Question

What is the angular diameter in arc seconds of a star like the sun (diameter= 1,400,000 km) located 8 light years (7.57 x 10^13 km) fom earth?
a. 0.038

b. 8

c. 0.0038

d.200,00

e. 1,400,000

Answer 1

Answer:

Option (C)

Explanation:

Here, Diameter= 1,400,000 km = 1.4 x 10^6 km

         Distance= 7.57 x 10^13 km

According to the formula, angular diameter= Diameter/Distance

$\alpha = \frac{Diameter}{Distance}$

$=\frac{1.4\times 10^6}{7.57}\times 10^{-13}$

$=0.18\times 10^{-7}\text{ radian}$

Since 1 radian=206265 arc second,

Therefore,

$206265\times 0.18\times 10^{-7}$

$=\frac{37127.7}{10^7}$

= .00371 arc second

Hence, the approximate answer is option (C) i.e. 0.0038 arc second.

Answer 2

Answer:

$\theta = 0.0038\ arc\ sec$

option c

Explanation:

given data:

Diameter of a star = 1,400,000 km

star distance from sun $= 7.57\times 10^{13} km$

we know that

Angle is given as

$\theta = \frac{d}{r} = \frac{14\times 10^5}{7.57\times 10^{13}}$

$\theta = 1.849\times 10^{-8} rad$

we know that

$1\ arc\ second = 4.84\times 10^{-6} rad$

$1 rad = \frac{1\ arc\ sec}{4.84\times 10^{-6}}$

SO, $\theta = \frac{1.849\times10^{-8}}{4.84\times10^{-6}}$

$\theta = 3.82\time 10^{-3} arc sec$

$\theta = 0.0038\ arc\ sec$

option c