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Two planets revolve around the sun such that their mean distances from the sun are d1 and d2 respectively and their time periods are T1 and T2. If n1 and n2 are their corresponding frequencies, which of the following is correct?
1. n1d12 = n2d22
2. n12d13 = n223. n1d1 = n2d2
4. n1d2 = n2d1

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Correct Answer - Option 2 : n12d13 = n2

The correct answer is option 2) i.e. n12d13 = n22d23

CONCEPT:

  • Kepler's laws of planetary motion
    • The first law (Law of orbits): All the planets revolve around the sun in elliptical orbits having the sun at one of the foci.
    • The second law (Law of areas): The radius vector drawn from the sun to the planet sweeps out equal areas in equal intervals of time.
    • The third law (Law of periods): The square of the time period of revolution of a planet around the sun in an elliptical orbit is directly proportional to the cube of its semi-major axis.

EXPLANATION:

From the third law of Kepler, the square of the time period (T) is directly proportional to the cube of the mean distance (d).

⇒ T2 ∝ d3

We also know that, time period, \(T =\frac{1}{frequency} = \frac{1}{n}\)

\(\Rightarrow\frac {1}{n^2} \propto d^3\)

For the two planets revolving around the sun, 

\(\Rightarrow\frac {n_2^2}{n_1^2} \propto \frac{d_1^3}{d_2^3}\)

\(\Rightarrow n_1^2d_1^3 = n_2^2d_2^3\)

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