#### Question

Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 × 10^{8} m. Show that the mass of Jupiter is about one-thousandth that of the sun.

#### Solution 1

Orbital period of `I_0, T_"Io" = 1.769 days = 1.769 xx 24 xx 60 xx 60 s`

Orbital radius of `I_0, R_"Io" = 4.22 xx 10^8` m

Satellite `I_0` is revolving around the Jupiter

Mass of the latter is given by the relation:

`M_J = (4pi^2R_(+-)^3)/(GT_"Io"^2)` ....(i)

Where

`M_J` = Mass of Jupiter

G = Universal gravitational constant

Orbital period of the earth

`T_e = 365.25 days = 365.25 xx 24 xx 60xx 60s`

Orbital radius of the Earth

`R_e = 1 AU = 1.496 xx 10^(11) m`

Mass of Sun is given as

`M_s = (4pi^2R_e^3)/(GT_e^2)` .. (ii)

`:. M_s/M_J = (4pi^2R_e^3)/(GT_e^2) xx (GT_(Io)^2)/(4pi^2R_"Io"^3) = (R_e^3)/R_"Io"^3xx(T_"Io"^2)/(T_e^2)`

=`((1.769xx24xx60xx60)/(365.25xx24xx60xx60))^2xx((1.496xx10^(11))/(4.22xx10^8))`

=1045.04

`:.M_s/M_J ~ 1000`

`M_s "~" 1000xx M_J`

Hence, it can be inferred that the mass of Jupiter is about one-thousandth that of the Sun.

#### Solution 2

For a satellite of Jupiter, orbital period, T_{1} = 1.769 days = 1.769 x 24 x 60 x 60 s Radius of the orbit of satellite, r_{1} = 4.22 x 10^{8} m

Mass of Jupiter, `M_1` is given by `M_1` = `(4pi^2r_1^3)/(GT_1^2) = (4pi^2xx(4.22xx10^8)^2)/(Gxx(1.769xx24xx60xx60)^2)`

We know that the orbital period of earth around the sun

T = 1 year = 365.25 x 24 x 60 x 60 s

orbital radius r = 1 A.U = `1.496 xx 10^(11) m`

Mass of sun is given by M = `(4pi^2r^3)/(GT^2) = (4pi^2xx(1.496xx10^11)^3)/(Gxx(1.769 xx 24 xx 60 xx 60)^2)` ...(ii)

Dividing equation ii by i we get

`M/M_1 = (4pi^2xx(1.496xx10^(11))^3)/(Gxx(1.769xx24xx60xx60)^2)xx (Gxx(1.769xx24xx60xx60)^2)/(4pi^2xx(4.22xx10^8)^3) = 1046`

or `M_1/M = 1/1046 = 1/1000 => M_1 = 1/1000 M`

Proved