Ncert Exercises & Solutions

8.4 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.

Orbital period of $I_{0},$
$T_{Io} = 1.769 days \times 24 \times 60 \times 60 s$

Orbital radius of $I_{o},$
$R_{Io} = 4.22 \times 10^{8} M$

Satellite $I_{o}$ is revolving around the Jupiter

Mass of the latter is given by the relation:

$M_{J} = \frac{4 π^{2} R^{3}}{T_{Io}^{2}} . . . (i)$

Where,

$M_{J}$ = Mass of Jupiter

G = Universal gravitational constant

The orbital radius of the Earth,

$T_{e} = 365.25 days = 365.25 \times 24 \times 60 \times 60 s$

The orbital radius of the Earth,

$R_{e} = 1 AU = 1.496 \times 10^{11} m$

Mass of sun is given as:

$M_{s} = \frac{4 π^{2} R^{3}}{{GT}_{e}^{2}} . . . (ii)$
$F r o m (i) a n d (i i),$
$\frac{M_{s}}{M_{J}} = \frac{4 π^{2} R_{e}^{3}}{{GT}_{e}^{2}} \times \frac{{GT}_{Io}^{2}}{4 π^{2} R_{Io}^{3}} = \frac{R_{e}^{3}}{R_{Io}} \times \frac{T_{Io}^{2}}{T_{e}^{2}}$
$= {(\frac{1.769 \times 24 \times 60 \times 60}{365.25 \times 24 \times 60 \times 60})}^{2} \times {(\frac{1.496 \times 10^{11}}{4.22 \times 10^{8}})}^{3}$
$= 1045.04$
$= 1000 (A p p r o x .)$

$\Rightarrow \frac{M_{s}}{M_{J}} = 1000$
$\Rightarrow M_{s} = 1000 \times M_{J}$

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