Question

The distance of Neptune and Saturn from the sun is nearly $10^{13}$m and $10^{12}$m respectively. Assuming that they move in circular orbits, their periodic times would be in the ratio of
A. 10
B. 100
C. 10$\sqrt{10}$
D. 1000

Answer 1

Hint: By Kepler's third law of planetary motion, the square of the time period is proportional to the cube of the length of the semi-major axis of the planet's orbit. In the case of circular motion, we can choose the radius of the orbit in place of the semi-major axis.

Formula used:
The time period of a circular orbit is given as:
$T = \sqrt{\dfrac{r^3}{GM}} $

Complete step-by-step solution:
We are given that $r = 10^{12}$ m is the distance of Saturn from the Sun and $r = 10^{13}$ m is the distance of Neptune from the Sun. These are nothing but their respective radii for circular orbits.
Keeping these values in the time period formula we get:
$T_1 = \sqrt{\dfrac{(10^{12})^3}{GM}} $
for the case of Saturn and
$T_2 = \sqrt{\dfrac{(10^{13})^3}{GM}} $
for the case of Neptune.
Taking ratios we get:
$\dfrac{T_1}{T_2} = \sqrt{\dfrac{(10^{13})^3}{(10^{12})^3}}$
Or
$\dfrac{T_1}{T_2} = \sqrt{1000} = 10 \sqrt{10}$
Therefore the correct answer is option (C).

Additional information:
If one does not remember the formula, one can equate the magnitude of centripetal force with the magnitude of gravitational force acting on the body for the case of circular motion:
$m \omega^2 r = \dfrac{GMm}{r^2}$
Gives
$ \omega^2 = \dfrac{GM}{r^3}$
We know that $\omega = 2 \pi / T$
So, this will help us in getting the same formula as the one we used in the solution.

Note: We already had the formula of the time period of a planet in case of the elliptical orbit. The area of an ellipse is $\pi ab$ and the area of a circle is $\pi r^2$. A circle is just an ellipse with zero eccentricity. Therefore, our formula can be easily guessed by the third law. And also we already know that the areal velocity of a planet is constant. The derivation for the third law comes from the second law only.