### Question 2.20

NCERT Class XI Physics

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 ?

### Physical Concept

In the first part we have to convert the given distance from light year to parsec.

In the second part the obtained parsec has to be used to find the parallax angle on the diametrically opposite positions of Earth in its path around the Sun.

### Video Explanation

Please watch this youtube video for a quick explanation of the solution:

### Solution in Detail

__PART I:__

Remember that:

$\displaystyle 1 \text{ parsec} \equiv 3.26 \text{ ly}$

$\displaystyle \therefore 4.29 \text{ ly} \equiv \frac{4.29}{3.26} = 1.316 \text{ parsec}$

$\displaystyle = 1.32 \text{ parsec} \:\underline{ Ans } \text{(first part)}$

(rounded to 3 SF because SF in 4.29 are 3 and the calculation is division type)

__PART II:__

$ \begin{aligned} &\text{by def. of parsec on earth-sun base}\\ &\text{distance of }1 \text{ parsec } \equiv 1 \rq\rq\\ &\therefore 1.32 \text{ parsec} = 1 \times 1.32 = 1.32 \rq\rq\\ &\text{but this is on earth-sun base}\\ &\therefore \text{parallax on dia } = 2 \times 1.32 = 2.64 \rq\rq\\ &= 2.64 \rq\rq\:\underline{Ans}\\ &\text{(SF in 1.32 are three, so SF }\\ &\text{in the answer will also be three)} \end{aligned} $

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