Physics Question

The Sun sets, fully disappearing over the horizon as you lie on the beach, your eyes 30 cm above the sand. You immediately jump up, your eyes now 170 cm above the sand, and you can again see the top of the Sun. If you count the number of seconds ( = t) until the Sun fully disappears again, you can estimate the radius of the Earth. Use the known radius of the Earth to calculate the time t.

Ok I have no idea how to approach this question any tips?

Drop out right now.

lrn2 Euclidean Geometry
no physics involved

if you have nothing better to do, you can add in a correction factor for the earth's orbit around the sun (earth is moving during the course of the experiment)

This is hard.
I don't get it.

22.5 seconds

you measure the time it take for the sun to set the second time after you stand up. from this you can find the amount the earth has rotated.

there are 3 points that make a right triange.
1) the center of the earch
2) your eye
3) this one is a bit harder to explain. if you think of the line connecting your eye and the top of the sun the point where it touched the earth is the last point.

the length from 1 -> 2 (adjacent side) & 1 -> 3 (hypotenuse) are known and the angle between them can be calculated (from the time the earth has rotated before it set).

so the only unknown is the time. and with some simple trig it can be calculated.

tl;dr: 5.8 seconds.

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Here's some diagrams I made for you.

http://rapidshare.com/files/147032660/Diagrams.zip.html

Diagram 1.

There are two concentric circles. The inner circle is Earth, the outer circle represents the path the Sun takes across the sky as the Earth rotates. Point A is where the observer is standing, Point B is the observer's eye when it is 30 cm from the ground (the ground is presumed to be sea level) and Point C is the observer's eye when it is 170 cm from the ground. Point D is the horizon seen from 30 cm, Point E is the horizon seen from 170 cm. Point F is the Sun's position as it meets the horizon at 30 cm, Point G is the Sun's position as it meets the horizon at 170 cm. I've marked the line of sight at 30 cm as green (Tangent Line D) and the line of sight at 170 cm as blue (Tangent Line E).

The ultimate goal here is to figure out how long it takes the Sun to travel from Point F to Point G, that is, how long it takes to traverse Arc FG (marked in red, with an arrow to show the direction the Sun is moving in).

Diagram 2.

We are given (implicitly) that it takes the Sun about 24 hours to make a complete journey across the sky back to where it started (once around the outer circle). To be precise, it takes 23 hours, 56 minutes, and 4.091 seconds. We are also allowed (after some research) that the Earth's radius is 6,378,137 meters at sea level at the equator. For this problem, we do not need to know the size nor distance of the sun. For our purposes, it is just a point orbiting at a constant distance and speed.

The Earth's center is Point H and Lines AH, DH, and EH are all radii.

Diagram 3.

Drawing Lines FH and GH, we create Sector FHG and Arc IJ (pink). Angle FHG (Arc FG) is congruent to Angle IHJ (Arc IJ).

Diagram 4.

It just so happens that Arc IJ is congruent to Arc DE (orange). This is not a coincidence. To prove it, we extend Radii DH and EH to the outer circle and make Points K and L. We then draw Tangent Lines I and J, which pass through Points K and L respectively. The reason this happens perfectly is because Arc IJ is actually defined by Tangent Lines D and E, so it will always be congruent to Arc DE.

Tangent Line D intersects with Line KH perpendicularly by definition. The same goes for Tangent Line E and Line LH. This means that Triangles BDH and CEH are Right Triangles. We also already know the lengths of two of the sides in each triangle. Sides DH and EH are just radii and Sides BH and CH are the radius plus their respective heights. Now that we know an angle and two sides of each of the triangles, we can figure out the other angles and sides as we please.

What we're interested in is Angle DHE. To get it, we can subtract Angle BHD from Angle CHE.

Angle BHD = arccos(Side DH / Side BH) = arccos(6378137 / 6378137.3) = 0.000306710536 radians = 0.0175732192 degrees.
Angle CHE = arccos(Side EH / Side CH) = arccos(6378137 / 6378138.7) = 0.000730117047 radians = 0.0418326253 degrees.

Angle DHE = Angle CHE - Angle BHD = 0.0418326253 - 0.0175732192 = 0.0242594061 degrees = Angle IHJ = Angle FHG

23 hours, 56 minutes, and 4.091 seconds = 86164.091 seconds

0.0242594061 degrees / 360 degrees * 86164.091 seconds = 5.80636021 seconds = t.

That's the answer.