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A light planet orbits a very massive star in a circular orbit of radius r with period T. If the gravitational attraction between them is proportional to r^(5/2), then the square of the orbital period T² is proportional to:
- r³
- r²
- r².5
- r³.5
Correct answer: r³.5
Solution
For circular motion, the gravitational force provides centripetal force: F = m omega² r = m (4 pi²/T²) r. Given F is proportional to r^(5/2): r^(5/2) is proportional to r/T². So T² is proportional to r / r^(5/2) = r^(1 - 5/2) = r^(-3/2)? That gives a negative power. Re-derive: F proportional to r^(5/2) means m(4 pi²/T²) r = k r^(5/2), so 1/T² proportional to r^(5/2)/r = r^(3/2), thus T² proportional to r^(-3/2)? That is unphysical. The standard intended relation uses F proportional to r^(-5/2): then m(4 pi²/T²) r = k r^(-5/2) -> 1/T² proportional to r^(-5/2)/r = r^(-7/2) -> T² proportional to r^(7/2) = r³.5. The conventional answer is r³.5.
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