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Question
A Saturn year is 29.5 times the earth year. How far is the Saturn from the sun if the earth is 1.50 ×108 km away from the sun?
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Solution 1
Distance of the Earth from the Sun, re = 1.5 × 108 km = 1.5 × 1011 m
Time period of the Earth = Te
Time period of Saturn, Ts = 29. 5 Te
Distance of Saturn from the Sun = rs
From Kepler’s third law of planetary motion, we have
T = `((4pi^2r^3)/"GM")^(1/2)`
For Saturn and Sun, we can write
`(r_s^3)/(r_e^3) = (T_s^2)/(T_e^2)`
`r_s = r_e (T_s/T_e)^(2/3)`
`=1.5 xx 10^11 ((29.5T_e)/T_e)^(2/3)`
`=1.5 xx 10^11 (29.5)^(2/3)`
`= 1.5 xx 10^11xx 9.55`
`=14.32 xx 10^11` m
Hence, the distance between Saturn and the Sun is `1.43 xx 10^(12)` m.
Solution 2
We know that `T^2 oo R^3`
`:.(T_s^2)/(T_e^2) = (R_s^3)/(R_e^3)`
where subscript s and e refer to the Saturn and Earth respectively.
Now `T_s/T_e = 29.5`[given]; `R_e = 1.50 xx 10^8` km
`(R_s/R_e)^3 = (T_s/T_e)`
`R_s = R_exx [(29.5)^2]^(1/3)`
`= 1.50 xx 10^8 xx (870.25)^"1.3" `km
`= 1.43 xx 10^9 km = 1.43 xx 10^12` m
∴Distance of saturn from Sum `= 1.43 xx 10^(12)` m
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