everyone's perspective!