If the distance of the earth from Sun is 1.5 × 10⁶ km. Then the distance of an imaginary planet from Sun, if its period of revolution is 2.83 years is :

Solution

Physics sir · 4 steps

See 2.83 as 2√2: square it to get 8, take the cube root, and the orbit radius just doubles.

1. Kepler's third law. For planets going round the same Sun, the square of the period is proportional to the cube of the orbit radius. Compare the planet with the Earth, which has T_E = 1 year.

   (T_P/T_E)² = (r_P/r_E)³

2. Put in numbers. T_P/T_E = 2.83 ≈ 2√2, so the left side is (2√2)² = 8. Then (r_P/r_E)³ = 8, which gives r_P/r_E = 2.
3. Scale the distance. Use the Earth distance as the question gives it, even though the real value is about 1.5 × 10⁸ km. Then r_P = 2 × 1.5 × 10⁶ km = 3 × 10⁶ km.
4. So. The planet is 3 × 10⁶ km from the Sun, which is option (d).

The idea to keep. For orbits round the same star, use ratios with T² ∝ r³ and take the Earth as 1 year. Watch for numbers like 2.83 (2√2) or 5.2 that are built to give perfect cubes.

Answer from the NTA final answer key. Solution by PrepWiser.