5. In astronomy, the sizes of celestial objects are often given in terms of thei
ID: 2910836 • Letter: 5
Question
5. In astronomy, the sizes of celestial objects are often given in terms of their angular diameter as seen from Earth, rather than their actual sizes, the idea being that for small angles 0 (and large distances d) the ar- clength s of the circle with radius d and central angle is very close to the height of an object h. Since these angular diameters are typically small, it is common to present them in arcseconds, or arcminutes, instead of degrees hes ur friend who is approximately 1.7 m tall, at a distance of 100 meters. What is her angular un is located 150 million kilometers from Earth and has a radius of 696000 kilometers. What is c How far away, in meters, would a dime (I centimeter) have to be so that its angular size is exactly one (d) How far away, in meters, would a dime (1 cm) have to be to cover the sun, in other words, to eclipse size in arcminutes? That is, what is the value of the central angle in minutes? (b) The s its angular diameter in arcminutes? arcsecond? the sun?Explanation / Answer
a)
tantheta = h/d
tan(t) = 1.7 / 100
t = arctan(1.7/100)
t = 0.973934436601 degrees
0.973934436601 * 60 or 58.436 arcminutes ---> theta
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b)
Distance = 150*10^6 km
Diameter of the sun = 696000 km
So, we have
tan(theta) = 696000 / (150*10^6)
tan(t) = 0.00464
t = 0.265850509067 deg
t = 0.265850509067 * 60 arcmin
t = 15.951 arcmin --> ANS
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c)
theta) = 1 arcsec = 1/3600 degree
tan(theta) = 1/x
tan(1/3600) = 1/x
x = 206264.7982101165870519cm
As in approx
2062.64 mtrs away ----> ANS
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