Question

(a) How many astronomical units (A.U.) make 1 parsec?

(b) Consider a sun like star at a distance of 2 parsecs. When it is seen through a telescope with 100 magnifications, what should be the angular size of the star? Sun appears to be (1/2)° from the earth. Due to atmospheric fluctuations, eye can’t resolve objects smaller than 1 arc minute.

(c) Mars has approximately half of the earth’s diameter. When it is closest to the earth it is at about 1/2 A.U. from the earth. Calculate what size it will appear when seen through the same telescope.(Comment : This is to illustrate why a telescope can magnify but not stars.)

Answer 1

By definition of parsec

∴ 1 parsec =\((\frac{1AU}{1arc\,sec})\)

1 deg = 3600 arc sec

∴ 1 parsec = \(\frac{\pi}{3600\times180}\)radians

∴ 1 parsec = \(\frac{3600\times180}{\pi}\)AU

= 206265AU ≈ 2 × 10⁵ AU

(b) At 1 AU distance, sun is (1/2) in diameter.

Therefore, at 1 parsec, star is \(\frac{1/2}{2\times 10^5}\) degree in diameter = 0.25 × 10^-5 arc min

With 100 magnification, it should look 0.25 × 10^-3 arc min. however, due to atmospheric fluctuations, it will still look of about 1 arc min.

∴ It can’t be magnified using telescope.

(c) \(\frac{D_{mars}}{D_{earth}}=\frac{1}{2},\)

\(\frac{D_{earth}}{D_{sun}}=\frac{1}{100}\)[Here, D = diameter]

\(\frac{D_{sun}}{D_{moon}}=400,\)

\(\frac{D_{earth}}{D_{moon}}=4\)

⇒ \(\frac{D_{mars}}{D_{sun}}=\frac{1}{2}\times\frac{1}{100}\)

At 1 AU sun is seen as ½ degree in diameter, and mars will be seen as 1/400 degree in diameter, i.e. mars diameter = \(\frac{1}{2}\times\frac{1}{200}=\frac{1}{400}\) at ½ AU. Mars diameter = \(\frac{1}{400}\times2°=(\frac{1}{200})°\)

At ½ AU mars will be seen as 1/400 degree in diameter. With 100 magnification mars will be seen mars diameter 

=\(\frac{1°}{200}\times100=(\frac{1}{2})°=30\)

This is larger than resolution limit due to atmospheric fluctuations. Hence, it looks magnified.