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.
Question:

Io, one of the satellites of Jupiter, has an orbital period of $1.769$ days and the radius of the orbit is $4.22 \times 10^{8} \mathrm{~m}$. Show that the mass of Jupiter is about one-thousandth that of the sun.

Solution:

Orbital period of $I_{0}, T_{l o}=1.769$ days $=1.769 \times 24 \times 60 \times 60 \mathrm{~s}$

Orbital radius of $I_{0}, R_{b}=4.22 \times 10^{8} \mathrm{~m}$

Satellite $I_{0}$ is revolving around the Jupiter

Mass of the latter is given by the relation:

$M_{J}=\frac{4 \pi^{2} R_{\pm}^{3}}{\mathrm{G} T_{l o}^{2}}$  $\ldots$ (i)

Where,

$M_{J}=$ Mass of Jupiter

G = Universal gravitational constant

Orbital period of the Earth,

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

Orbital radius of the Earth,

$R_{c}=1 \mathrm{AU}=1.496 \times 10^{11} \mathrm{~m}$

Mass of Sun is given as:

$M_{\mathrm{s}}=\frac{4 \pi^{2} R_{\mathrm{c}}^{3}}{\mathrm{G} T_{c}^{2}}$ (ii)

$\therefore \frac{M_{s}}{M_{j}}=\frac{4 \pi^{2} R_{e}^{3}}{\mathrm{G} T_{c}^{2}} \times \frac{\mathrm{G} T_{b}^{2}}{4 \pi^{2} R_{l o}^{3}}=\frac{R_{e}^{3}}{R_{b o}^{3}} \times \frac{T_{l o}^{2}}{T_{c}^{2}}$

$=\left(\frac{1.769 \times 24 \times 60 \times 60}{365.25 \times 24 \times 60 \times 60}\right)^{2} \times\left(\frac{1.496 \times 10^{11}}{4.22 \times 10^{8}}\right)^{3}$

$=1045.04$

$\therefore \frac{M_{S}}{M_{1}} \sim 1000$

$M_{s} \sim 1000 \times M_{j}$

Hence, it can be inferred that the mass of Jupiter is about one-thousandth that of the Sun.

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