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Mathematical Applications

# Mathematical Applications - Mathematical Applications Fo m...

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12/24/11 Mathematical Applications 1/6 ZZZ.uni.edu/morgans/astro/course/Notes/section1/math1.html FoUmXla - InWUodXcWion Wo Whe Sk\ Small Angle FoUmXla - ThiV iV a UelaWion beWZeen diVWance, Vi]e and angXlaU Vi]e, and iW onl\ ZoUkV Zell foU Vmall angleV, Zhich VoUW of e[plainV iWV name. FoUmXla: S = 0.0175 R ZheUe: S = Vi]e of Whe objecW (XVXall\ Whe diameWeU) R = diVWance of Whe objecW = angXlaU Vi]e of Whe objecW, meaVXUed in degUeeV S and R aUe XniWV of lengWh, and can be in an\ XniW of meaVXUe, hoZeYer Whe\ need Wo be in Whe Vame XniWV. So if S iV meaVXUed in cm, Vo iV R. If R iV meaVXUed in incheV, Vo iV S. The conVWanW iV WheUe Wo accoXnW foU Whe degUeeV. ThiV foUmXla onl\ ZoUkV foU angleV leVV Whan 10 degUeeV. T\pical Problems 1. An objecW ZiWh a diameWeU of 38 cm iV locaWed 688 m aZa\. WhaW iV iWV angXlaU Vi]e? SolXWion: YoX'll need Wo UeaUUange Whe foUmXla, Vince \oX haYe Wo VolYe foU S = 0.0175 R = S/(0.0175 R) Since S and R mXVW be in Whe Vame XniWV of meaVXUe, Whe YalXe of S Zill be conYeUWed Wo meWeUV, Vo S = 38 cm = 0.38 m = 0.38 / (0.0175 [ 688) = 0.032 degrees . 2. An objecW iV locaWed 45 km aZa\. If iWV angXlaU Vi]e iV 2.0 degUeeV, ZhaW iV iWV acWXal Vi]e? SolXWion: YoX don'W haYe Wo UeaUUange Whe foUmXla Vince \oX aUe looking foU Whe YalXe of S, WheUefoUe S = 0.0175 R S = 0.0175 [ 45 [ 2.0 = 1.6 km 3. If an objecW iV 25 cm in Vi]e, and iW haV an angXlaU Vi]e of 15 aUcminXWeV, hoZ faU aZa\ iV iW? SolXWion: YoX'll haYe Wo UeaUUange Whe foUmXla Wo VolYe foU R.

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Mathematical Applications - Mathematical Applications Fo m...

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