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

Answer: (C)

if T is time period of the planet and r the distance of the planet from the sun,

then from kepler's third law

$T^2=k{r}^3$ ,

Let, Time period of jupiter be $T_1$ and for earth be $T_2$ , distance of jupiter from sun be $r_1$ and of earth from sun be $r_2$ ,

$r_1$ = $r_2(T_1/T_2{)}^{3/2}$

$T_1/T_2=12$

$G=$$6.67\times {10}^{-11}$ and $r_2$ = $1.4\times {10}^{11}m$ , $M$ =$1.97\times {10}^{30}kg$

So acceleration of jupiter is a = $GM/r_1$ = $2.2\times {10}^{-4}m/s^2$