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The period of revolution of planet A around the Sun is 8 times that of B. The distance of A from the Sun is how many times greater than that of B from the Sun?

1. 2
2. 3
3. 4
4. 5

Correct answer: 4

Solution

According to Kepler's third law, the square of the period of revolution is proportional to the cube of the semi-major axis of the orbit. If the period of A is 8 times that of B, then (T_A / T_B)^2 = (r_A / r_B)^3. Substituting T_A / T_B = 8, we get 8^2 = (r_A / r_B)^3, which simplifies to r_A / r_B = 4.

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