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How far above the earth? - June 3-7, 1996 Next Problem || Previous Problem || Contents || All Problems || Search POWs Recall from last week that the radius of the earth (at the equator) is 6374 km. If you wanted to go out into space and take a picture of the earth, how far from the earth (above the equator) would you have to go to make sure that you got 1/4 of the equator in your picture? How about 1/3? Extra: How far out would you have to go to get 1/2 of the equator in your picture? Please explain your answers thoroughly! The heavyweights weighed in this week - see the solutions by Brian Gordon, Dartmouth '92, and Ken Duisenberg, a Stanford grad working for Hewlett Packard - but Eric Jonas sent us a solution in the form of his own Web page. Check it out! Here are the answers Brian and Ken sent us: Brian Gordon Grade: Dartmouth 1992 The general answer to the question of: How high do you need to be above the ground to capture "x" amount of the equator in the picture is given by the formula: h = 6374 * ( sec (180x) - 1) meters So for 1/4 of the equator, it's 6374 * (sqrt(2) - 1) = 2640.2 m for 1/3 of the equator, it's 6374 * (2 - 1) = 6374 m for 1/2 of the equator, it's 6374 * (infinity - 1) = infinity. Here's the derivation of the formula: Imagine a cross section of the earth at the equator. Put your camera at some point above the earth. Draw the tangents from this point to the earth. The amount of arc between them will be the amount of equator you're able to capture. Now, here's how we'll determine the height of the camera corresponding to a given amount of equator. The angle between the two radii of tangency represents the amount of the equator. That is, a 120 degree angle gives you 1/3 the equator, and a 90 degree angle gives you 1/4 the equator. (This is simply a statement that arc measure equals the central angle.) A formula for this angle is 360 * the amount of arc. Okay, now connect the camera to the center of the earth. This divides the central angle of the arc in half, so that the two triangles formed each contain an angle of degree measure half the central angle, or 180 * the amount of arc. The two triangles are right triangles (radii are perpendicular to points of tangency), so we can express the distance from the earth's center to the camera via trig ratios: the cosine of (180 * amount of arc) = 6374/dist. from camera to earth Solving for the distance we get that the distance from the camera to the earth is equal to 6374/(cos (180 * amount of arc)). To eliminate the trig function in the denominator, I used the fact that secant = 1/cosine to rewrite this, using x=amount of arc, as d = 6374 * sec (180x) Now since this distance includes the distance from the center of the earth to its surface, let's subtract that. h = d - 6374 = 6374 * ( sec (180x) - 1 ) This is cool because it shows why you can't get half the equator. 180 * 1/2 is 90, and sec 90 is infinity. Hope that was pretty clear. --bri Ken Duisenberg Grade: Post-Master's School: Stanford University, currently an HP Engineer Answer: 1/4: 6374*(sqrt(2)-1) km. (approx. 2640.2 km) 1/3: 6374 km. 1/2: Infinitely far. Solution: Model the Earth as a circle with center O and radius R. For 1/4, divide the circle into quarters, through points ABCD on the circle. Draw tangent lines at A and B, intersecting at external point E. AOB, OAE, and OBE are all right angles, and OA and OB are of length R, so OAEB is a square. The distance from O to E is R*sqrt(2), so the distance from E to the circle is R*(sqrt(2)-1). Or the height to view 1/4 of the equator is 6374*(sqrt(2)-1) km, or 2640.2 km. For 1/3, divide the circle into thirds, through points ABC on the circle. Draw tangent lines at A and B, intersecting at external point E. AOB is a 120 degree angle, and OAE and OBE are right angles, so OAE is a 60 degree angle and triangle OAE is a 30-60-90 triangle. OA is of length R, so OE is of length 2R, and the distance from E to the circle is R. Or the height to view 1/3 of the equator is 6374 km. For 1/2, divide the circle into halves, through points A and B. Draw tangent lines to A and B, noticing that they are parallel and thus never intersect, so there is no point where half the equator could be viewed. [Privacy Policy] [Terms of Use] Home || The Math Library || Quick Reference || Search || Help  © 1994-2008 Drexel University. All rights reserved. http://mathforum.org/ The Math Forum is a research and educational enterprise of the Drexel School of Education.