Use these lines to form a right triangle: one whose length is the distance from the observer to the center of the star's orbit; one whose length is the distance from the observer to a point at the radius of the orbit; and one whose length is the orbital radius itself. With these we can find the half-angle, which is the angle subtended by the orbital radius. The whole angle (the angular size of the orbit) is just twice this.
Let D be the distance to the center (the long leg of the right triangle); let R be the orbital radius (the short leg).
The half-angle is then tanˉ¹(R/D).
Convert kpc to km: (9.10 kpc)(3.08 × 10¹⁶ km/kpc) = 2.80 × 10²⁰ km = D, so the half-angle θ is
tanˉ¹((1.43 × 10¹² km)/(2.80 × 10²⁰ km)) = 2.926 × 10ˉ⁷ degrees. Multiply this by 2 to get the whole angle:
5.85 × 10ˉ⁷ degrees.
Multiply this by 3600 to get it in arcseconds: 2.11 × 10ˉ³ arcseconds.
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Verified answer
1 kpc (kiloparsec) is not 3.08 x 10^16. It is 3.086 x 10^19.
Use these lines to form a right triangle: one whose length is the distance from the observer to the center of the star's orbit; one whose length is the distance from the observer to a point at the radius of the orbit; and one whose length is the orbital radius itself. With these we can find the half-angle, which is the angle subtended by the orbital radius. The whole angle (the angular size of the orbit) is just twice this.
Let D be the distance to the center (the long leg of the right triangle); let R be the orbital radius (the short leg).
The half-angle is then tanˉ¹(R/D).
Convert kpc to km: (9.10 kpc)(3.08 × 10¹⁶ km/kpc) = 2.80 × 10²⁰ km = D, so the half-angle θ is
tanˉ¹((1.43 × 10¹² km)/(2.80 × 10²⁰ km)) = 2.926 × 10ˉ⁷ degrees. Multiply this by 2 to get the whole angle:
5.85 × 10ˉ⁷ degrees.
Multiply this by 3600 to get it in arcseconds: 2.11 × 10ˉ³ arcseconds.