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 x 108 m. Show that the mass of Jupiter is about one-thousat
that of the sun.​

Answer 1

hi mate here is your solution

Orbital period of I0 , TI0 = 1.769 days = 1.769 × 24 × 60 × 60 s

here solution so you can find

Orbital radius of I0 , RI0 = 4.22 × 108 m

Satellite I0 is revolving around the Jupiter

Mass of the latter is given by the relation:

MJ = 4π2RI03 / GTI02 …..(i)

Where, you can find mate

MJ = Mass of Jupiter

G = Universal gravitational constant

Orbital period of the earth,

Te = 365.25 days = 365.25 × 24 × 60 × 60 s

Orbital radius of the Earth,

Re = 1 AU = 1.496 × 1011 m

Mass of sun is given as:

Ms = 4π2Re3 / GTe2 ……(ii)

∴ Ms / MJ = (4π2Re3 / GTe2) × (GTI02 / 4π2RI03) = (Re3 × TI02) / (RI03 × Te2)

Substituting the values, we get:

= (1.769 × 24 × 60 × 60 / 365.25 × 24 × 60 × 60)2 × (1.496 × 1011 / 4.22 × 108)3

= 1045.04

∴ Ms / MJ ~ 1000

Ms ~ 1000 × MJ

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

mate your question is not correct there is thousandth not thousat

i hope mate it helpfull to you..

Answer 2

For the satellite around jupiter

$\sf{T_j\:=\:1.769\,days,\:R_j\:=\:4.22\times 10^8m}$

⠀⠀⠀⠀⠀⠀

For earth around sun

$\sf{T_e\:=\:365.25\,days,\:R_e\:=\:1.496\times 10^{11}m}$

⠀⠀⠀⠀⠀⠀

we know that,

$\sf{M\:=\: \dfrac{4\pi\,R_{e}^3}{GT^2}}$

⠀⠀⠀⠀⠀⠀

$\sf{\implies\dfrac{M_s}{M_j}\:=\: \left(\dfrac{1.496\times 10^{11}}{4.22\times 10^8}\right)^3 \times\left(\dfrac{1.769}{365.25}\right)^2}$

⠀⠀⠀⠀⠀⠀

$\sf\underline{=1045.0.39\approx 1000}$

⠀⠀⠀⠀⠀⠀

Thus mass of Sun is about Thousand times mass of Jupiter