Question

a) Using s = rθ, calculate the arc length (in meters) subtended by the sun when it moves an angular
distance of 2 x10-6 °. The distance between the Earth and Sun is called an AU (astronomical unit). 1
AU = 1.5 x1011 m. (Don’t forget to convert to radians).

As you can see from the above calculation, a small change in angle can correspond to a large distance.
In astronomy angles are typically measured in arcminutes or arcseconds
1 arcmin = 1/60 °
1 arcsec = (1/60) arcmin = 1/(602) °
In addition, when discussing distances that are trillions of miles away it is more sensible to units of
distance that are on the scale of the solar system, i.e. light years or AU’s.
Light year: the distance traveled by light in one year

b) Using the speed of light c = 3 x 108 m/s, convert the distance unit of 1 ly (light year) to meters

c) Given that SgrA* lies at a distance of 26,000 ly away from Earth, find the distance subtended by star
S2 when our telescopes detect that S2 moves 1 arcsec. Remember to convert angles into radians,
and express your answer in ld (light days). This ratio of 1 arcsec = __ ld will be needed for
calculations in Lab 10.

Answer 1

Sol) ca we are given. S = ro where o is in radians. we are given or 2x10-60 We know that 1° Po rade so 2x10-6o = I x 2x10-6 rad. trono radi (6) We are given na IAU = 1.58101 in 8o are length is sanoa (lostio my co x106) Ise 2618 m) One light year is the distance that light covered in 1-year so Distance (Skeed) (Time) 2 (3x108) ( 1 year) = 3x108 m/s) (365+24X60X60 sec) & qu.61X Cothm

(c) We are given era 26000 ly = 26000) (94-61x104m) by and the star moved I arc secondo we know that os lare second chance toe - tection rad. so are length is sa roz (26000) 94.6181014) I eg) E3600 (180) loc Is= 11.926x1014 ml Now we know that distance - (Speed) ( Times (11.926 X1014) (3x108mls) (Time) Т. 1. с дето“ м - 2-975ж to see 3x108 mis days converting to To 46 days so distance is 46 light days

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