Question

The Jupiter’s period of revolution around the Sun is $12$ times that of the earth. Assuming the planetary orbits to be circular, find the acceleration of Jupiter in the heliocentric reference frame.
(A) $2 \times {10^{ - 4}}m/{s^2}$
(B) $4.2 \times {10^{ - 4}}m/{s^2}$
(C) $2.2 \times {10^{ - 4}}m/{s^2}$
(D) $4 \times {10^{ - 4}}m/{s^2}$

Answer 1

Hint:Here first we have to find the time period and velocity for Jupiter with the help of force and then we can find the acceleration. Often known as Newton’s law of gravitation, the gravitational force is described as the magnitude of force between two points.We shall use this formula to solve this question.

Complete step by step answer:
Given,
Jupiter's period of revolution around the Sun is $12$ times that of the earth.
Let velocity of Jupiter be ${V_J}$ and velocity of earth be ${V_E}$.Let the force between Jupiter and Sun be ${F_1}$ and the force between Sun and Earth be ${F_2}$.Also, let the mass of Sun be ${M_S}$ and mass of Jupiter be ${M_J}$.Let $G$ be the gravitational constant.Let us consider ${R_J}$ as the radius of Jupiter.
Now force ${F_1}$ is given by-
${F_1} = \dfrac{{G{M_s}{M_J}}}{{{R_J}^2}} \\$
$\Rightarrow\dfrac{{{M_J}{V_J}^2}}{{{R_J}}} = \dfrac{{G{M_s}{M_J}}}{{{R_J}^2}} \\$
$\Rightarrow{V_J}^2 = \dfrac{{G{M_s}}}{{{R_j}}} \\$
$\Rightarrow{V_J} = \sqrt {\dfrac{{G{M_s}}}{{{R_J}}}} \\$
Similarly, the velocity of Earth can be found as-
${V_E} = \sqrt {\dfrac{{G{M_s}}}{{{R_E}}}} $
Now, let the time period of revolution of Jupiter be ${T_J}$ and earth be ${T_E}$.
According to question,
${T_J} = 12{T_E} \\$
$\Rightarrow\dfrac{{{T_J}}}{{{T_E}}} = 12 \\$
Also,
${T_J} = \dfrac{{2\pi {R_J}}}{{{V_J}}} \\$
$\Rightarrow{T_E} = \dfrac{{2\pi {R_E}}}{{{V_E}}} \\ $
Then,
$\dfrac{{{T_J}}}{{{T_E}}} = \dfrac{{{R_J}}}{{{R_E}}}\left( {\dfrac{{{V_E}}}{{{v_J}}}} \right) \\$
$\Rightarrow\dfrac{{{T_J}}}{{{T_E}}}= \dfrac{{{R_J}}}{{{R_E}}}\left( {\sqrt {\dfrac{{{R_J}}}{{{R_E}}}} } \right)\\$
$\Rightarrow\dfrac{{{T_J}}}{{{T_E}}}= {\left( {\dfrac{{{R_J}}}{{{R_E}}}} \right)^{3/2}} \\ $
Therefore,
$\dfrac{{{R_J}}}{{{R_E}}} = {\left( {\dfrac{{{T_J}}}{{{T_E}}}} \right)^{2/3}} = {(12)^{2/3}} \\$
$\Rightarrow{R_J} = 5.24\,{R_E} \\$

Also, velocity of Jupiter, ${V_J}^2 = \dfrac{{G{M_S}}}{{{R_J}}}$
And we know that
${R_J} = \dfrac{{{T_J}{V_J}}}{{2\pi }}$
$\Rightarrow{V_J}^2 = \dfrac{{G{M_S}}}{{{R_J}}} = \dfrac{{G{M_S}}}{{\dfrac{{T{V_J}}}{{2\pi }}}} \\$
$\Rightarrow{V_J}^3 = \dfrac{{G{M_S}2\pi }}{{{T_J}}} \\$
$\Rightarrow{V_J} = {\left( {\dfrac{{G{M_S}2\pi }}{{{T_J}}}} \right)^{1/3}} \\$
In heliocentric reference frame-
${T_J} = 12{T_E}$
$\Rightarrow{T_J} = 12$ years
Therefore,
${V_J} = {\left( {\dfrac{{6.67 \times {{10}^{ - 11}} \times 1.98 \times {{10}^{30}} \times 2\pi }}
{{12 \times 365 \times 24 \times 3600}}} \right)^{1/3}} \\$
$\Rightarrow{V_J}= 12989.08\,m/s \\$
$\Rightarrow{V_J}= 12.98\,km/s \\$
Therefore, acceleration of Jupiter
$\dfrac{{G{M_S}}}{{{R_J}^2}} \\$
$\Rightarrow \dfrac{{G{M_S}}}{{{{\left( {\dfrac{{T{V_J}}}{{2\pi }}} \right)}^2}}} \\$
$\Rightarrow \dfrac{{4{\pi ^2}{M_S}}}{{{T^2}{V_J}^2}} \\$
$\Rightarrow \dfrac{{{{\left( {2\pi } \right)}^2}G{M_S}}}{{{T^2}
{{\left( {\dfrac{{2\pi G{M_S}}}{{{T_S}}}} \right)}^{1/3}}}} \\$
$\Rightarrow\dfrac{{{{\left( {2\pi } \right)}^{5/3}}{{\left( {G{M_S}} \right)}^{2/3}}}}
{{{T_J}^{5/3}}} \\$
$\therefore 2.15 \times {10^{ - 4}}m/{s^2} \\$

Hence, option C is correct.

Additional information:
Heliocentrism is a cosmical paradigm in which the Sun, while the Earth and other bodies revolve around it, is believed to lie at or near a focal point (e.g. of the solar system or the universe).

Note:In this solution we have to be very careful while putting the values in each and every equation as putting wrong values will lead to a different answer. Also, the time period and radius values cannot be mixed up.