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If the radius of Earth's orbit is reduced by 75%, the duration of an year will-
1. increase by 65%
2. decrease by 12.5%
3. decrease by 87.5%
4. increase by 8%

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Correct Answer - Option 3 : decrease by 87.5%

The correct answer is option 3) i.e. decrease by 87.5%


  • 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.


From the third law of Kepler, the square of the time period (T) is directly proportional to the cube of the semi-major axis (R).

⇒ T2 ∝ R3

Let the radius and time period of Earth be R and T respectively.

If R' and T' are the new radius and time period,

When the radius of Earth's orbit is reduced by 75%, R' = 25% of R = 0.25 R

Then the new time period will be:

\(\Rightarrow( \frac{T'}{T} )^2= (\frac {R'}{R})^3\)

\(\Rightarrow( \frac{T'}{T} )^2= (\frac {0.25R}{ R})^3\)

\(\Rightarrow( \frac{T'}{T} )^2= 0.25^3\)

\(\Rightarrow T' = \sqrt{0.25^3} T= 0.125T\)

\(\Rightarrow T' = 12.5\%T\)

Therefore, the new time period gets decreased by 87.5% (100% - 12.5% = 87.5%)

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