The angular diameter of the sun is 1920". If the distance of the sun from the earth is `1.5xx10^(11)`m, what is the linear diameter of the sun ?

Answer

To find the linear diameter of the Sun given its angular diameter and distance from the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Convert Angular Diameter from Arcseconds to Radians:**
The angular diameter of the Sun is given as \( 1920 \) arcseconds. We need to convert this to radians.
\[ \text{1 degree} = 3600 \text{ arcseconds} \quad \text{and} \quad \text{1 radian} = \frac{180}{\pi} \text{ degrees} \]
First, convert arcseconds to degrees:
\[ \text{Degrees} = \frac{1920 \text{ arcseconds}}{3600} = 0.5333 \text{ degrees} \]
Now convert degrees to radians:
\[ \text{Radians} = 0.5333 \times \frac{\pi}{180} \approx 0.0093 \text{ radians} \]

2. **Use the Angular Diameter Formula:**
The relationship between the angular diameter \( \theta \) (in radians), the linear diameter \( D \), and the distance \( d \) is given by:
\[ \theta = \frac{D}{d} \]
Rearranging this formula to find the linear diameter \( D \):
\[ D = \theta \times d \]

3. **Substitute the Values:**
We know \( \theta \approx 0.0093 \) radians and the distance \( d = 1.5 \times 10^{11} \text{ m} \).
\[ D = 0.0093 \times (1.5 \times 10^{11}) \]

4. **Calculate the Linear Diameter:**
Performing the multiplication:
\[ D \approx 0.0093 \times 1.5 \times 10^{11} \approx 1.395 \times 10^{10} \text{ m} \]
Rounding this gives:
\[ D \approx 1.4 \times 10^{10} \text{ m} \]

### Final Answer:
The linear diameter of the Sun is approximately \( 1.4 \times 10^{10} \text{ m} \).

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