Respuesta :
0.05 arc-second = 1 degree/72000 = (pi radians)/(180*72000) = 2.424 x 10^(-7) radians
The distance is roughly:
R*(theta) = (600 light-years)*2.424 x 10^(-7) = 0.00014544 light-years = 1.275 light-hours = (3600 seconds)*(3 x 10^8 m/s)*(1.275) = 1.38 x 10^12 meters.
Answer:
Diameter of the star, [tex]d=1.38\times 10^{12}\ m[/tex]
Step-by-step explanation:
Given that,
Angular diameter of the star, angle, [tex]\theta=0.05\ arcsecond=2.42\times 10^{-7}\ radian[/tex]
Distance, [tex]D=600\ ly=5.67\times 10^{18}\ m[/tex]
We need to find the diameter of the supergiant star Betelgeuse. The relationship between the diameter of star and angle subtended is given by :
[tex]\theta=\dfrac{d}{D}[/tex], d = diameter
[tex]d=\theta\times D[/tex]
[tex]d=2.42\times 10^{-7}\times 5.67\times 10^{18}[/tex]
[tex]d=1.37214\times 10^{12}\ m[/tex]
or
[tex]d=1.38\times 10^{12}\ m[/tex]
Hence, this is the required solution.