Question
The distances of two planets from the sun are $${10^{13}}m$$ and $${10^{12}}m$$ respectively. The ratio of time periods of these two planets is
A.
$$\frac{1}{{\sqrt {10} }}$$
B.
$$100$$
C.
$$10\sqrt {10} $$
D.
$$\sqrt {10} $$
Answer :
$$10\sqrt {10} $$
Solution :
According to Kepler's third law (or law of periods) the square of the time taken to complete the orbit (time period $$T$$) is proportional to the cube of the semi-major axis $$\left( r \right)$$ of the elliptical orbit i.e. $${T^2} \propto {r^3}$$
Here, $${r_1} = {10^{13}}m,\,{r_2} = {10^{12}}m$$
$$\eqalign{
& \therefore \frac{{T_1^2}}{{T_2^2}} = \frac{{r_1^3}}{{r_2^3}} = \frac{{{{\left( {{{10}^{13}}} \right)}^3}}}{{{{\left( {{{10}^{12}}} \right)}^3}}} \cr
& {\text{or}}\,\,\frac{{T_1^2}}{{T_2^2}} = \frac{{{{10}^{39}}}}{{{{10}^{36}}}} = {10^3} \cr
& {\text{or}}\,\,\frac{{{T_1}}}{{{T_2}}} = 10\sqrt {10} \cr} $$