(a) The earth- moon distance is about 60 earth radius. What wil be the diameter of the earth (approximately in degress) as seen from the moon ? (b) Moon is seen to be of `(1//2)^@` diameter from the earth. What must be the relative size compared to the earth ? (c ) From parallax measurement, the sun is found to be at a distance of about 400 times the earth. moon distance. Estimate the ratio of sun-earth diameters.

Explanation

By definition,
1 parsec=Distance at which 1 AU long arc subtends an angle of 1 s.
`therefore1 "parsec "=(("1AU")/("1 arc sec"))`
1 deg=3600 arc sec
`therefore1"parsec "=pi/(3600xx180)rad`
`therefore 1"parsec "=(3600xx180)/pi"AU"=206265 "AU"approx2xx10^(5)"AU"`
(b) Sun's diameter is `(1/2)^(@)` at 1 AU.
Therefore,at 1 parsec, star is `(1//2)/(2xx10^(5))` degree in diameter =`15xx10^(-5)` arc min.
With 100 magnification, it should look `15xx10^(-3)` arcmin. However, due to atmospheric fluctuation, it will still look of about 1 arcmin. It cannot be magnified using telescope.
(c) Given that `(D_("mars"))/(D_("earth"))=1/2" "...(i)`
where D represents diameter.
From answer 25 (e)
we know that, `(D_("earth"))/(D_(sun))=1/100`
`therefore (D_("mars"))/(D_(sun))=1/2xx1/100" "["from Eq.(i)"]`
At 1 AU sun's diameter =`(1/2)^(@)`
`therefore"mar's diameter"=1/2xx1/200=1/400`
`At1/2"AU,mar's diameter"=1/400xx2^(@)=(1/200)^(@)`
With 100 magnification, Mar's diameter `=1/200xx100^(@)=(1/2)^(@)=30'`
This is larger than resolution limit due to atmospheric flutuations. Hence, it looks magnified.