Question

The distances of two planets (Neptune and Saturn) from the sun are ${10^{13}}\,m\,and\,{10^{12}}\,m$ respectively. The ratio of time periods of the planets is:
A. $100:1$
B. $1:\sqrt {10}$
C. $\sqrt {10} :1$
D. $10\sqrt {10} :1$

Answer 1

In order to this question, to find the ratio of time periods of the given planets Neptune and Saturn, first we will rewrite the given facts of the question, and then we will apply Kepler’s 3rd law of planetary motion. And then we will also compare between Copernicus and Kepler’s 3rd law of motion, whether which one is better.

Complete answer:
Distance of Neptune from the Sun, ${d_{neptune}} = {10^{13}}\,m$.
Distance of Saturn from the Sun, ${d_{saturn}} = {10^{12}}\,m$
Now, we will apply Kepler’s third Law of Planetary Motion: Kepler's Third Law, often known as the 3rd Law of Kepler, is a fundamental law of physics that deals with the period of a satellite's revolution and how it is affected by the radius of its orbit.

So, according to Kepler’s 3rd law: the ratio of square of time period is directly proportional to the cube of distance of planets.
${(\dfrac{{{T_{neptune}}}}{{{T_{saturn}}}})^2} = {(\dfrac{{{d_{neptune}}}}{{{d_{saturn}}}})^3}$
where, ${T_{neptune}}$ is the Time Period of Neptune.
${T_{Saturn}}$ is the Time Period of Saturn.

\Rightarrow \dfrac{{{T_{neptune}}}}{{{T_{saturn}}}} = \sqrt {\dfrac{{1000}}{1}} \\\ \therefore \dfrac{{{T_{neptune}}}}{{{T_{saturn}}}} = 10\sqrt {10} :1 $$ Therefore, the ratio of time periods of the planets is $10\sqrt {10} :1$. **Hence, the correct option is D.** **Note:** To know which is better to apply between Copernicus and Kepler’s Law of motion, so the eccentricities of the orbits of the planets known to Copernicus and Kepler are tiny, the preceding principles provide reasonable approximations of planetary motion, but Kepler's laws better fit the data than Copernicus' model. The planetary orbit is an ellipse, not a circle.