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Suppose the gravitational force varies inversely as the nth power of distance. Then the time period of a planet in a circular orbit of radius R around the sun will be proportional to

Asked in AIEEE 2004 · Modified force laws

Answer: (1) R^(n+1)/2

Step-by-step solution

A circular orbit means the assumed force is exactly what is needed to keep the planet turning.

k/(Rⁿ)=mRω², so ω²=k/(mRⁿ⁺¹).

Since T=(2π)/ω, T²∝ Rⁿ⁺¹ and T∝ R^(n+1)/2.

Setting n=2 gives T∝ R^3/2, the familiar Kepler result, which confirms the exponent.

Why the other options are wrong

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