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The distance of Neptune and Saturn from the sun is nearly 10¹³ and 10¹² meter respectively. Assuming that they move in circular orbits, their periodic times will be in the ratio

1. 10
2. 100
3. 10√10
4. 1000

Correct answer: 1000

Solution

According to Kepler's third law, the square of the orbital period (T) is proportional to the cube of the semi-major axis (r). Thus, (T₁/T₂)² = (r₁/r₂)³. Substituting r₁ = 10¹³ m and r₂ = 10¹² m, we get (T₁/T₂)² = (10¹³/10¹²)³ = 10³. Taking the square root, T₁/T₂ = 10^(3/2) = 1000. Hence, the ratio is 1000.

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