Physics, asked by PragyaTbia, 1 year ago

Io, one of the satellites of Jupiter has an orbital period of 1.769 days and the radius of the orbit is 4.22 × 10⁸ m. Show that the mass of Jupiter is about one-thousandth that of the Sun.

Answers

Answered by gadakhsanket
3
Hii dear,

◆ Explained answer-
# Given-
Ti = 1.769 days
Ri = 4.22×10^8 m
Te = 365.25 days
Re = 1.5×10^11 m

# Formula-
Io is revolving around the Jupiter, hence mass of the Jupiter is given by-
Mj = 4π^2Ri^3 / GTi^2

Earth is revolving around the Sun, hence mass of the Sun is given by-
Ms = 4π^2Re^3 / GTe^2

Taking ratio -
Ms/Mj = (4π^2Re^3 / GTe^2) /(4π^2Ri^3 / GTi^2)

Substituting the values, we get:
Ms/Mj = (1.769 / 365.25)^2 × (1.496×10^11 / 4.22×10^8)^3
Ms/Mj = 1045.04
∴ Ms ~ 1000 Mj

Therefore, we can say that the mass of Jupiter is about one-thousandth that of the Sun.

Hope that was useful...
Answered by jack6778
4

Answer:

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

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,

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.

Similar questions