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.