Question

The distance between the Sun and Earth is \( R \). The duration of a year if the distance between the Sun and Earth becomes \( 3R \) will be:

• A. \( \sqrt{3} \) years
• B. \( 3 \) years
• C. \( 9 \) years
• D. \( 3\sqrt{3} \) years

Hint:

Kepler’s third law states: \[ T^2 \propto R^3 \] which helps determine orbital periods when the radius changes.

Answer 1

The Correct Option is D

Step 1: {Kepler’s third law}
\[ T^2 \propto R^3 \] Step 2: {Find new time period}
\[ \left( \frac{T_2}{T_1} \right)^2 = \left( \frac{R_2}{R_1} \right)^3 \] Substituting values: \[ T_2 = \left( \frac{3R}{R} \right)^{3/2} \times 1 \] \[ = 3\sqrt{3} { years} \] Thus, the correct answer is \( 3\sqrt{3} \) years.

Answer 2

Step 1: According to Kepler’s Third Law:
\( T^2 \propto R^3 \), where:
- \( T \) is the time period (orbital duration)
- \( R \) is the radius (mean distance from the Sun)

Step 2: Let the current time period be \( T_1 = 1 \) year for distance \( R \).
Let the new time period be \( T_2 \) when the distance becomes \( 3R \).

Step 3: Use the ratio form of Kepler’s law:
\( \left( \frac{T_2}{T_1} \right)^2 = \left( \frac{R_2}{R_1} \right)^3 \)
\( \left( \frac{T_2}{1} \right)^2 = (3)^3 = 27 \)
\( T_2 = \sqrt{27} = 3\sqrt{3} \) years

Final Answer: \( 3\sqrt{3} \) years