Question

Suppose that, while lying on a beach near the equator of a far-off planet 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 = 1.12 m, and stop the watch when the top of the sun again disappears. If the elapsed time is t = 10.8 s, what is the radius r of the planet to two significant figures? Notice that duration of a solar day at the far-off planet is the same that is on Earth.

Answer 1

In 10.8 seconds the earth rotates by θ = 10.8 / (24*60*60)*360 deg

                                                        = 0.045deg

In that time a triangle has been created with the 2 longer sides having lengths r and r+1.12

Cos(0.045) = r/(r+1.12)
0.999999691 = r/(r+1.12)
r(1 - 0.999999691) = 1.12*0.999999691
r = 3624594.35 m
r = 3624 km

Therefore radius of the planet is 3.624*10^6 m