it takes planet jupiter 12 years to orbit the sun once.Whats the average distance from jupiter to the sun?
(orbital distance=1.5×10^11 m)

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Solution

$T h e t i m e p e r i o d o f j u p i t e r o r b i t t i n g r o u n d t h e s u n i s, T_{J} = 2 π \sqrt{\frac{{R_{J}}^{3}}{G M_{s}}} - - - - (1) a n d t i m e p e r i o d o f r e v o l u t i o n o f e a r t h r o u n d t h e s u n i s, T_{e} = 2 π \sqrt{\frac{{R_{e}}^{3}}{G M_{s}}} - - - - (2) d i v i d i n g e q n (1) b y e q n (2), w e g e t, \frac{T_{J}}{T_{e}} = \sqrt{\frac{{R_{J}}^{3}}{{R_{e}}^{3}}} B u t w e a r e g i v e n \frac{T_{J}}{T_{e}} = \frac{12}{1}, s o \frac{12}{1} = \sqrt{\frac{{R_{J}}^{3}}{{R_{e}}^{3}}} o r 144 = \frac{{R_{J}}^{3}}{{R_{e}}^{3}} o r R_{J} = R_{e} \times {(144)}^{\frac{1}{3}} o r R_{J} = 1.5 \times 10^{11} \times {(144)}^{\frac{1}{3}} b y \log c a l c u l a t i o n w e g e t, \log x = \frac{1}{3} \log 144 = \frac{1}{3} \times 2.1584 = 0.7195 o n t a k i n g a n t i \log, w e g e t, x = 5.242 s o R_{J} = 1.5 \times 10^{11} \times 5.242 = 7.863 \times 10^{11} m .$

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