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.

Asked by Pragya Singh | 1 year ago |  110

1 Answer

Solution :-

Given,

Orbital period of Io, TI0 = 1.769 days  

=  1.769 × 24 × 60 × 60 s

Orbital radius of Io, RI0 = 4.22 × 108 m

We know the mass of Jupiter:

MJ = 4π2RI03 / GTI02. . . . . . . . . . . . (1)

Where;

MJ = Mass of Jupiter

G = Universal gravitational constant

Also,

The orbital period of the earth,

T= 365.25 days = 365.25 × 24 × 60 × 60 s

Orbital radius of the Earth, R= 1 AU = 1.496 × 1011 m

We know that the mass of sun is:

MS =\( \dfrac{4\;\pi ^{2}\;R_{E}^{3}}{G\;T_{E}^{2}}\). . . . . . . . . . . . .  (2)

Therefore,  \( \\\dfrac{M_{S}}{M_{J}} = \dfrac{4\;\pi ^{2}\;R_{E}^{3}}{G\;T_{E}^{2}}\;\times \;\dfrac{G\;T_{10}^{2}}{4\;\pi ^{2}\;R_{10}^{3}}\)

\( = \dfrac{R_{E}^{3}}{T_{E}^{2}}\;\times \;\dfrac{T_{10}^{2}}{R_{10}^{3}}\)

Now, on substituting the values, we will get:

\( \left [ \dfrac{1.769\times 24\times 60\times 60}{365.25\times 24\times 60\times 60} \right ]^{2}\)

\( \times \left [ \dfrac{1.496\times 10^{11}}{4.22\times 10^{8}} \right ]^{3}\)

Therefore, \( \dfrac{M_{S}}{M_{J}}​​ ~ 1000\)

M~ 1000 × MJ     

[Which proves that, the Sun’s mass is 1000 times that of Jupiter’s]

Answered by Abhisek | 1 year ago

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