Question

Thus, fors mallangles we canapproximatetan φ 4. I parsec is defined as the distance, d, along a right triangle with the other side equal to Earth's orbital distance of 1 AU and an angle p of 1 arcsecond (1/3600th of a degree), or using that small angle approximation: 1 pc 1 AU/1 arcsecond -see figure below. If a star is closer, that angle gets larger, while a star farther away forms a smaller angle. Equation: d [pel-1 JAUJ/p larcsecondsl Earth's position in July 1 AU 本 本 "Close" star Background Stars Earth's position in January 本 本 本 本 January View from Earth July View from Earth Tan(p) = ifp = 0.00021°, p = arcseconds, and d =-parsecs :

Answer 1

According to the question,

1 AU 1 arcsecond pc

Where, AU stands for astronomical unit.

$1 \, \, arcsecond = \left ( \frac{1}{3600} \right )^{\circ}$

pc stands for parsec

From the given figure, it is clear that

1 AU Tan(P) =

and if p = 0.00021⁰ than

  $Tan (p) = Tan (0.00021^{\circ}) = 3.66 \times 10^{-6}$

And as

$1 \, \, arcsecond = \left ( \frac{1}{3600} \right )^{\circ}$

So,

$p = 0.00021^{\circ} = \frac{0.00021^{\circ}}{(1/3600)^{\circ}} \, \, arcsecond = 0.756 \,\, arcsecond$

Now, from the given diagram and for p = 0.00021⁰ = 0.756 arcsecond , we can write

  $d = \frac{1 \, \, AU}{0.756 \, \, arcsecond} = 1.322 \, \, pc \, (parsecs)$ ( used the approximation " tan (0.00021⁰) ~ 0.00021^{0 "})

For any doubt please comment and please give an up vote. Thank you.