Question

If the distance of earth and sun reduces to one fourth of its present value, then the length of the year will become:
A) $\dfrac{1}{6}$of present year
B) $\dfrac{1}{8}$ of present year
C) $\dfrac{1}{4}$ of present year
D) $\dfrac{1}{2}$ of present year

Answer 1

Hint
One year is the time period of Earth orbiting around the Sun. The relation between the distance between 2 objects under mutual gravity and the time periods of their orbits is given by Kepler’s law. We will use it to find the value of a new time period or year with respect to the original year.

Complete step by step answer
In this question we are given that distance between earth and sun is reduced by a factor of 4, kepler’s law of planetary motion states that:
${T^2} \propto {R^3}$
${T^2}\, = \,k{R^3}$ --- (1)
Where k is a constant, when the value of R reduces to one fourth of its original value, then
${T_N}^2\, = \,k\dfrac{{{R^3}}}{{{4^3}}}$ ---(2)
Dividing (2) by (1) we get,
$ \dfrac{{{T_N}^2}}{{{T^2}}}\, = \,\dfrac{{{R^3}}}{{{4^3}{R^3}}} \\
{T_N}\, = \,\dfrac{T}{{{4^{3/2}}}} \\
{T_N}\, = \,\dfrac{T}{8} \\ $
Therefore the option with the correct answer is option B.

Note
Remembering the Kepler's third law can be difficult. So a student can easily derive it by equating the centrifugal force and the Gravitational force on Earth
$ \Rightarrow \dfrac{{GMm}}{{{r^2}}} = m\,{\omega ^2}r$
$ \Rightarrow \dfrac{{GMm}}{{{r^3}}} = m\,{\left( {\dfrac{{2\pi }}{T}} \right)^2}$
$ \Rightarrow {r^3} \propto {T^2}$
The value of constant k in this relation is equal to $\dfrac{{4{\pi ^2}}}{{Gm}}$, where G is the gravitational constant and m is the mass of planet around which the other is revolving, in this case it should be sun. This can be found by equating the centripetal force between two planetary objects to the force between 2 bodies.