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MHT CET 2021 (24 Sep, Shift 1) · Kepler's Laws & Planetary Motion

Kepler's Laws & Planetary Motion question from MHT CET 2021 (24 Sep, Shift 1)

Q17MHT CET 2021 · 24 SEP · SHIFT 1Easy

Kepler's Laws & Planetary Motion

The period of revolution of planet A\text{A} around the sun is 8 times that of planet B\text{B}. How many times the distance of A from the sun is greater than that of B from the sun?

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Answer: (c)

  1. Kepler's third law: r∝T2/3r \propto T^{2/3}.
  2. rArB=82/3=4\dfrac{r_A}{r_B} = 8^{2/3} = 4, so rA=4rBr_A = 4r_B.
  3. rAr_A exceeds rBr_B by 4rB−rB=3rB4r_B - r_B = 3r_B.
  4. So the answer is (c) 3 times.
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