Question

- The largest width (diameter) of the ring in parsecs is 0.48 parsecs. - The smallest width (diameter) of the ring: 53.76 pixels a) convert pixels in to archseconds b) Use the small angle formula to calculate the smallest width (diameter) in parsecs. c) Substitute the value for the largest width (in pc) and the smallest width (in pc) into the inclination formula and calulcate i. Note that sin^-1 is the nverse sine function (Hint: your answer will be greater than 10 degrees, bu less then 50 degrees)

Answer 1

The largest width (diameter) of the ring, D_large = 0.48 parsecs

The smallest width (diameter) of the ring, D_small = 53.76 pixels

(a) To convert pixels into arcseconds which will be given as -

D_small = [(53.76 pixels) (0.025 arcseconds/pixels)]

D_small = 1.344 arcseconds

(b) The smallest width (diameter) of the ring in parsecs which will be given as -

From small angle formula, we have

D_small = 1 / (1.344 arcseconds)

D_small = 0.74 parsecs

(c) An inclination angle which will be given by -

sin = (D_large / D_small)

= sin^-1 [(0.48 pc) / (0.74 pc)]

= sin^-1 (0.6486)

= 40.4 degree