The Jupiter's period of revolution around the Sun is 12 times that of the Earth. Assuming the planetary orbits to be circular, find:
(a) how many times the distance between the Jupiter and the Sun exceeds that between the Earth and the Sun,
(b) the velocity and the acceleration of Jupiter in the heliocentric reference frame.

Explanation

For any planet
`MRomega^2=(gammaMm_s)/(R^2)` or `omega=sqrt((gammam_s)/(R^3))`
So, `T=(2pi)/(omega)=2piR^(3//2)//sqrt(gammam_s)`
(a) Thus `T_J/T_E=(R_J/R_E)^(3//2)`
So `R_J/R_E=(T_J//T_E)^(2//3)=(12)^(2//3)=5*24`.
(b) `V_J^2=(gammam_s)/(R_J)`, and `R_J=(T(sqrt(gammam_s))/(2pi))^(2//3)`
So `V_J^2=((gammam_s)^(2//3)(2pi)^(2//3))/(T^(2//3))` or, `V_J=((2pigammam_s)/(T))^(2//3)`
where `T=12` years. `m_s=` mass of the Sun.
Putting the values we get `V_J=12*97 km//s`
Acceleration `=(v_J^2)/(R_J)=((2pigammam_s)/(T))^(2//3)xx((2pi)/(Tsqrt(gammam_s)))^(2//3)`
`=((2pi)/(T))^(4//3)(gammam_s)^(1//3)`
`=2*15xx10^-4 km//s^2`