Question

Define astronomical unit and parsec.

Answer 1

$\textsf{\Large{\underline {Astronomical Unit ( A. U.) }}}$ : It is the average distance between the sun and the earth.

⚫ It is used to measure distances of planets.

$\boxed{\mathsf{1\:AU\:=\:1.496 *{10}^{11}m}}$

$\textsf{\Large{\underline {Parallactic Seconds ( Parsec) }}}$ : It is the distance at which an arc of length 1 Astronomical Unit subtends an angle of 1 second.

$\textsf{\underline {Proof}}$ :

Angle = $\mathsf{\dfrac{l} {r}}$

Where l = Length of arc.

r = Radius

$\mathsf{r\:=\:{\dfrac{l} {\theta}}}$ = $\mathsf{\dfrac{1\:AU}{1"}}$

[ 1" = 1 second ]

1 Parsec = $\mathsf{\dfrac{1.496*{10}^{11}}{\dfrac{\pi}{\:3600*180°}}}$

$\boxed{\mathsf {1\: Parsec \:=\:3.08*{10}^{16}m}}$

⚫Parsec is the $\tt{\underline {largest \:unit\:of\:distance.}}$

Answer 2

Explanation:

Astronomical Unit ( A. U.)

: It is the average distance between the sun and the earth.

⚫ It is used to measure distances of planets.

\boxed{\mathsf{1\:AU\:=\:1.496 *{10}^{11}m}}

1AU=1.496∗10

11

m

\textsf{\Large{\underline {Parallactic Seconds ( Parsec) }}}

Parallactic Seconds ( Parsec)

: It is the distance at which an arc of length 1 Astronomical Unit subtends an angle of 1 second.

\textsf{\underline {Proof}}

Proof

:

Angle = \mathsf{\dfrac{l} {r}}

r

l

Where l = Length of arc.

r = Radius

\mathsf{r\:=\:{\dfrac{l} {\theta}}} r=

θ

l

= \mathsf{\dfrac{1\:AU}{1"}}

1"

1AU

[ 1" = 1 second ]

1 Parsec = \mathsf{\dfrac{1.496*{10}^{11}}{\dfrac{\pi}{\:3600*180°}}}

3600∗180°

π

1.496∗10

11

\boxed{\mathsf {1\: Parsec \:=\:3.08*{10}^{16}m}}

1Parsec=3.08∗10

16

m

⚫Parsec is the \tt{\underline {largest \:unit\:of\:distance.}}

largestunitofdistance.