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
Given,
Orbital period (T) = 1.769 days
= 1.769 × 24 × 3600
= 1.528 × 10^5 s
Radius of orbit ( r) = 4.22 × 10^8m
Mass of the sun (Ms) = 2×10^30 Kg
Let mass of Jupiter 's satellite is m . and mass of Jupiter is Mj
Centripital force = Gravitational force
mv²/r = GMjm/r²
v² = GMj/r
We know,
v = wr and w = 2π/T
So, v = 2πr/T [ use this here,
4π²r²/T² = GMj/r
Mj = 4π²r³/T²G
Put the values of r , T, G .
Mj = 4× (3.14)² × (4.22×10^8)³/(1.528×10^5)²×6.67×10^-11
= 1.9 × 10^27 Kg ≈ 2×10^27 Kg
Mj/Ms = 2 × 10^27/2×10^30 = 1/1000
Mj = Ms/1000
Hence , the mass of Jupiter is about one thousandth that of sun.
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.