Question

The period of the revolution of a planet A around the sun is 8 times that that of the B. The distance of A from the sun is how many greater than that of the M from the sun?
A) 4
B) 5
C) 2
D) 3

Answer 1

Hint
It can be solved with the help of Kepler's third law. Kepler’s third law states that- “ The square of the Time of period of the revolution is directly proportional to the cube of the radius of the revolution of the planet. ”
Numerically it can be expressed as-
$T^2 \propto R^3$.

Complete step by step answer
As in the given question:
$\left( {\dfrac{{{T_a}}}{{{T_b}}}} \right)\sqrt {{{\left( {\dfrac{{{R_a}}}{{{R_b}}}} \right)}^3} + {{\left( {\dfrac{{{a_1}}}{{{a_2}}}} \right)}^{\dfrac{3}{2}}}} \sqrt {{a^2} + \,\,\,\,{2^{3 \times \dfrac{2}{3}}}} T_a = 8{T_b}$
To find the distance we will use the Kepler’s law
$T^2 \propto R^3$.
Or it can also be written as
$\Rightarrow$ ${\left( {\dfrac{{{T_a}}}{{{T_b}}}} \right)^2}$= ${\left( {\dfrac{{{R_a}}}{{{R_b}}}} \right)^3}$
$\Rightarrow$ $\left( {\dfrac{{{R_a}}}{{{R_b}}}} \right)$ = ${\left( {\dfrac{{{T_a}}}{{{T_b}}}} \right)^{\dfrac{2}{3}}}$
$\Rightarrow$ $\left( {\dfrac{{{R_a}}}{{{R_b}}}} \right)$ = ${\left( {\dfrac{{8{T_b}}}{{{T_b}}}} \right)^{\dfrac{2}{3}}}$
$\Rightarrow$ $\left( {\dfrac{{{R_a}}}{{{R_b}}}} \right)$ = ${2^{3 \times \dfrac{2}{3}}}$
$\Rightarrow$ $\left( {\dfrac{{{R_a}}}{{{R_b}}}} \right) = 22$
$\Rightarrow$ $\left( {\dfrac{{{R_a}}}{{{R_b}}}} \right) = 4$
Hence,
${R_a}$ =$4{R_b}$.
Hence the correct answer is 4 times and the correct option is (A).

Note
Kepler’s three laws with the planetary motion:
The orbit are ellipses with focal points ${F_1}$ and $F_2$ for the first planet and $F_1$ and $F_3$ for $t$ second planet The two shaded sectors $A_1$ and $A_2$ have the same surface area and the time for the planet 1 to covers the $A_1$ is equal to the time to cover the segment $A_2$.
The total orbit times for the planet 1 and planet have a ratio-
${\left( {\dfrac{{{a_1}}}{{{a_2}}}} \right)^{\dfrac{3}{2}}}$
The value of G is constant at any point in the universe therefore it is called the universal constant at any point. The value of acceleration due to gravity is $9.8 \dfrac{m}{{s}^{2}}$; its value differs in other planes because it depends on the mass.