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Q.
Earth is revolving around the sun. If the distance of the earth from the sun is reduced to 1/4th of the present distance, then the length of present day will be reduced by:
BHUBHU 2005
Solution:
From Kepler's third law of planetary motion, the square of period of revolution of any planet around the sun is directly proportional to the cube of the semi-major axis of its elliptical orbit.
$ {{T}^{2}}\propto {{a}^{3}} $
Given, $ a'=\frac{a}{4} $
$ \therefore $ $ \frac{{{T}^{2}}}{T{{'}^{2}}}=\frac{{{a}^{3}}}{{{\left( \frac{a}{4} \right)}^{3}}}={{(4)}^{3}} $
$ \Rightarrow $ $ T'=\frac{T}{{{(4)}^{3/2}}}=\frac{T}{8} $