Question

The distance of the Sun from Earth is \(1.5 × 10^{11}\) \(m\), and its angular diameter is \((2000)\) \(s\) when observed from the earth. The diameter of the Sun will be :

• A. \(2.45 × 10^{10 }\) \(m\)
• B. \(1.45 × 10^{10}\; m\)
• C. \(1.45 × 10^9 \;m\)
• D. \(0.14 × 10^9 \;m\)

Answer 1

The Correct Option is C

To find the diameter of the Sun, we can use the concept of angular diameter. The angular diameter \( \theta \) in radians is given by the formula:

\(\theta = \frac{D}{d}\)

where:

• \(D\) is the actual diameter of the object.
• \(d\) is the distance to the object.

We need to convert the angular diameter from arcseconds to radians. We know that:

• 1 degree = 60 minutes
• 1 minute = 60 seconds
• 1 degree = \(\frac{\pi}{180}\) radians

So, converting the angular diameter from arcseconds to radians:

\(\theta = \frac{2000}{3600} \times \frac{\pi}{180}\)

Thus, the angular diameter in radians is:

\(\theta = \frac{2000 \times \pi}{648000}\)

Now, substituting the values into the formula to find the diameter \(D\):

\(D = \theta \times d\)

where the distance \((d)\) from Earth to the Sun is \(1.5 \times 10^{11}\;m\).

Using the calculated angular diameter in radians, the formula becomes:

\(D = \left(\frac{2000 \times \pi}{648000}\right) \times (1.5 \times 10^{11})\)

Calculating this gives:

\(D \approx 1.45 \times 10^9\;m\)

Therefore, the diameter of the Sun is \(1.45 \times 10^9\;m\).