Question

The angular diameter of a planet measured from earth is \(90''\). If the diameter of the planet is \(x \times 10^4\) m, then its distance from the earth is

• A. \(3.6\times10^9\text{ m}\)
• B. \(7.2\times10^9\text{ m}\)
• C. \(3.6\times10^8\text{ m}\)
• D. \(7.2\times10^8\text{ m}\)
• E. \(1.8\times10^8\text{ m}\)

Hint:

For very small angles: \[ \theta(\text{in rad})=\frac{\text{diameter}}{\text{distance}} \] and \[ 1''=\frac{1}{206265}\text{ rad} \]

Answer 1

The Correct Option is B

For small angles: \[ \theta = \frac{d}{D} \] Here angular diameter: \[ 90'' = \frac{90}{206265}\text{ rad} \approx 4.36\times10^{-4}\text{ rad} \] Diameter of planet: \[ d=\pi\times 10^6 \text{ m} \] So distance: \[ D=\frac{d}{\theta} \approx \frac{\pi\times10^6}{4.36\times10^{-4}} \approx 7.2\times10^9\text{ m} \] Hence, \[ \boxed{(B)\ 7.2\times10^9\text{ m}} \]