Question

Two sailors are at the top of their ships' masts in the open ocean. The mast of ship A is twice as high as that of ship B. How much farther can sailor A see than sailor B?

Answer 1

a) no farther
b) a little more than 40% farther
c) twice as far
d) almost three times as far
e) four times as far

Answer 2

R is the radius of the Earth, r is the line of sight up to the curvature of the Earth, h is the height of the ship and mast.

Answer 3

It seems to me, unless I'm doing it wrong, that, to find r (the line of sight), I need to have h (the height of the mast), but to find h, I need r. I can't find one without having the other.

Answer 4

Or am I just going about this the wrong way?

Answer 5

I am given the radius of the Earth as 6.37 * 10^6 m, but I am not given a height.

Answer 6

But it isn't really asking for the height, it just wants to know how much farther the ship with the twice-as-tall mast can see.

Answer 7

\[d_1^2=(R_E+h_1)^2-R_E^2=2R_Eh_1+h_1^2=h_1(2R_E+h_1)\]\[d_2^2=(R_E+h_2)^2-R_E^2=2R_Eh_2+h_2^2=h_2(2R_E+h_2)\]We may assume that \[h_1<<R_E \rightarrow d_1^2\approx 2h_1R_E\]\[h_2<<R_E \rightarrow d_2^2\approx 2h_2R_E\]Thus\[\frac{ d_1^2 }{ d_2^2}\approx \frac{ 2h_1R_E }{ 2h_2R_E }=\frac{ h_1 }{ h_2 }=\frac{ 2h_2 }{ h_2 }=2\]\[\frac{ d_1^2 }{ d_2^2 }=\left( \frac{ d_1 }{ d_2 } \right)^2\approx 2 \rightarrow \frac{ d_1 }{ d_2}\approx \sqrt{2} \approx 1.4142\]
\[\frac{ d_1-d_2 }{ d_2 }=\frac{ d_1 }{ d_2 }-1 \approx 1.4142-1=0.4142\] which is a little more than a 40%