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
Explanation:
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.
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.