Browse by Stream
Discuss Doubts
Home
Search for Exam, Articles, Questions
get app
Go To Your Study Dashboard
  1. Home
  2. The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star show when viewed from two locations of the Earth six months apart in its orbit around the Sun ?

Filters

Select Stream
Select Type
Select Exam
Select Subject
Select Chapter
Q&A - Ask Doubts and Get Answers
Q
NCERT

The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star show when viewed from two locations of the Earth six months apart in its orbit around the Sun ?

Q 2.20: The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun?

Answers (1)
Views
S safeer

Given, Distance of the star from the solar system = 4.29 ly (light years)
1 light year is the distance travelled by light in one year.

(Note: Light year is a measurement of distance and not time!)
(a) 1 ly = (3 \times 10^8)ms^{-1} \times (365 \times 24 \times 60 \times 60)s = 94608 \times 10^{11} m
4.29 ly = 405868.32 \times 10^{11} m
We know, 1 parsec = 3.08 \times 10^{16} m

\therefore 4.29 ly = \frac{405868.32\times10^{11}}{3.08\times10^{16}}= 1.32 parsec

Now,

(b) ? = d/D\Theta = d/D

& d = 3 \times 10^{11} m; D = 405868.32 \times 10^{11} m

\theta = (3\times10^{11})/(405868.32\times10^{11} )= 7.39\times10^{-6} rad

Also, We know 1\ sec = 4.85 \times10^{-6} rad

\therefore 7.39\times10^{-6} rad = \frac{7.39\times10^{-6}}{4.85 \times10^{-6}} = 1.52''

Similar Questions

Related Chapters

Exams
Articles
Questions