Question

$\Large{\underline{\underline{\bf{\red{Question:-}}}}}$
Saturn year is 29.5 times the Earth year. How far is the Saturn from the Sun if the Earth is 1.50 x 10⁸ km away from the Sun?

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Answer 1

${\large{\underbrace{\underline{\bf{Question:-}}}}}$

Q)) A Saturn year is 29.5 times the Earth year. How far is the Saturn from the Sun if the Earth is $1.50\times10^8$ km away from Sun ?

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${\underline{\underline{\frak{\orange\dag\:Given:-}}}}$

• Time Period of Saturn = 29.5 Time period of Earth,
• $\sf \: t _{ \: s} = 29.5 \times t _{ \: e}$
• Distance of Earth from Sun ,
• $\sf \: r _{ \: e} = 1.50 \times {10}^{8} \: km$

${\underline{\underline{\frak{\orange\dag\:To \:Find:-}}}}$

• Distance of Saturn from Sun,
• $\sf \: r _{ \: s} = \: ?$

${\underline{\underline{\frak{\orange\dag\:Required\:Solution:-}}}}$

We know,

• $\sf \: t _{ \: s} = 29.5 \times t _{ \: e}$

So,

• $\displaystyle \implies \: \sf \frac{t _{ \: s} }{t _{ \: e} } = 29.5$

${\textit{\textbf{According\:to\:Kepler's\:3rd\:Law}}}$

$\sf \: t {}^{2} \: \alpha \: {r}^{3}$

So,

$\displaystyle \mapsto \sf \: \frac{ {t _{ \: s} }^{2} }{t _{ \: e} {}^{2} } = \frac{ {r _{ \: s} }^{3} }{r_{ \: e} {}^{3} } \\ \\ \mapsto \sf \: {r _{ \: s} }^{3} = {r _{ \: e} }^{3} \times {( \frac{t _{ \: s} }{t _{ \: e}}) }^{2} \\ \\ \mapsto \sf \: r _{ \: s} = \sqrt{ {r _{ \: e} }^{3} \times {( \frac{t _{ \: s} }{t _{ \: e} } )}^{2} } \\ \\ \mapsto \sf \: r _{ \: s} = r _{ \: e}\times {( \frac{t _{ \: s}}{t _{ \: e}}) }^{ \frac{2}{3} } \\ \\ \tt \: put \: the \: values \\ \\ \mapsto \sf \: r _{ \: s} = 1.5 \times {10}^{8} \times {29.5}^{ \frac{2}{3} } \\ \\ \mapsto \: \underline{ \boxed{ \tt{ \pink{r _{ \: s} = \purple{ \bf 1.43 \times {10}^{9} \: km}}}}}$

So,

The distance of Saturn from Sun would be $\bf{1.43\times10^9\:km}$ .

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Answer 2

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