Assuming that the orbit of the planet Mercury around the sun to be a circle, Copernicus detrmined the orbital radius to be `0.38AU`. From this determine the angle of maximum elongation for Mercury and its distance from the earth when the elongation is maximum .

Answer

`22.3^@ ; 1.384xx10^8km`

Explanation

Here, `r_(ps) =0.38A U , r_(es)= 1A U`
The angle of maximum elongation `"in"`is given
by
`sin "in" = (r_(ps))/(r_(es)) = (0.38)/(1)`
Now,`r_(pe) =r_(es)xxcos "in"`
`1.496xx10^(11) cos (22.3^@)`
`=1.496xx10^(11)xx0.9252 = 1.384xx10^(11)m`
`=1.384xx10^8km`