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What is the maximum height we can see a certain rocket as an extensive body watching a space launch? An intersting interdisciplinary problem.

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In this article, I investigated what the maximum height someone can still see a certain space rocket after being launched, from a safe distance, having a criterion the human visual acuity of one minute of arc. After the analitical solution, I apply
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  What is the maximum height we can see a certain rocket as an extensive bodywatching a space launch?An intersting interdisciplinary problem. Neri Luiz von HollebenNovember 1st, 2015 Abstract In this article, I investigated what the maximum height someone can still see a certain space rocket afterbeing launched, from a safe distance, having a criterion the human visual acuity of one minute of arc. After the analitical solution, I apply the result to the particular case of the launch of the space rocket Atlas V 401 by an observer in the safe distance of 3700 meters. This problem has an innovative pedagogical character because contain in its essence, its main condiction, a restriction of physiological nature, something that the student do not usually see in the common textbooks of mathmatics. As an example, this problem could be worked physics and biology class where the methodology of the PBL (problem-based learning) is applied. Many subjects canunfold from this problem as the involvement of students increase, providing, in this way, a valuable theme for teacher work. Key words : visual acuity, visual range, trigonometry, interdiciplinarity, PBL. Introdution A precise calculation to determine what the height is possible someone to see a rocket in a real situation canbe highly complex, becouse we must to know the localweather conditions like a, luminosity, brightness of jet,and so on. The proposal in this article is to know the range of human vision considering only a normal humam vision acuity of one minute of arc. This mean that twopoints can not be distinguished when a angle betweenthe points with the vertex in the observer eye can not be less of one minute of arc according to Fig. (1). This limit is due the existing physical distance between thephotoreceptor cells of our retina. I propose, therefore,to assign such distinguished points as one belonging torocket tail and other belonging to rocket tip in inicialsituation and, after a certain height from the ground,reach the indistiguishable situation that two points be- came one point. In other words, when a rocket become apoint, by definition, we can say that the maximum limit of human vision resolution is reached. Figure 1:  Vision acuity.Fonte:  http://canonicalmomentum.tumblr.com/post/ 86342049687/how-to-write-your-name-on-the-moon  1  Results and discussion The problem consists in to determinate what the ma-ximum height  h  is possible to someone still see a certainspace rocket of high  a , from a safe distance  D . I started by setting the inicial situation as shown in Fig(2a) andthen setting the final state with the object in the limit of a vision as shown in the Fig.(2b). αDa (a)  Inicial set. θ 1 Daθ 2 hα  = 1 ′ (b)  Final set in visual acuity limit. Figure 2:  Inicial set and final set of problem.Source: Author. From Fig.(2a) we obtain the relation of eq.(1): tan α  =  aD,  (1)and from Fig.(2b): tan θ 1  =  hD,  (2) tan θ 2  =  h  +  aD .  (3)To solve we use the main restriction of problem: θ 2 − θ 1  = 1  .  (4) In this way, we enploy the trigonometric identity to arcs difference. tan( θ 2 − θ 1 ) = tan θ 2 − tan θ 1 1 + tan θ 1  tan θ 2 .  (5)Replacing (4), (2) e (3) in equation (5), we obtain: tan1  = h + aD  −  hD 1 +  hDh + aD .  (6)Solving (6) to  h , gives: (tan1  ) h 2 + (tan1  a ) h  + ( D 2 tan1  − aD ) = 0 h  =   a 2 − 4  D 2 −  aD tan1   − a 2  ,  (7) where we can despise the negative solution for  h . Thus, we have two conditions to the true set of   h . a 2 >  4  D 2 −  aD tan1    a 2 − 4  D 2 −  aD tan1    > a However, we can rewrite a solution (7) in function of  initial angle  α  given by equation (1) to give h  =   1 −  4tan 2 α  +  4tan α tan1   − 12 /a .  (8) In this case, we obtain some conditions for true set of   h : 1  >  4tan 2 α  + 4tan α tan1    1 −  4tan 2 α  + 4tan α tan1   >  1 .  (9) Taking as example the launch of the Atlas V 401 rocket  1 , we obtain the following data:Height  a  = 58 . 3  m (10)Safe distance  D  = 3700  m (11) Figure 3:  Lauching of the Atlas V 401.Source:  https://en.wikipedia.org/wiki/Atlas_V  Replacing of data (10) and (11) in (7), we obtain: h  =   58 . 3 2 − 4  3700 2 −  85 . 3 37002 . 9088810 − 4  − 58 . 32 h  = 27  km (12) 1 Local of launch: Kennedy Space Center Launch Complex 39 of NASA, in Merritt Island, Flórida, EUA 2  Conclusion The result of the eq.(8) is interesting because we could generalize to the case of any object moving away from us. One can say that it is a description of how the resolution of the human eye varies when objects move away. From the result obtained in eq.(12), we can conclude that, in principle, we can see clearly the space rocket Atlas V until it reach 27 km from the ground. This correspond to a second layer of atmospheric 2 , called stratosphere, that is situated between  10000m  and  50000m  from the ground. 2 Source:  http://www.physicalgeography.net/fundamentals/7b.html 3
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