Mathematics 5 Worksheet 14 The Horizon

Transcription

1 Mathematics 5 Worksheet 14 The Horizon For the problems below, we will assume that the Earth is a sphere whose radius is 4,000 miles. Note that there are 5,280 feet in one mile. Problem 1. If a line intersects a circle at precisely one point, P, then the line is perpendicular to the line that contains P and the center of the circle. Draw a picture that describes this relationship. Solution. In the following figure, the center of the circle is the point O, and the line AP is perpendicular to line OP, which passes through the center of the circle and the point P. P O A Problem 2. Draw a sphere and mark two points on the sphere. A plane that intersects the two points on the sphere and the center of the sphere will divide the sphere into two even hemispheres. The intersection of the sphere with this plane is a great circle, a circle whose radius is the radius of the sphere. Convince yourself that this is correct by drawing a picture. Solution. The picture is as follows: 1

2 Problem 3. Suppose that you stand 49 feet above sea level and you look out across the ocean. You see a small dot that sits directly on the horizon. Draw a cross section of the earth that contains the dot, the point at which you stand on the earth, and the center of the earth. This cross section should be a disk whose boundary is a great circle on the surface of the earth. Solution. Again, the picture is below. In addition to the dot representing our position and the small dot on the horizon, we have included some additional lines, which represent our line of sight and segments joining the interesting points to the center of the Earth. These will prove helpful in problem 4, below. 49 ft our position small dot on the horizon 4000 mi α 4000 mi line of sight Problem 4. Suppose that you stand 49 feet above sea level and you look out across the ocean. What is the length of your line of sight? Make sure that you answer give the distance in miles. Solution. We are begin asked to determine the distance to the horizon, i.e. the distance between our position and the small dot on the horizon. The line of sight is tangent to the surface of the Earth, which justifies our assertion in the above figure that the angle between the line of sight and a radius is a right angle. But then we can apply the Pythagorean theorem in order to conclude that distance to horizon = 4000 mi + 49 ft) 2 (4000 mi) mi + 49 ft 1mi = 5280 ft ) 2 (4000 mi) 2 = ) mi 8.6 mi. (use a calculator) Therefore our line of sight is about eight and a half miles. 2

3 Problem 5. In Problem 4, you obtained an exact distance. Find a good estimate that you could use to quickly calculate this distance for relatively small heights. Solution. Let h represent our height above the ground, measured in feet. Then the distance to the horizon is given by the formula distance to horizon = 4000 mi + h ) mi (4000 mi) 2 = ) 5280 h + h mi 8000 = 5280 h + h mi. But if h is small, then h 2 is even smaller, and h 2 / is even smaller still, hence we can ignore this term in our estimate. Thus, we have 8000 distance to horizon 5280 h mi h mi 5 h mi. 4 For example, if we are 49 ft up (as in the previous problems), then the distance to the horizon is about = =8.75 mi. 4 4 This is pretty close to the exact answer, which was approximately 8.6 miles, which indicates that this is likely a good estimate. Problem 6. In Problem 4, you calculated the distance to the horizon. This is not the distance one would have to travel on the earth to get to the horizon. Can you estimate this distance? This is to say, at most how far off is the length of the line of sight from length of the shortest arc that starts at your position and goes to where you see the horizon? Hint: Youll need to draw a picture and think geometrically, remembering a critical fact about triangles. Solution. Let s zoom in on the interesting part of the picture from problem 3: B A C 3

4 In this picture, our position is the point B, the dot on the horizon is the point C, and the point on the ground directly below us is A. The distance along the surface of the Earth is shown as a heavy line joining A and C (this is the arc > AC). In a circle, the length of an arc between two points is greater than the length of the segment between those two points. Thus l( > AC) l(ac), where, via a slight abuse of notation, l(x) denotes the length of some object X. Next, recall that no one side of a triangle can be longer than the sum of the other two sides. Applied to ABC, this means that l(bc) l(ab)+l(ac)= l(ac) l(bc) l(ab). Combining the inequalities, we have l( > AC) l(bc) l(ab), which gives us a lower bound for the length of > AC. That is, we have found something that is smaller than this length, and which we can compute (why? do you remember what l(bc) andl(ab) represent?). Now we want to find an upper bound for l( > AC), with the goal of squeezing this quantity between two things that are very close together. We claim that l(bc) l( > AC). (Do you believe this claim? Can you come up with an argument for whyitmustbetrue?) Butthenl(BC) is exactly the upper bound we require. Combining all of our inequalities, we have l(bc) l( > AC) l(bc) l(ab). But recall that if our height above the ground is h feet, then ell(bc) our distance from the horizon is approximately 5 4 h miles. On the other hand, l(ab) = our height = h ft = h 5280 mi, which is a very small number compared to our distance to the horizon. That is, l(bc) l(ab) (this notation means that l(bc) ismuch larger than l(ab) ). This implies that from which we conclude that l(bc) l(ab) l(bc), l(bc) l( > AC) l(bc) l(ab) l(bc)= l( > AC) l(bc). 4

5 Remember that l(ac) represents how far we would have to walk along the ground to reach a point on the horizon, and that l(bc) represents our straight-line distance to the horizon. Since these two distances are approximately equal to each other, we conclude that if we are at some point above the ground and can see a point on the horizon, the distance we would have to walk on the ground to get to that point is about the same as the straight-line distance along our line of sight. Note, however, that this argument was made using an estimate that relied on h being relatively small it may not hold up for large values of h. 5